{"id":2178,"date":"2018-04-11T23:52:09","date_gmt":"2018-04-12T03:52:09","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/neutralization-reactions\/"},"modified":"2019-05-14T13:57:40","modified_gmt":"2019-05-14T17:57:40","slug":"neutralization-reactions","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/chapter\/neutralization-reactions\/","title":{"raw":"6.3 Acid-Base Reactions","rendered":"6.3 Acid-Base Reactions"},"content":{"raw":"<div class=\"section\" id=\"ball-ch04_s05\" lang=\"en\">\r\n<div class=\"bcc-box bcc-highlight\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Identify common acids and bases<\/li>\r\n \t<li>Define acid-base reactions<\/li>\r\n \t<li>Recognize and identify examples of acid-base reactions<\/li>\r\n \t<li><span style=\"font-size: 1em\">Predict the products of acid-base reactions.<\/span><\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n<h2>Acids and Bases<\/h2>\r\n<div class=\"section\" id=\"ball-ch04_s05\" lang=\"en\">\r\n<p class=\"para block\">The definition of an <span class=\"margin_term\"><a class=\"glossterm\">acid<\/a><\/span>\u00a0is often cited as: any compound that increases the amount of hydrogen ion (H<sup class=\"superscript\">+<\/sup>) in an aqueous solution. The chemical opposite of an acid is a base. The equivalent definition of a base is that a <span class=\"margin_term\"><a class=\"glossterm\">base<\/a><\/span>\u00a0is a compound that increases the amount of hydroxide ion (OH<sup class=\"superscript\">\u2212<\/sup>) in an aqueous solution. These original definitions were proposed by Arrhenius (the same person who proposed ion dissociation) in 1884, so they are referred to as the <strong class=\"emphasis bold\">Arrhenius definition<\/strong> of an acid and a base, respectively.<\/p>\r\n<p id=\"ball-ch04_s05_p02\" class=\"para block\">You may recognize that, based on the description of a hydrogen atom, an H<sup class=\"superscript\">+<\/sup> ion is a hydrogen atom that has lost its lone electron; that is, H<sup class=\"superscript\">+<\/sup> is simply a proton. Do we really have bare protons moving about in aqueous solution? No. What is more likely is that the H<sup class=\"superscript\">+<\/sup> ion has attached itself to one (or more) water molecule(s). To represent this chemically, we define the <span class=\"margin_term\"><a class=\"glossterm\">hydronium ion\u00a0H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup><\/a><span class=\"glossdef\"><span class=\"inlineequation\">(aq)<\/span>, a water molecule with an extra hydrogen ion attached to it.<\/span><\/span> as H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>, which represents an additional proton attached to a water molecule. We use the hydronium ion as the more logical way a hydrogen ion appears in an aqueous solution, although in many chemical reactions H<sup class=\"superscript\">+<\/sup> and H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup> are treated equivalently.<\/p>\r\n<p id=\"fs-idp89436016\">For purposes of this brief introduction, we will consider only the more common types of acid-base reactions that take place in aqueous solutions. In this context, an <strong>acid<\/strong> is a substance that will dissolve in water to yield hydronium ions, H<sub>3<\/sub>O<sup>+<\/sup>. As an example, consider the equation shown here:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm54028336\" style=\"text-align: center\">$latex \\text{HCl}(aq) + \\text{H}_2 \\text{O}(aq) \\longrightarrow \\text{Cl}^{-}(aq) + \\text{H}_3 \\text{O}^{+}(aq)$<\/div>\r\n<p id=\"fs-idm10390992\">The process represented by this equation confirms that hydrogen chloride is an acid. When dissolved in water, H<sub>3<\/sub>O<sup>+<\/sup> ions are produced by a chemical reaction in which H<sup>+<\/sup> ions are transferred from HCl molecules to H<sub>2<\/sub>O molecules (<a href=\"#CNX_Chem_04_02_HClsoln\" class=\"autogenerated-content\">Figure 1<\/a>).<\/p>\r\n\r\n\r\n[caption id=\"attachment_1414\" align=\"aligncenter\" width=\"400\"]<a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_HClsoln-3.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_HClsoln-3-300x170.jpg\" alt=\"\" width=\"400\" height=\"227\" class=\"wp-image-1414\" \/><\/a> <strong>Figure 1.<\/strong> When hydrogen chloride gas dissolves in water, (a) it reacts as an acid, transferring protons to water molecules to yield (b) hydronium ions (and solvated chloride ions).[\/caption]\r\n<figure id=\"CNX_Chem_04_02_HClsoln\"><figcaption><\/figcaption><\/figure>\r\nThe nature of HCl is such that its reaction with water as just described is essentially 100% efficient: Virtually every HCl molecule that dissolves in water will undergo this reaction. Acids that completely react in this fashion are called <strong>strong acids<\/strong>, and HCl is one among just a handful of common acid compounds that are classified as strong (<a href=\"#fs-idp55395904\" class=\"autogenerated-content\">Table 1<\/a>).\r\n<div class=\"section\" id=\"ball-ch04_s05\" lang=\"en\">\r\n<table id=\"fs-idp55395904\" class=\"span-all\" summary=\"This table contains two columns and seven rows. The columns are labeled, \u201cCompound Formula,\u201d and, \u201cName in Aqueous Solution.\u201d Under the column, \u201cCompound Formula,\u201d are: \u201cH B r,\u201d \u201cH C l,\u201d \u201cH I,\u201d \u201cH N O subscript 3,\u201d \u201cH C l O subscript 4,\u201d and, \u201cH subscript 2 S O subscript 4.\u201d Under the column, \u201cName in Aqueous Solution,\u201d are: \u201chydrobromic acid,\u201d \u201chydrochloric acid,\u201d \u201chydroiodic acid,\u201d \u201cnitric acid,\u201d \u201cperchloric acid,\u201d and, \u201csulfuric acid.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th>Compound Formula<\/th>\r\n<th>Name in Aqueous Solution<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td>HBr<\/td>\r\n<td>hydrobromic acid<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>HCl<\/td>\r\n<td>hydrochloric acid<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>HI<\/td>\r\n<td>hydroiodic acid<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>HNO<sub>3<\/sub><\/td>\r\n<td>nitric acid<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>HClO<sub>4<\/sub><\/td>\r\n<td>perchloric acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>HClO<sub>3<\/sub><\/td>\r\n<td>chloric acid<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td>H<sub>2<\/sub>SO<sub>4<\/sub><\/td>\r\n<td>sulfuric acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td colspan=\"3\"><strong>Table 1.<\/strong> Common Strong Acids<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idp2912576\">\u00a0A far greater number of compounds behave as <strong>weak acids<\/strong> and only partially react with water, leaving a large majority of dissolved molecules in their original form and generating a relatively small amount of hydronium ions.<\/p>\r\n\r\n<\/div>\r\n<div class=\"section\" id=\"ball-ch04_s05\" lang=\"en\">\r\n<table id=\"fs-idp55395904\" class=\"span-all\" style=\"width: 368px\" summary=\"This table contains two columns and seven rows. The columns are labeled, \u201cCompound Formula,\u201d and, \u201cName in Aqueous Solution.\u201d Under the column, \u201cCompound Formula,\u201d are: \u201cH B r,\u201d \u201cH C l,\u201d \u201cH I,\u201d \u201cH N O subscript 3,\u201d \u201cH C l O subscript 4,\u201d and, \u201cH subscript 2 S O subscript 4.\u201d Under the column, \u201cName in Aqueous Solution,\u201d are: \u201chydrobromic acid,\u201d \u201chydrochloric acid,\u201d \u201chydroiodic acid,\u201d \u201cnitric acid,\u201d \u201cperchloric acid,\u201d and, \u201csulfuric acid.\u201d\">\r\n<thead>\r\n<tr valign=\"top\">\r\n<th style=\"width: 161px\">Compound Formula<\/th>\r\n<th style=\"width: 207px\">Name in Aqueous Solution<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr valign=\"top\">\r\n<td style=\"width: 161px\">HF<\/td>\r\n<td style=\"width: 207px\">hydrofluoric acid<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"width: 161px\">HCN<\/td>\r\n<td style=\"width: 207px\">hydrocyanic acid<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"width: 161px\">HC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><\/td>\r\n<td style=\"width: 207px\">acetic acid<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"width: 161px\">HNO<sub>2<\/sub><\/td>\r\n<td style=\"width: 207px\">nitrous acid<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"width: 161px\">HClO<\/td>\r\n<td style=\"width: 207px\">hypochlorous acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 161px\">HClO<sub><span style=\"font-size: small\">2<\/span><\/sub><\/td>\r\n<td style=\"width: 207px\">chlorous acid<\/td>\r\n<\/tr>\r\n<tr valign=\"top\">\r\n<td style=\"width: 161px\">H<sub>2<\/sub>SO<sub><span style=\"font-size: small\">3<\/span><\/sub><\/td>\r\n<td style=\"width: 207px\">sulfurous acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 161px\">H<sub>2<\/sub>CO<sub>3<\/sub><\/td>\r\n<td style=\"width: 207px\">carbonic acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 161px\">H<sub>3<\/sub>PO<sub>4<\/sub><\/td>\r\n<td style=\"width: 207px\">phosphoric acid<\/td>\r\n<\/tr>\r\n<tr>\r\n<td style=\"width: 368px\" colspan=\"2\"><strong>Table 2.<\/strong> Common Weak Acids<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<p id=\"fs-idp2912576\">Weak acids are commonly encountered in nature, being the substances partly responsible for the tangy taste of citrus fruits, the stinging sensation of insect bites, and the unpleasant smells associated with body odor. A familiar example of a weak acid is acetic acid, the main ingredient in food vinegars:<\/p>\r\n\r\n<\/div>\r\n<div class=\"equation\" id=\"fs-idp23273024\" style=\"text-align: center\">$latex \\text{CH}_3 \\text{CO}_2 \\text{H}(aq) + \\text{H}_2 \\text{O}(l) \\leftrightharpoons \\text{CH}_3 {\\text{CO}_2}^{-}(aq) + \\text{H}_3 \\text{O}^{+}(aq)$<\/div>\r\n<div><\/div>\r\n<div class=\"equation\" style=\"text-align: left\"><span style=\"text-align: justify;font-size: 14pt\">When dissolved in water under typical conditions, only about 1% of acetic acid molecules are present in the ionized form, $latex \\text{CH}_3 {\\text{CO}_2}^{-} $(<\/span><a href=\"#CNX_Chem_04_02_Citrus\" class=\"autogenerated-content\" style=\"text-align: justify;font-size: 14pt\">Figure 2<\/a><span style=\"text-align: justify;font-size: 14pt\">). The use of a double-arrow in the equation above denotes the partial reaction aspect of this process, a concept addressed fully in the chapters on chemical equilibrium.)<\/span><\/div>\r\n<div>\r\n<div class=\"section\" id=\"ball-ch04_s05\" lang=\"en\">\r\n\r\n[caption id=\"attachment_1415\" align=\"aligncenter\" width=\"300\"]<a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_Citrus-2.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_Citrus-2-300x267.jpg\" alt=\"\" width=\"300\" height=\"267\" class=\"wp-image-1415 size-medium\" \/><\/a> <strong>Figure 2.<\/strong> (a) Fruits such as oranges, lemons, and grapefruit contain the weak acid citric acid. (b) Vinegars contain the weak acid acetic acid. (credit a: modification of work by Scott Bauer; credit b: modification of work by Br\u00fccke-Osteuropa\/Wikimedia Commons)[\/caption]\r\n\r\n<\/div>\r\n<\/div>\r\n<figure id=\"CNX_Chem_04_02_Citrus\"><\/figure>\r\n<\/div>\r\n<div class=\"section\" id=\"ball-ch04_s05\" lang=\"en\">\r\n\r\nA <strong>base<\/strong> is a substance that will dissolve in water to yield hydroxide ions, OH<sup>\u2212<\/sup>. The most common bases are ionic compounds composed of alkali or alkaline earth metal cations (groups 1 and 2) combined with the hydroxide ion\u2014for example, NaOH and Ca(OH)<sub>2<\/sub>. When these compounds dissolve in water, hydroxide ions are released directly into the solution. For example, KOH and Ba(OH)<sub>2<\/sub> dissolve in water and dissociate completely to produce cations (K<sup>+<\/sup> and Ba<sup>2+<\/sup>, respectively) and hydroxide ions, OH<sup>\u2212<\/sup>. These bases, along with other hydroxides that completely dissociate in water, are considered <strong>strong bases<\/strong>.\r\n<p id=\"fs-idp74282160\">Consider as an example the dissolution of lye (sodium hydroxide) in water:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm62931696\" style=\"text-align: center\">$latex \\text{NaOH}(s) \\longrightarrow \\text{Na}^{+}(aq) + \\text{OH}^{-}(aq)$<\/div>\r\n<p id=\"fs-idp8724736\">This equation confirms that sodium hydroxide is a base. When dissolved in water, NaOH dissociates to yield Na<sup>+<\/sup> and OH<sup>\u2212<\/sup> ions. This is also true for any other ionic compound containing hydroxide ions. Since the dissociation process is essentially complete when ionic compounds dissolve in water under typical conditions, NaOH and other ionic hydroxides are all classified as strong bases.<\/p>\r\n<p id=\"fs-idm50199792\">Unlike ionic hydroxides, some compounds produce hydroxide ions when dissolved by chemically reacting with water molecules. In all cases, these compounds react only partially and so are classified as <strong>weak bases<\/strong>. These types of compounds are also abundant in nature and important commodities in various technologies. For example, global production of the weak base ammonia is typically well over 100 metric tons annually, being widely used as an agricultural fertilizer, a raw material for chemical synthesis of other compounds, and an active ingredient in household cleaners (<a href=\"#CNX_Chem_04_02_ammonia\" class=\"autogenerated-content\">Figure 3<\/a>). When dissolved in water, ammonia reacts partially to yield hydroxide ions, as shown here:<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm9327664\" style=\"text-align: center\">$latex \\text{NH}_3(aq) + \\text{H}_2 \\text{O}(l) \\rightleftharpoons {\\text{NH}_4}^{+}(aq) + \\text{OH}^{-}(aq)$<\/div>\r\n<p id=\"fs-idm73811808\">Under typical conditions, only about 1% of the dissolved ammonia is present as NH<sub>4<\/sub><sup>+<\/sup> ions.<\/p>\r\n\r\n<figure id=\"CNX_Chem_04_02_ammonia\"><figcaption>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]<a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_02_ammonia.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_ammonia-2.jpg\" alt=\"This photograph shows a large agricultural tractor in a field pulling a field sprayer and a large, white cylindrical tank which is labeled \u201cCaution Ammonia.\u201d\" width=\"975\" height=\"316\" \/><\/a> <strong>Figure 3.<\/strong> Ammonia is a weak base used in a variety of applications. (a) Pure ammonia is commonly applied as an agricultural fertilizer. (b) Dilute solutions of ammonia are effective household cleansers. (credit a: modification of work by National Resources Conservation Service; credit b: modification of work by pat00139)[\/caption]\r\n\r\n<\/figcaption><\/figure>\r\n<h2>Acid-Base Reactions<\/h2>\r\n<p id=\"fs-idm1255344\">An <strong>acid-base reaction<\/strong> is one in which a hydrogen ion, H<sup>+<\/sup>, is transferred from one chemical species to another. Such reactions are of central importance to numerous natural and technological processes, ranging from the chemical transformations that take place within cells and the lakes and oceans, to the industrial-scale production of fertilizers, pharmaceuticals, and other substances essential to society. The subject of acid-base chemistry, therefore, is worthy of thorough discussion.<\/p>\r\n<p id=\"ball-ch04_s05_p03\" class=\"para editable block\">The reaction between an acid and a base is called an acid-base reaction or a <strong><span class=\"margin_term\"><a class=\"glossterm\">neutralization reaction<\/a><\/span><\/strong>. Although acids and bases have their own unique chemistries, the acid and base cancel each other\u2019s chemistry to produce a rather innocuous substance\u2014water. In fact, the <strong>general\u00a0<\/strong><strong>acid-base reaction<\/strong> is<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">acid +\u00a0base $latex \\longrightarrow$ water +\u00a0salt<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s05_p04\" class=\"para editable block\">where the term <strong><span class=\"margin_term\"><a class=\"glossterm\">salt<\/a><\/span><\/strong>\u00a0is used to define any ionic compound (soluble or insoluble) that is formed from a reaction between an acid and a base. In chemistry, the word <em class=\"emphasis\">salt<\/em> refers to more than just table salt. For example, the balanced chemical equation for the reaction between HCl(aq) and KOH(aq) is<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">HCl(aq) +\u00a0KOH(aq) $latex \\longrightarrow$ H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0KCl(aq)<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s05_p05\" class=\"para editable block\">where the salt is KCl. By counting the number of atoms of each element, we find that only one water molecule is formed as a product. However, in the reaction between HCl(aq) and Mg(OH)<sub class=\"subscript\">2<\/sub>(aq), additional molecules of HCl and H<sub class=\"subscript\">2<\/sub>O are required to balance the chemical equation:<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 HCl(aq) +\u00a0Mg(OH)<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$ 2 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0MgCl<sub class=\"subscript\">2<\/sub>(aq)<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s05_p06\" class=\"para editable block\">Here, the salt is MgCl<sub class=\"subscript\">2<\/sub>. This is one of several reactions that take place when a type of antacid\u2014a base\u2014is used to treat stomach acid.<\/p>\r\nThere are acid-base reactions that do not follow the \"general acid-base\" equation given above. \u00a0For example,\u00a0, the balanced chemical equation for the reaction between HCl(aq) and NH<sub>3<\/sub>(aq) is\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">HCl(aq) + NH<sub>3<\/sub>(aq) $latex \\longrightarrow$ NH<sub>4<\/sub>Cl(aq)<\/span><\/span><\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 1<\/h3>\r\n<p id=\"ball-ch04_s05_p07\" class=\"para\">Write the neutralization reactions between each acid and base.<\/p>\r\n<p class=\"para\">a) HNO<sub class=\"subscript\">3<\/sub>(aq) and Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 b)H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>(aq) and Ca(OH)<sub class=\"subscript\">2(aq)<\/sub><\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch04_s05_p08\" class=\"para\">First, we will write the chemical equation with the formulas of the reactants and the expected products; then we will balance the equation.<\/p>\r\n<p class=\"para\">a) The expected products are water and barium nitrate, so the initial chemical reaction is<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">HNO<sub class=\"subscript\">3<\/sub>(aq) +\u00a0Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$ H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Ba(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq)<\/span><\/span>\r\n<p id=\"ball-ch04_s05_p09\" class=\"para\">To balance the equation, we need to realize that there will be two H<sub class=\"subscript\">2<\/sub>O molecules, so two HNO<sub class=\"subscript\">3<\/sub> molecules are required:<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">2HNO<sub class=\"subscript\">3<\/sub>(aq) +\u00a0Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$ 2H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Ba(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq)<\/span><\/span>\r\n<p id=\"ball-ch04_s05_p10\" class=\"para\">This chemical equation is now balanced.<\/p>\r\n<p class=\"para\">b) The expected products are water and calcium phosphate, so the initial chemical equation is<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0Ca(OH)<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$ H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Ca<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span>\r\n<p id=\"ball-ch04_s05_p11\" class=\"para\">According to the solubility rules, Ca<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub> is insoluble, so it has an (s) phase label. To balance this equation, we need two phosphate ions and three calcium ions; we end up with six water molecules to balance the equation:<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">2 H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>(aq) +\u00a03 Ca(OH)<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$ 6 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Ca<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span>\r\n<p id=\"ball-ch04_s05_p12\" class=\"para\">This chemical equation is now balanced.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s05_p13\" class=\"para\">Write the neutralization reaction between H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) and Sr(OH)<sub class=\"subscript\">2<\/sub>(aq).<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s05_p14\" class=\"para\">H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0Sr(OH)<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$ 2 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0SrSO<sub class=\"subscript\">4<\/sub>(aq)<\/p>\r\n\r\n<\/div>\r\n<p id=\"ball-ch04_s05_p15\" class=\"para editable block\">Neutralization reactions are one type of chemical reaction that proceeds even if one reactant is not in the aqueous phase. For example, the chemical reaction between HCl(aq) and Fe(OH)<sub class=\"subscript\">3<\/sub>(s) still proceeds according to the equation<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">3 HCl(aq) +\u00a0Fe(OH)<sub class=\"subscript\">3<\/sub>(s) $latex \\longrightarrow$ 3 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0FeCl<sub class=\"subscript\">3<\/sub>(aq)<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s05_p16\" class=\"para editable block\">even though Fe(OH)<sub class=\"subscript\">3<\/sub> is not soluble. When one realizes that Fe(OH)<sub class=\"subscript\">3<\/sub>(s) is a component of rust, this explains why some cleaning solutions for rust stains contain acids\u2014the neutralization reaction produces products that are soluble and wash away. Washing with acids like HCl is one way to remove rust and rust stains, but HCl must be used with caution!<\/p>\r\n<p id=\"ball-ch04_s05_p17\" class=\"para editable block\">Complete and net ionic reactions for neutralization reactions will depend on whether the reactants and products are soluble, even if the acid and base react. For example, in the reaction of HCl(aq) and NaOH(aq),<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">HCl(aq) +\u00a0NaOH(aq) $latex \\longrightarrow$ H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0NaCl(aq)<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s05_p18\" class=\"para editable block\">the complete ionic reaction is<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s05_p19\" class=\"para editable block\">The Na<sup class=\"superscript\">+<\/sup>(aq) and Cl<sup class=\"superscript\">\u2212<\/sup>(aq) ions are spectator ions, so we can remove them to have<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s05_p20\" class=\"para editable block\">as the net ionic equation. If we wanted to write this in terms of the hydronium ion, H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq), we would write it as<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ 2H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s05_p21\" class=\"para editable block\">With the exception of the introduction of an extra water molecule, these two net ionic equations are equivalent.<\/p>\r\n<p id=\"ball-ch04_s05_p22\" class=\"para editable block\">However, for the reaction between HCl(aq) and Cr(OH)<sub class=\"subscript\">2<\/sub>(s), because chromium(II) hydroxide is insoluble, we cannot separate it into ions for the complete ionic equation:<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 H<sup class=\"superscript\">+<\/sup>(aq) +\u00a02 Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Cr(OH)<sub class=\"subscript\">2<\/sub>(s) $latex \\longrightarrow$ 2 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Cr<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\r\n<p id=\"ball-ch04_s05_p23\" class=\"para editable block\">The chloride ions are the only spectator ions here, so the net ionic equation is<\/p>\r\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cr(OH)<sub class=\"subscript\">2<\/sub>(s) $latex \\longrightarrow$ 2 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Cr<sup class=\"superscript\">2+<\/sup>(aq)<\/span><\/span><\/p>\r\n\r\n<div class=\"key_takeaways editable block\" id=\"ball-ch04_s05_n04\"><section id=\"fs-idp128853312\">\r\n<div class=\"textbox shaded\" id=\"fs-idm49295040\">\r\n<h3 id=\"fs-idm22209232\">Example 2<\/h3>\r\nWrite balanced chemical equations for the acid-base reactions described here:\r\n<p id=\"fs-idp55337856\">a) the <strong>weak acid<\/strong> hydrogen hypochlorite reacts with water<\/p>\r\n<p id=\"fs-idm22923872\">b) a solution of barium hydroxide is neutralized with a solution of nitric acid<\/p>\r\n&nbsp;\r\n<p id=\"fs-idp20180224\"><strong>Solution<\/strong>\r\na) The two reactants are provided, HOCl and H<sub>2<\/sub>O. Since the substance is reported to be an acid, its reaction with water will involve the transfer of H<sup>+<\/sup> from HOCl to H<sub>2<\/sub>O to generate hydronium ions, H<sub>3<\/sub>O<sup>+<\/sup> and hypochlorite ions, OCl<sup>\u2212<\/sup>.<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idm141852368\" style=\"text-align: center\">$latex \\text{HOCl}(aq) + \\text{H}_2 \\text{O}(l) \\rightleftharpoons \\text{OCl}^{-}(aq) + \\text{H}_3 \\text{O}^{+}(aq)$<\/div>\r\n<p id=\"fs-idp73299936\">A double-arrow is appropriate in this equation because it indicates the HOCl is a weak acid that has not reacted completely.<\/p>\r\n<p id=\"fs-idm23237280\">b) The two reactants are provided, Ba(OH)<sub>2<\/sub> and HNO<sub>3<\/sub>. Since this is a neutralization reaction, the two products will be water and a salt composed of the cation of the ionic hydroxide (Ba<sup>2+<\/sup>) and the anion generated when the acid transfers its hydrogen ion (NO<sup>3\u2212<\/sup>).<\/p>\r\n\r\n<div class=\"equation\" id=\"fs-idp42762464\" style=\"text-align: center\">$latex \\text{Ba(OH)}_2(aq) + 2\\text{HNO}_3(aq) \\longrightarrow \\text{Ba(NO}_3)_2(aq) + 2\\text{H}_2 \\text{O}(l)$<\/div>\r\n&nbsp;\r\n<p id=\"fs-idm20270192\"><em><strong>Test Yourself<\/strong><\/em>\r\nWrite the net ionic equation representing the neutralization of any strong acid with an ionic hydroxide. Hint: Consider the ions produced when a strong acid is dissolved in water.<\/p>\r\n&nbsp;\r\n\r\n<em><strong>Answer<\/strong><\/em>\r\n\r\n$latex \\text{H}_3 \\text{O}^{+}(aq) + \\text{OH}^{-}(aq) \\longrightarrow 2\\text{H}_2 \\text{O}(l)$\r\n\r\n<\/div>\r\n<div class=\"textbox shaded\">\r\n<h3 class=\"title\">Example 3<\/h3>\r\n<p id=\"ball-ch04_s05_p24\" class=\"para\">Oxalic acid, H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>(s), and Ca(OH)<sub class=\"subscript\">2<\/sub>(s) react very slowly. What is the net ionic equation between these two substances if the salt formed is insoluble? The anion in oxalic acid is the oxalate ion, C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\r\n<p id=\"ball-ch04_s05_p25\" class=\"para\">The products of the neutralization reaction will be water and calcium oxalate:<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>(s) +\u00a0Ca(OH)<sub class=\"subscript\">2<\/sub>(s) $latex \\longrightarrow$ 2 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0CaC<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>(s)<\/span><\/span>\r\n<p id=\"ball-ch04_s05_p26\" class=\"para\">Because nothing is dissolved, there are no substances to separate into ions, so the net ionic equation is the equation of the three solids and one liquid.<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s05_p27\" class=\"para\">What is the net ionic equation between HNO<sub class=\"subscript\">3<\/sub>(aq) and Ti(OH)<sub class=\"subscript\">4<\/sub>(s)?<\/p>\r\n&nbsp;\r\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\r\n<p id=\"ball-ch04_s05_p28\" class=\"para\">4 H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Ti(OH)<sub class=\"subscript\">4<\/sub>(s) $latex \\longrightarrow$ 4 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Ti<sup class=\"superscript\">4+<\/sup>(aq)<\/p>\r\n\r\n<\/div>\r\n<div id=\"fs-idp164722352\" class=\"textbox shaded\">\r\n\r\n<img src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Interactive_200DPI-4-3.png\" alt=\"\u00a0\" width=\"132\" height=\"82\" class=\"alignleft\" \/>\r\n\r\n&nbsp;\r\n<p id=\"fs-idm40317104\">Explore the microscopic <a href=\"http:\/\/openstaxcollege.org\/l\/16AcidsBases\">view<\/a> of strong and weak acids and bases.<\/p>\r\n&nbsp;\r\n\r\n<\/div>\r\n<\/section><section id=\"fs-idm51820592\" class=\"summary\">\r\n<h2>Gas-forming Acid-Base reactions<\/h2>\r\nA driving force for certain acid-base reactions is the formation of a gas. Common gases formed are\u00a0 H<sub>2<\/sub>, O<sub>2<\/sub>, and CO<sub>2<\/sub>.\r\n\r\nFor example:\r\n<p style=\"text-align: center\">2HCl(aq) + Na<sub>2<\/sub>CO<sub>3<\/sub>(aq) $latex \\longrightarrow$ H<sub>2<\/sub>CO<sub>3<\/sub>(aq) + 2NaCl(aq) $latex \\longrightarrow$ CO<sub>2<\/sub>(g) + H<sub>2<\/sub>O(l) + 2NaCl(aq)<\/p>\r\nThe above example can be viewed as an acid-base reaction followed by a decomposition. The driving force in this case is the gas formation. \u00a0The decomposition of H<sub>2<\/sub>CO<sub>3\u00a0<\/sub>into CO<sub>2\u00a0<\/sub>and H<sub>2<\/sub>O is a very common reaction. Both Na<sub>2<\/sub>CO<sub>3<\/sub> and NaHCO<sub>3<\/sub> mixed with acid result in a gas-forming acid-base reaction.\r\n<p style=\"text-align: center\">HCl(aq) + NaHCO<sub>3<\/sub>(aq) $latex \\longrightarrow$ H<sub>2<\/sub>CO<sub>3<\/sub>(aq) + NaCl(aq) $latex \\longrightarrow$ CO<sub>2<\/sub>(g) + H<sub>2<\/sub>O(l) + NaCl(aq)<\/p>\r\n\r\n<div class=\"textbox shaded\">\r\n<div class=\"callout block\" id=\"ball-ch04_s06_n04\">\r\n<h3 class=\"title\">Food and Drink App: Acids in Foods<\/h3>\r\n<p id=\"ball-ch04_s06_p67\" class=\"para\">Many foods and beverages contain acids. Acids impart a sour note to the taste of foods, which may add some pleasantness to the food. For example, orange juice contains citric acid, H<sub class=\"subscript\">3<\/sub>C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">7<\/sub>. Note how this formula shows hydrogen atoms in two places; the first hydrogen atoms written are the hydrogen atoms that can form H<sup class=\"superscript\">+<\/sup> ions, while the second hydrogen atoms written are part of the citrate ion, C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">7<\/sub><sup class=\"superscript\">3\u2212<\/sup>. Lemons and limes contain much more citric acid\u2014about 60 times as much\u2014which accounts for these citrus fruits being more sour than most oranges. Vinegar is essentially a ~5% solution of acetic acid (HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>) in water. Apples contain malic acid (H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">5<\/sub>; the name <em class=\"emphasis\">malic acid<\/em> comes from the apple\u2019s botanical genus name, <em class=\"emphasis\">malus<\/em>), while lactic acid (HC<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">3<\/sub>) is found in wine and sour milk products, such as yogurt and some cottage cheeses.<\/p>\r\n<p id=\"ball-ch04_s06_p68\" class=\"para\"><a class=\"xref\" href=\"#ball-ch04_s06_t01\">Table 3 \"Various Acids Found in Food and Beverages\"<\/a> lists some acids found in foods, either naturally or as an additive. Frequently, the salts of acid anions are used as additives, such as monosodium glutamate (MSG), which is the sodium salt derived from glutamic acid. As you read the list, you should come to the inescapable conclusion that it is impossible to avoid acids in food and beverages.<\/p>\r\n\r\n<div class=\"table\" id=\"ball-ch04_s06_t01\">\r\n<table style=\"border-spacing: 0px\" cellpadding=\"0\">\r\n<thead>\r\n<tr style=\"height: 38px\">\r\n<th style=\"height: 38px\">Acid Name<\/th>\r\n<th style=\"height: 38px\">Acid Formula<\/th>\r\n<th style=\"height: 38px\">Use and Appearance<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">acetic acid<\/td>\r\n<td style=\"height: 24px\">HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><\/td>\r\n<td style=\"height: 24px\">flavouring; found in vinegar<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">adipic acid<\/td>\r\n<td style=\"height: 24px\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">8<\/sub>O<sub class=\"subscript\">4<\/sub><\/td>\r\n<td style=\"height: 24px\">flavouring; found in processed foods and some antacids<\/td>\r\n<\/tr>\r\n<tr style=\"height: 19px\">\r\n<td style=\"height: 19px\">alginic acid<\/td>\r\n<td style=\"height: 19px\">various<\/td>\r\n<td style=\"height: 19px\">thickener; found in drinks, ice cream, and weight loss products<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">ascorbic acid<\/td>\r\n<td style=\"height: 24px\">HC<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">7<\/sub>O<sub class=\"subscript\">6<\/sub><\/td>\r\n<td style=\"height: 24px\">antioxidant, also known as vitamin C; found in fruits and vegetables<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">benzoic acid<\/td>\r\n<td style=\"height: 24px\">HC<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>CO<sub class=\"subscript\">2<\/sub><\/td>\r\n<td style=\"height: 24px\">preservative; found in processed foods<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">citric acid<\/td>\r\n<td style=\"height: 24px\">H<sub class=\"subscript\">3<\/sub>C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">7<\/sub><\/td>\r\n<td style=\"height: 24px\">flavouring; found in citrus fruits<\/td>\r\n<\/tr>\r\n<tr style=\"height: 38px\">\r\n<td style=\"height: 38px\">dehydroacetic acid<\/td>\r\n<td style=\"height: 38px\">HC<sub class=\"subscript\">8<\/sub>H<sub class=\"subscript\">7<\/sub>O<sub class=\"subscript\">4<\/sub><\/td>\r\n<td style=\"height: 38px\">preservative, especially for strawberries and squash<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">erythrobic acid<\/td>\r\n<td style=\"height: 24px\">HC<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">7<\/sub>O<sub class=\"subscript\">6<\/sub><\/td>\r\n<td style=\"height: 24px\">antioxidant; found in processed foods<\/td>\r\n<\/tr>\r\n<tr style=\"height: 19px\">\r\n<td style=\"height: 19px\">fatty acids<\/td>\r\n<td style=\"height: 19px\">various<\/td>\r\n<td style=\"height: 19px\">thickener and emulsifier; found in processed foods<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">fumaric acid<\/td>\r\n<td style=\"height: 24px\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub><\/td>\r\n<td style=\"height: 24px\">flavouring; acid reactant in some baking powders<\/td>\r\n<\/tr>\r\n<tr style=\"height: 38px\">\r\n<td style=\"height: 38px\">glutamic acid<\/td>\r\n<td style=\"height: 38px\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">7<\/sub>NO<sub class=\"subscript\">4<\/sub><\/td>\r\n<td style=\"height: 38px\">flavouring; found in processed foods and in tomatoes, some cheeses, and soy products<\/td>\r\n<\/tr>\r\n<tr style=\"height: 38px\">\r\n<td style=\"height: 38px\">lactic acid<\/td>\r\n<td style=\"height: 38px\">HC<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">3<\/sub><\/td>\r\n<td style=\"height: 38px\">flavouring; found in wine, yogurt, cottage cheese, and other sour milk products<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">malic acid<\/td>\r\n<td style=\"height: 24px\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">5<\/sub><\/td>\r\n<td style=\"height: 24px\">flavouring; found in apples and unripe fruit<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">phosphoric acid<\/td>\r\n<td style=\"height: 24px\">H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub><\/td>\r\n<td style=\"height: 24px\">flavouring; found in some colas<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">propionic acid<\/td>\r\n<td style=\"height: 24px\">HC<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">2<\/sub><\/td>\r\n<td style=\"height: 24px\">preservative; found in baked goods<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">sorbic acid<\/td>\r\n<td style=\"height: 24px\">HC<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">7<\/sub>O<sub class=\"subscript\">2<\/sub><\/td>\r\n<td style=\"height: 24px\">preservative; found in processed foods<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">stearic acid<\/td>\r\n<td style=\"height: 24px\">HC<sub class=\"subscript\">18<\/sub>H<sub class=\"subscript\">35<\/sub>O<sub class=\"subscript\">2<\/sub><\/td>\r\n<td style=\"height: 24px\">anticaking agent; found in hard candies<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">succinic acid<\/td>\r\n<td style=\"height: 24px\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">4<\/sub><\/td>\r\n<td style=\"height: 24px\">flavouring; found in wine and beer<\/td>\r\n<\/tr>\r\n<tr style=\"height: 24px\">\r\n<td style=\"height: 24px\">tartaric acid<\/td>\r\n<td style=\"height: 24px\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">6<\/sub><\/td>\r\n<td style=\"height: 24px\">flavouring; found in grapes, bananas, and tamarinds<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n<strong><span class=\"title-prefix\">Table 3.<\/span><\/strong>\u00a0Various Acids Found in Food and Beverages\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2>Key Concepts and Summary<\/h2>\r\nChemical reactions are classified according to similar patterns of behaviour. Acid-base reactions involve the transfer of hydrogen ions between reactants.\r\n\r\nGeneral acid-base reactions, also called neutralization reactions can be summarized with the following reaction equation:\r\n<p style=\"text-align: center\">ACID(aq) + BASE(aq)\u00a0$latex \\longrightarrow$ H<sub>2<\/sub>O(l) + SALT(aq) or (s)<\/p>\r\nThe DRIVING FORCE for a general acid-base reaction is the formation of water.\r\n\r\nGas-forming acid-base reactions can be summarized with the following reaction equation:\r\n<p style=\"text-align: center\">ACID(aq) + NaHCO<sub>3<\/sub> or Na<sub>2<\/sub>CO<sub>3<\/sub>(aq)\u00a0$latex \\longrightarrow$ H<sub>2<\/sub>O(l) + CO<sub>2<\/sub>(g) + SALT(aq) or (s)<\/p>\r\nThe DRIVING FORCE for a gas-forming acid-base reaction is the formation of gas.\u00a0There are three ways of\r\n\r\nThere are three ways of representing a neutralization reaction, using a molecular equation, complete ionic equation or net ionic equation, as described in section 6.1.\r\n<div class=\"key_takeaways editable block\" id=\"ball-ch04_s05_n04\">\r\n<div class=\"bcc-box bcc-info\">\r\n<h3>Exercises<\/h3>\r\n<div class=\"qandaset block\" id=\"ball-ch04_s05_qs01\">\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s05_qs01_p1\" class=\"para\">1. What is the Arrhenius definition of an acid?<\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">2. What is the Arrhenius definition of a base?<\/span><\/p>\r\n<p class=\"para\"><span style=\"font-size: 1em\">3. Predict the products of each acid-base combination listed. Assume that a neutralization reaction occurs.<\/span><\/p>\r\n\r\n<\/div>\r\na) \u00a0HCl and KOH\r\n\r\nb) \u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> and KOH\r\n\r\nc) \u00a0H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> and Ni(OH)<sub class=\"subscript\">2<\/sub>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s05_qs01_p7\" class=\"para\">4. \u00a0Write a balanced chemical equation for each neutralization reaction in Exercise 3.<\/p>\r\n\r\n<\/div>\r\n<span style=\"font-size: 1em\">5. \u00a0Write a balanced chemical equation for the neutralization reaction between each given acid and base. Include the proper phase labels.<\/span>\r\n<div class=\"question\">\r\n\r\na) \u00a0HI(aq) +\u00a0KOH(aq) $latex \\longrightarrow$ ?\r\n\r\nb) \u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$ ?\r\n\r\n<\/div>\r\n<div class=\"question\">\r\n<p id=\"ball-ch04_s05_qs01_p10\" class=\"para\"><span style=\"font-size: 1em\">6. \u00a0Write the net ionic equation for each neutralization reaction in Exercise 7.<\/span><\/p>\r\n\r\n<\/div>\r\n<span style=\"font-size: 1em\">7. \u00a0Write the complete and net ionic equations for the neutralization reaction between HClO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">(aq) and Zn(OH)<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(s). Assume the salt is soluble.<\/span>\r\n\r\n<span style=\"font-size: 1em\">8. \u00a0Explain why the net ionic equation for the neutralization reaction between HCl(aq) and KOH(aq) is the same as the net ionic equation for the neutralization reaction between HNO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">(aq) and RbOH.<\/span>\r\n\r\n<span style=\"font-size: 1em\">9. \u00a0Write the complete and net ionic equations for the neutralization reaction between HCl(aq) and KOH(aq) using the hydronium ion in place of H<\/span><sup class=\"superscript\">+<\/sup><span style=\"font-size: 1em\">. What difference does it make when using the hydronium ion?<\/span>\r\n\r\n10.\u00a0Complete and balance the following acid-base equations:\r\n<p id=\"fs-idp63939984\">a) HCl gas reacts with solid Ca(OH)<sub>2<\/sub>(<em>s<\/em>).<\/p>\r\n<p id=\"fs-idm54386864\">b) A solution of Sr(OH)<sub>2<\/sub> is added to a solution of HNO<sub>3<\/sub>.<\/p>\r\n11. Complete and balance the equations for the following acid-base neutralization reactions. If water is used as a solvent, write the reactants and products as aqueous ions. In some cases, there may be more than one correct answer, depending on the amounts of reactants used.\r\n<p id=\"fs-idm49795408\">a) $latex \\text{Mg(OH)}_2(s) + \\text{HClO}_4(aq) \\longrightarrow $<\/p>\r\n<p id=\"fs-idp205785072\">b) $latex \\text{SrO}(s) + \\text{H}_2 \\text{SO}_4(l) \\longrightarrow $<\/p>\r\n12. Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:\r\n<p id=\"fs-idp89463616\">a) $latex \\text{Ca(OH)}_2(s) + \\text{H}_2 \\text{S}(g) \\longrightarrow $<\/p>\r\n<p id=\"fs-idp2916112\">b) $latex \\text{Na}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{S}(g) \\longrightarrow $<\/p>\r\n\r\n<\/div>\r\n&nbsp;\r\n\r\n<b>Answers<\/b>\r\n\r\n<span style=\"font-size: 1em\">1. An Arrhenius acid increases the amount of H<\/span><sup class=\"superscript\">+<\/sup><span style=\"font-size: 1em\"> ions in an aqueous solution.<\/span>\r\n<div class=\"answer\">\r\n<p class=\"para\">2.\u00a0<span style=\"font-size: 1em\">An Arrhenius base increases the amount of OH<\/span><sup class=\"superscript\">-<\/sup><span style=\"font-size: 1em\"> ions in an aqueous solution.<\/span><\/p>\r\n\r\n<\/div>\r\n3.\u00a0a) \u00a0KCl and H<sub class=\"subscript\">2<\/sub>O\r\n<div class=\"answer\">\r\n\r\nb) \u00a0K<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> and H<sub class=\"subscript\">2<\/sub>O\r\n\r\nc) \u00a0Ni<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub> and H<sub class=\"subscript\">2<\/sub>O\r\n\r\n<\/div>\r\n4.\u00a0a) \u00a0 \u00a0HCl +\u00a0KOH $latex \\longrightarrow$ KCl +\u00a0H<sub class=\"subscript\">2<\/sub>O\r\n\r\nb) \u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> +\u00a02 KOH $latex \\longrightarrow$ K<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O\r\n\r\nc) \u00a02 H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> +\u00a03 Ni(OH)<sub class=\"subscript\">2<\/sub> $latex \\longrightarrow$ Ni<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O\r\n\r\n<span style=\"font-size: 1em\">5.\u00a0<\/span><span style=\"font-size: 1em\">a) \u00a0HI(aq) +\u00a0KOH(aq) $latex \\longrightarrow$ KCl(aq) +\u00a0H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O(\u2113)<\/span>\r\n<div class=\"answer\">\r\n\r\nb) \u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) $latex \\longrightarrow$ BaSO<sub class=\"subscript\">4<\/sub>(s) +\u00a02 H<sub class=\"subscript\">2<\/sub>O(\u2113)\r\n\r\n<\/div>\r\n<span style=\"font-size: 1em\">6.\u00a0<\/span><span style=\"font-size: 1em\">a) \u00a0H<\/span><sup class=\"superscript\">+<\/sup><span style=\"font-size: 1em\">(aq) +\u00a0OH<\/span><sup class=\"superscript\">\u2212<\/sup><span style=\"font-size: 1em\">(aq) $latex \\longrightarrow$ H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O(\u2113)<\/span>\r\n<div class=\"answer\">\r\n\r\nb) \u00a02 H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq) +\u00a0Ba<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 OH<sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ BaSO<sub class=\"subscript\">4<\/sub>(s) +\u00a02 H<sub class=\"subscript\">2<\/sub>O(\u2113)\r\n\r\n7. \u00a0Complete ionic equation:\r\n\r\n<\/div>\r\n<span style=\"font-size: 1em\">2 H<\/span><sup class=\"superscript\">+<\/sup><span style=\"font-size: 1em\">(aq) +\u00a02 ClO<\/span><sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup><span style=\"font-size: 1em\">(aq) +\u00a0Zn<\/span><sup class=\"superscript\">2+<\/sup><span style=\"font-size: 1em\">(aq) +\u00a02 OH<\/span><sup class=\"superscript\">\u2212<\/sup><span style=\"font-size: 1em\">(aq) $latex \\longrightarrow$ Zn<\/span><sup class=\"superscript\">2+<\/sup><span style=\"font-size: 1em\">(aq) +\u00a02 ClO<\/span><sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup><span style=\"font-size: 1em\">(aq) +\u00a02 H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O(\u2113)<\/span>\r\n<div class=\"answer\">\r\n<p id=\"ball-ch04_s05_qs01_p15_ans\" class=\"para\">Net ionic equation:<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">2 H<sup class=\"superscript\">+<\/sup>(aq) +\u00a02 OH<sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ 2 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span>\r\n\r\n<\/div>\r\n<span style=\"font-size: 1em\">8. \u00a0Because the salts are soluble in both cases, the net ionic reaction is just H<\/span><sup class=\"superscript\">+<\/sup><span style=\"font-size: 1em\">(aq) +\u00a0OH<\/span><sup class=\"superscript\">\u2212<\/sup><span style=\"font-size: 1em\">(aq) $latex \\longrightarrow$ H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O(\u2113).<\/span>\r\n\r\n<span style=\"font-size: 1em\">9. \u00a0Complete ionic equation:<\/span>\r\n<div class=\"answer\">\r\n\r\n<span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ 2 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span>\r\n<p id=\"ball-ch04_s05_qs01_p25_ans\" class=\"para\">Net ionic equation:<\/p>\r\n<span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) $latex \\longrightarrow$ 2 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span>\r\n<p id=\"ball-ch04_s05_qs01_p26_ans\" class=\"para\">The difference is simply the presence of an extra water molecule as a product.<\/p>\r\n10.\u00a0<span style=\"font-size: 1em\">a) $latex 2\\text{HCl}(g) + \\text{Ca(OH)}_2(s) \\longrightarrow \\text{CaCl}_2(s) + 2\\text{H}_2 \\text{O}(l)$;<\/span>\r\n<p id=\"fs-idm141927648\">b) $latex \\text{Sr(OH)}_2(aq) + 2\\text{HNO}_3(aq) \\longrightarrow \\text{Sr(NO}_3)_2(aq) + 2\\text{H}_2 \\text{O}(l)$;<\/p>\r\n<p id=\"fs-idp32868368\">11.\u00a0a) $latex \\text{Mg(OH)}_2(s) + 2\\text{HClO}_4(aq) \\longrightarrow \\text{Mg}^{2+}(aq) + 2{\\text{ClO}_4}^{-}(aq) + 2\\text{H}_2 \\text{O}(l); $\r\nb) $latex \\text{SrO}(s) + \\text{H}_2 \\text{SO}_4(l) \\longrightarrow \\text{SrSO}_4(s) + \\text{H}_2 \\text{O}$<\/p>\r\n<p id=\"fs-idp19195056\">12.\u00a0a) $latex \\text{Ca(OH)}_2(s) + \\text{H}_2 \\text{S}(g) \\longrightarrow \\text{CaS}(s) + 2\\text{H}_2\\text{O}(l);$\r\nb) $latex \\text{Na}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{S}(g) \\longrightarrow \\text{Na}_2 \\text{S}(aq) + \\text{CO}_2(g) + \\text{H}_2 \\text{O}(l)$<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<h2>Glossary<\/h2>\r\n<strong>acid:\u00a0<\/strong>substance that produces H<sub>3<\/sub>O<sup>+<\/sup> when dissolved in water\r\n\r\n<strong>acid-base reaction:\u00a0<\/strong>reaction involving the transfer of a hydrogen ion between reactant species\r\n\r\n<strong>base:\u00a0<\/strong>substance that produces OH<sup>\u2212<\/sup> when dissolved in water\r\n\r\n<strong>neutralization reaction:\u00a0<\/strong>reaction between an acid and a base to produce salt and water\r\n\r\n<strong>salt:\u00a0<\/strong>ionic compound that can be formed by the reaction of an acid with a base that contains a cation and an anion other than hydroxide or oxide\r\n\r\n<strong>strong acid:\u00a0<\/strong>acid that reacts completely when dissolved in water to yield hydronium ions\r\n\r\n<strong>strong base:\u00a0<\/strong>base that reacts completely when dissolved in water to yield hydroxide ions\r\n\r\n<strong>weak acid:\u00a0<\/strong>acid that reacts only to a slight extent when dissolved in water to yield hydronium ions\r\n\r\n<strong>weak base:\u00a0<\/strong>base that reacts only to a slight extent when dissolved in water to yield hydroxide ions\r\n\r\n<\/section><\/div>\r\n<\/div>","rendered":"<div class=\"section\" id=\"ball-ch04_s05\" lang=\"en\">\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Identify common acids and bases<\/li>\n<li>Define acid-base reactions<\/li>\n<li>Recognize and identify examples of acid-base reactions<\/li>\n<li><span style=\"font-size: 1em\">Predict the products of acid-base reactions.<\/span><\/li>\n<\/ul>\n<\/div>\n<\/div>\n<h2>Acids and Bases<\/h2>\n<div class=\"section\" id=\"ball-ch04_s05\" lang=\"en\">\n<p class=\"para block\">The definition of an <span class=\"margin_term\"><a class=\"glossterm\">acid<\/a><\/span>\u00a0is often cited as: any compound that increases the amount of hydrogen ion (H<sup class=\"superscript\">+<\/sup>) in an aqueous solution. The chemical opposite of an acid is a base. The equivalent definition of a base is that a <span class=\"margin_term\"><a class=\"glossterm\">base<\/a><\/span>\u00a0is a compound that increases the amount of hydroxide ion (OH<sup class=\"superscript\">\u2212<\/sup>) in an aqueous solution. These original definitions were proposed by Arrhenius (the same person who proposed ion dissociation) in 1884, so they are referred to as the <strong class=\"emphasis bold\">Arrhenius definition<\/strong> of an acid and a base, respectively.<\/p>\n<p id=\"ball-ch04_s05_p02\" class=\"para block\">You may recognize that, based on the description of a hydrogen atom, an H<sup class=\"superscript\">+<\/sup> ion is a hydrogen atom that has lost its lone electron; that is, H<sup class=\"superscript\">+<\/sup> is simply a proton. Do we really have bare protons moving about in aqueous solution? No. What is more likely is that the H<sup class=\"superscript\">+<\/sup> ion has attached itself to one (or more) water molecule(s). To represent this chemically, we define the <span class=\"margin_term\"><a class=\"glossterm\">hydronium ion\u00a0H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup><\/a><span class=\"glossdef\"><span class=\"inlineequation\">(aq)<\/span>, a water molecule with an extra hydrogen ion attached to it.<\/span><\/span> as H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>, which represents an additional proton attached to a water molecule. We use the hydronium ion as the more logical way a hydrogen ion appears in an aqueous solution, although in many chemical reactions H<sup class=\"superscript\">+<\/sup> and H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup> are treated equivalently.<\/p>\n<p id=\"fs-idp89436016\">For purposes of this brief introduction, we will consider only the more common types of acid-base reactions that take place in aqueous solutions. In this context, an <strong>acid<\/strong> is a substance that will dissolve in water to yield hydronium ions, H<sub>3<\/sub>O<sup>+<\/sup>. As an example, consider the equation shown here:<\/p>\n<div class=\"equation\" id=\"fs-idm54028336\" style=\"text-align: center\">[latex]\\text{HCl}(aq) + \\text{H}_2 \\text{O}(aq) \\longrightarrow \\text{Cl}^{-}(aq) + \\text{H}_3 \\text{O}^{+}(aq)[\/latex]<\/div>\n<p id=\"fs-idm10390992\">The process represented by this equation confirms that hydrogen chloride is an acid. When dissolved in water, H<sub>3<\/sub>O<sup>+<\/sup> ions are produced by a chemical reaction in which H<sup>+<\/sup> ions are transferred from HCl molecules to H<sub>2<\/sub>O molecules (<a href=\"#CNX_Chem_04_02_HClsoln\" class=\"autogenerated-content\">Figure 1<\/a>).<\/p>\n<figure id=\"attachment_1414\" aria-describedby=\"caption-attachment-1414\" style=\"width: 400px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_HClsoln-3.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_HClsoln-3-300x170.jpg\" alt=\"\" width=\"400\" height=\"227\" class=\"wp-image-1414\" srcset=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_HClsoln-3-300x170.jpg 300w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_HClsoln-3-768x436.jpg 768w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_HClsoln-3-65x37.jpg 65w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_HClsoln-3-225x128.jpg 225w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_HClsoln-3-350x199.jpg 350w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_HClsoln-3.jpg 975w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><\/a><figcaption id=\"caption-attachment-1414\" class=\"wp-caption-text\"><strong>Figure 1.<\/strong> When hydrogen chloride gas dissolves in water, (a) it reacts as an acid, transferring protons to water molecules to yield (b) hydronium ions (and solvated chloride ions).<\/figcaption><\/figure>\n<figure id=\"CNX_Chem_04_02_HClsoln\"><figcaption><\/figcaption><\/figure>\n<p>The nature of HCl is such that its reaction with water as just described is essentially 100% efficient: Virtually every HCl molecule that dissolves in water will undergo this reaction. Acids that completely react in this fashion are called <strong>strong acids<\/strong>, and HCl is one among just a handful of common acid compounds that are classified as strong (<a href=\"#fs-idp55395904\" class=\"autogenerated-content\">Table 1<\/a>).<\/p>\n<div class=\"section\" id=\"ball-ch04_s05\" lang=\"en\">\n<table id=\"fs-idp55395904\" class=\"span-all\" summary=\"This table contains two columns and seven rows. The columns are labeled, \u201cCompound Formula,\u201d and, \u201cName in Aqueous Solution.\u201d Under the column, \u201cCompound Formula,\u201d are: \u201cH B r,\u201d \u201cH C l,\u201d \u201cH I,\u201d \u201cH N O subscript 3,\u201d \u201cH C l O subscript 4,\u201d and, \u201cH subscript 2 S O subscript 4.\u201d Under the column, \u201cName in Aqueous Solution,\u201d are: \u201chydrobromic acid,\u201d \u201chydrochloric acid,\u201d \u201chydroiodic acid,\u201d \u201cnitric acid,\u201d \u201cperchloric acid,\u201d and, \u201csulfuric acid.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th>Compound Formula<\/th>\n<th>Name in Aqueous Solution<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td>HBr<\/td>\n<td>hydrobromic acid<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>HCl<\/td>\n<td>hydrochloric acid<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>HI<\/td>\n<td>hydroiodic acid<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>HNO<sub>3<\/sub><\/td>\n<td>nitric acid<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>HClO<sub>4<\/sub><\/td>\n<td>perchloric acid<\/td>\n<\/tr>\n<tr>\n<td>HClO<sub>3<\/sub><\/td>\n<td>chloric acid<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td>H<sub>2<\/sub>SO<sub>4<\/sub><\/td>\n<td>sulfuric acid<\/td>\n<\/tr>\n<tr>\n<td colspan=\"3\"><strong>Table 1.<\/strong> Common Strong Acids<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idp2912576\">\u00a0A far greater number of compounds behave as <strong>weak acids<\/strong> and only partially react with water, leaving a large majority of dissolved molecules in their original form and generating a relatively small amount of hydronium ions.<\/p>\n<\/div>\n<div class=\"section\" id=\"ball-ch04_s05\" lang=\"en\">\n<table id=\"fs-idp55395904\" class=\"span-all\" style=\"width: 368px\" summary=\"This table contains two columns and seven rows. The columns are labeled, \u201cCompound Formula,\u201d and, \u201cName in Aqueous Solution.\u201d Under the column, \u201cCompound Formula,\u201d are: \u201cH B r,\u201d \u201cH C l,\u201d \u201cH I,\u201d \u201cH N O subscript 3,\u201d \u201cH C l O subscript 4,\u201d and, \u201cH subscript 2 S O subscript 4.\u201d Under the column, \u201cName in Aqueous Solution,\u201d are: \u201chydrobromic acid,\u201d \u201chydrochloric acid,\u201d \u201chydroiodic acid,\u201d \u201cnitric acid,\u201d \u201cperchloric acid,\u201d and, \u201csulfuric acid.\u201d\">\n<thead>\n<tr valign=\"top\">\n<th style=\"width: 161px\">Compound Formula<\/th>\n<th style=\"width: 207px\">Name in Aqueous Solution<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr valign=\"top\">\n<td style=\"width: 161px\">HF<\/td>\n<td style=\"width: 207px\">hydrofluoric acid<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"width: 161px\">HCN<\/td>\n<td style=\"width: 207px\">hydrocyanic acid<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"width: 161px\">HC<sub>2<\/sub>H<sub>3<\/sub>O<sub>2<\/sub><\/td>\n<td style=\"width: 207px\">acetic acid<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"width: 161px\">HNO<sub>2<\/sub><\/td>\n<td style=\"width: 207px\">nitrous acid<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"width: 161px\">HClO<\/td>\n<td style=\"width: 207px\">hypochlorous acid<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 161px\">HClO<sub><span style=\"font-size: small\">2<\/span><\/sub><\/td>\n<td style=\"width: 207px\">chlorous acid<\/td>\n<\/tr>\n<tr valign=\"top\">\n<td style=\"width: 161px\">H<sub>2<\/sub>SO<sub><span style=\"font-size: small\">3<\/span><\/sub><\/td>\n<td style=\"width: 207px\">sulfurous acid<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 161px\">H<sub>2<\/sub>CO<sub>3<\/sub><\/td>\n<td style=\"width: 207px\">carbonic acid<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 161px\">H<sub>3<\/sub>PO<sub>4<\/sub><\/td>\n<td style=\"width: 207px\">phosphoric acid<\/td>\n<\/tr>\n<tr>\n<td style=\"width: 368px\" colspan=\"2\"><strong>Table 2.<\/strong> Common Weak Acids<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idp2912576\">Weak acids are commonly encountered in nature, being the substances partly responsible for the tangy taste of citrus fruits, the stinging sensation of insect bites, and the unpleasant smells associated with body odor. A familiar example of a weak acid is acetic acid, the main ingredient in food vinegars:<\/p>\n<\/div>\n<div class=\"equation\" id=\"fs-idp23273024\" style=\"text-align: center\">[latex]\\text{CH}_3 \\text{CO}_2 \\text{H}(aq) + \\text{H}_2 \\text{O}(l) \\leftrightharpoons \\text{CH}_3 {\\text{CO}_2}^{-}(aq) + \\text{H}_3 \\text{O}^{+}(aq)[\/latex]<\/div>\n<div><\/div>\n<div class=\"equation\" style=\"text-align: left\"><span style=\"text-align: justify;font-size: 14pt\">When dissolved in water under typical conditions, only about 1% of acetic acid molecules are present in the ionized form, [latex]\\text{CH}_3 {\\text{CO}_2}^{-}[\/latex](<\/span><a href=\"#CNX_Chem_04_02_Citrus\" class=\"autogenerated-content\" style=\"text-align: justify;font-size: 14pt\">Figure 2<\/a><span style=\"text-align: justify;font-size: 14pt\">). The use of a double-arrow in the equation above denotes the partial reaction aspect of this process, a concept addressed fully in the chapters on chemical equilibrium.)<\/span><\/div>\n<div>\n<div class=\"section\" id=\"ball-ch04_s05\" lang=\"en\">\n<figure id=\"attachment_1415\" aria-describedby=\"caption-attachment-1415\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_Citrus-2.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_Citrus-2-300x267.jpg\" alt=\"\" width=\"300\" height=\"267\" class=\"wp-image-1415 size-medium\" srcset=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_Citrus-2-300x267.jpg 300w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_Citrus-2-65x58.jpg 65w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_Citrus-2-225x200.jpg 225w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_Citrus-2-350x312.jpg 350w, https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_Citrus-2.jpg 650w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><figcaption id=\"caption-attachment-1415\" class=\"wp-caption-text\"><strong>Figure 2.<\/strong> (a) Fruits such as oranges, lemons, and grapefruit contain the weak acid citric acid. (b) Vinegars contain the weak acid acetic acid. (credit a: modification of work by Scott Bauer; credit b: modification of work by Br\u00fccke-Osteuropa\/Wikimedia Commons)<\/figcaption><\/figure>\n<\/div>\n<\/div>\n<figure id=\"CNX_Chem_04_02_Citrus\"><\/figure>\n<\/div>\n<div class=\"section\" id=\"ball-ch04_s05\" lang=\"en\">\n<p>A <strong>base<\/strong> is a substance that will dissolve in water to yield hydroxide ions, OH<sup>\u2212<\/sup>. The most common bases are ionic compounds composed of alkali or alkaline earth metal cations (groups 1 and 2) combined with the hydroxide ion\u2014for example, NaOH and Ca(OH)<sub>2<\/sub>. When these compounds dissolve in water, hydroxide ions are released directly into the solution. For example, KOH and Ba(OH)<sub>2<\/sub> dissolve in water and dissociate completely to produce cations (K<sup>+<\/sup> and Ba<sup>2+<\/sup>, respectively) and hydroxide ions, OH<sup>\u2212<\/sup>. These bases, along with other hydroxides that completely dissociate in water, are considered <strong>strong bases<\/strong>.<\/p>\n<p id=\"fs-idp74282160\">Consider as an example the dissolution of lye (sodium hydroxide) in water:<\/p>\n<div class=\"equation\" id=\"fs-idm62931696\" style=\"text-align: center\">[latex]\\text{NaOH}(s) \\longrightarrow \\text{Na}^{+}(aq) + \\text{OH}^{-}(aq)[\/latex]<\/div>\n<p id=\"fs-idp8724736\">This equation confirms that sodium hydroxide is a base. When dissolved in water, NaOH dissociates to yield Na<sup>+<\/sup> and OH<sup>\u2212<\/sup> ions. This is also true for any other ionic compound containing hydroxide ions. Since the dissociation process is essentially complete when ionic compounds dissolve in water under typical conditions, NaOH and other ionic hydroxides are all classified as strong bases.<\/p>\n<p id=\"fs-idm50199792\">Unlike ionic hydroxides, some compounds produce hydroxide ions when dissolved by chemically reacting with water molecules. In all cases, these compounds react only partially and so are classified as <strong>weak bases<\/strong>. These types of compounds are also abundant in nature and important commodities in various technologies. For example, global production of the weak base ammonia is typically well over 100 metric tons annually, being widely used as an agricultural fertilizer, a raw material for chemical synthesis of other compounds, and an active ingredient in household cleaners (<a href=\"#CNX_Chem_04_02_ammonia\" class=\"autogenerated-content\">Figure 3<\/a>). When dissolved in water, ammonia reacts partially to yield hydroxide ions, as shown here:<\/p>\n<div class=\"equation\" id=\"fs-idm9327664\" style=\"text-align: center\">[latex]\\text{NH}_3(aq) + \\text{H}_2 \\text{O}(l) \\rightleftharpoons {\\text{NH}_4}^{+}(aq) + \\text{OH}^{-}(aq)[\/latex]<\/div>\n<p id=\"fs-idm73811808\">Under typical conditions, only about 1% of the dissolved ammonia is present as NH<sub>4<\/sub><sup>+<\/sup> ions.<\/p>\n<figure id=\"CNX_Chem_04_02_ammonia\"><figcaption>\n<figure style=\"width: 975px\" class=\"wp-caption aligncenter\"><a href=\"https:\/\/opentextbc.ca\/chemistry\/wp-content\/uploads\/sites\/150\/2016\/05\/CNX_Chem_04_02_ammonia.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Chem_04_02_ammonia-2.jpg\" alt=\"This photograph shows a large agricultural tractor in a field pulling a field sprayer and a large, white cylindrical tank which is labeled \u201cCaution Ammonia.\u201d\" width=\"975\" height=\"316\" \/><\/a><figcaption class=\"wp-caption-text\"><strong>Figure 3.<\/strong> Ammonia is a weak base used in a variety of applications. (a) Pure ammonia is commonly applied as an agricultural fertilizer. (b) Dilute solutions of ammonia are effective household cleansers. (credit a: modification of work by National Resources Conservation Service; credit b: modification of work by pat00139)<\/figcaption><\/figure>\n<\/figcaption><\/figure>\n<h2>Acid-Base Reactions<\/h2>\n<p id=\"fs-idm1255344\">An <strong>acid-base reaction<\/strong> is one in which a hydrogen ion, H<sup>+<\/sup>, is transferred from one chemical species to another. Such reactions are of central importance to numerous natural and technological processes, ranging from the chemical transformations that take place within cells and the lakes and oceans, to the industrial-scale production of fertilizers, pharmaceuticals, and other substances essential to society. The subject of acid-base chemistry, therefore, is worthy of thorough discussion.<\/p>\n<p id=\"ball-ch04_s05_p03\" class=\"para editable block\">The reaction between an acid and a base is called an acid-base reaction or a <strong><span class=\"margin_term\"><a class=\"glossterm\">neutralization reaction<\/a><\/span><\/strong>. Although acids and bases have their own unique chemistries, the acid and base cancel each other\u2019s chemistry to produce a rather innocuous substance\u2014water. In fact, the <strong>general\u00a0<\/strong><strong>acid-base reaction<\/strong> is<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">acid +\u00a0base [latex]\\longrightarrow[\/latex] water +\u00a0salt<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p04\" class=\"para editable block\">where the term <strong><span class=\"margin_term\"><a class=\"glossterm\">salt<\/a><\/span><\/strong>\u00a0is used to define any ionic compound (soluble or insoluble) that is formed from a reaction between an acid and a base. In chemistry, the word <em class=\"emphasis\">salt<\/em> refers to more than just table salt. For example, the balanced chemical equation for the reaction between HCl(aq) and KOH(aq) is<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">HCl(aq) +\u00a0KOH(aq) [latex]\\longrightarrow[\/latex] H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0KCl(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p05\" class=\"para editable block\">where the salt is KCl. By counting the number of atoms of each element, we find that only one water molecule is formed as a product. However, in the reaction between HCl(aq) and Mg(OH)<sub class=\"subscript\">2<\/sub>(aq), additional molecules of HCl and H<sub class=\"subscript\">2<\/sub>O are required to balance the chemical equation:<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 HCl(aq) +\u00a0Mg(OH)<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex] 2 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0MgCl<sub class=\"subscript\">2<\/sub>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p06\" class=\"para editable block\">Here, the salt is MgCl<sub class=\"subscript\">2<\/sub>. This is one of several reactions that take place when a type of antacid\u2014a base\u2014is used to treat stomach acid.<\/p>\n<p>There are acid-base reactions that do not follow the &#8220;general acid-base&#8221; equation given above. \u00a0For example,\u00a0, the balanced chemical equation for the reaction between HCl(aq) and NH<sub>3<\/sub>(aq) is<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">HCl(aq) + NH<sub>3<\/sub>(aq) [latex]\\longrightarrow[\/latex] NH<sub>4<\/sub>Cl(aq)<\/span><\/span><\/p>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 1<\/h3>\n<p id=\"ball-ch04_s05_p07\" class=\"para\">Write the neutralization reactions between each acid and base.<\/p>\n<p class=\"para\">a) HNO<sub class=\"subscript\">3<\/sub>(aq) and Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 b)H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>(aq) and Ca(OH)<sub class=\"subscript\">2(aq)<\/sub><\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch04_s05_p08\" class=\"para\">First, we will write the chemical equation with the formulas of the reactants and the expected products; then we will balance the equation.<\/p>\n<p class=\"para\">a) The expected products are water and barium nitrate, so the initial chemical reaction is<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">HNO<sub class=\"subscript\">3<\/sub>(aq) +\u00a0Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex] H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Ba(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p09\" class=\"para\">To balance the equation, we need to realize that there will be two H<sub class=\"subscript\">2<\/sub>O molecules, so two HNO<sub class=\"subscript\">3<\/sub> molecules are required:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">2HNO<sub class=\"subscript\">3<\/sub>(aq) +\u00a0Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex] 2H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Ba(NO<sub class=\"subscript\">3<\/sub>)<sub class=\"subscript\">2<\/sub>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p10\" class=\"para\">This chemical equation is now balanced.<\/p>\n<p class=\"para\">b) The expected products are water and calcium phosphate, so the initial chemical equation is<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0Ca(OH)<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex] H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Ca<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p11\" class=\"para\">According to the solubility rules, Ca<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub> is insoluble, so it has an (s) phase label. To balance this equation, we need two phosphate ions and three calcium ions; we end up with six water molecules to balance the equation:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">2 H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub>(aq) +\u00a03 Ca(OH)<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex] 6 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Ca<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub>(s)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p12\" class=\"para\">This chemical equation is now balanced.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s05_p13\" class=\"para\">Write the neutralization reaction between H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) and Sr(OH)<sub class=\"subscript\">2<\/sub>(aq).<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s05_p14\" class=\"para\">H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0Sr(OH)<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex] 2 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0SrSO<sub class=\"subscript\">4<\/sub>(aq)<\/p>\n<\/div>\n<p id=\"ball-ch04_s05_p15\" class=\"para editable block\">Neutralization reactions are one type of chemical reaction that proceeds even if one reactant is not in the aqueous phase. For example, the chemical reaction between HCl(aq) and Fe(OH)<sub class=\"subscript\">3<\/sub>(s) still proceeds according to the equation<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">3 HCl(aq) +\u00a0Fe(OH)<sub class=\"subscript\">3<\/sub>(s) [latex]\\longrightarrow[\/latex] 3 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0FeCl<sub class=\"subscript\">3<\/sub>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p16\" class=\"para editable block\">even though Fe(OH)<sub class=\"subscript\">3<\/sub> is not soluble. When one realizes that Fe(OH)<sub class=\"subscript\">3<\/sub>(s) is a component of rust, this explains why some cleaning solutions for rust stains contain acids\u2014the neutralization reaction produces products that are soluble and wash away. Washing with acids like HCl is one way to remove rust and rust stains, but HCl must be used with caution!<\/p>\n<p id=\"ball-ch04_s05_p17\" class=\"para editable block\">Complete and net ionic reactions for neutralization reactions will depend on whether the reactants and products are soluble, even if the acid and base react. For example, in the reaction of HCl(aq) and NaOH(aq),<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">HCl(aq) +\u00a0NaOH(aq) [latex]\\longrightarrow[\/latex] H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0NaCl(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p18\" class=\"para editable block\">the complete ionic reaction is<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Na<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p19\" class=\"para editable block\">The Na<sup class=\"superscript\">+<\/sup>(aq) and Cl<sup class=\"superscript\">\u2212<\/sup>(aq) ions are spectator ions, so we can remove them to have<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p20\" class=\"para editable block\">as the net ionic equation. If we wanted to write this in terms of the hydronium ion, H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq), we would write it as<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] 2H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p21\" class=\"para editable block\">With the exception of the introduction of an extra water molecule, these two net ionic equations are equivalent.<\/p>\n<p id=\"ball-ch04_s05_p22\" class=\"para editable block\">However, for the reaction between HCl(aq) and Cr(OH)<sub class=\"subscript\">2<\/sub>(s), because chromium(II) hydroxide is insoluble, we cannot separate it into ions for the complete ionic equation:<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 H<sup class=\"superscript\">+<\/sup>(aq) +\u00a02 Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0Cr(OH)<sub class=\"subscript\">2<\/sub>(s) [latex]\\longrightarrow[\/latex] 2 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Cr<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p23\" class=\"para editable block\">The chloride ions are the only spectator ions here, so the net ionic equation is<\/p>\n<p style=\"text-align: center\"><span class=\"informalequation block\"><span class=\"mathphrase\">2 H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cr(OH)<sub class=\"subscript\">2<\/sub>(s) [latex]\\longrightarrow[\/latex] 2 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Cr<sup class=\"superscript\">2+<\/sup>(aq)<\/span><\/span><\/p>\n<div class=\"key_takeaways editable block\" id=\"ball-ch04_s05_n04\">\n<section id=\"fs-idp128853312\">\n<div class=\"textbox shaded\" id=\"fs-idm49295040\">\n<h3 id=\"fs-idm22209232\">Example 2<\/h3>\n<p>Write balanced chemical equations for the acid-base reactions described here:<\/p>\n<p id=\"fs-idp55337856\">a) the <strong>weak acid<\/strong> hydrogen hypochlorite reacts with water<\/p>\n<p id=\"fs-idm22923872\">b) a solution of barium hydroxide is neutralized with a solution of nitric acid<\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idp20180224\"><strong>Solution<\/strong><br \/>\na) The two reactants are provided, HOCl and H<sub>2<\/sub>O. Since the substance is reported to be an acid, its reaction with water will involve the transfer of H<sup>+<\/sup> from HOCl to H<sub>2<\/sub>O to generate hydronium ions, H<sub>3<\/sub>O<sup>+<\/sup> and hypochlorite ions, OCl<sup>\u2212<\/sup>.<\/p>\n<div class=\"equation\" id=\"fs-idm141852368\" style=\"text-align: center\">[latex]\\text{HOCl}(aq) + \\text{H}_2 \\text{O}(l) \\rightleftharpoons \\text{OCl}^{-}(aq) + \\text{H}_3 \\text{O}^{+}(aq)[\/latex]<\/div>\n<p id=\"fs-idp73299936\">A double-arrow is appropriate in this equation because it indicates the HOCl is a weak acid that has not reacted completely.<\/p>\n<p id=\"fs-idm23237280\">b) The two reactants are provided, Ba(OH)<sub>2<\/sub> and HNO<sub>3<\/sub>. Since this is a neutralization reaction, the two products will be water and a salt composed of the cation of the ionic hydroxide (Ba<sup>2+<\/sup>) and the anion generated when the acid transfers its hydrogen ion (NO<sup>3\u2212<\/sup>).<\/p>\n<div class=\"equation\" id=\"fs-idp42762464\" style=\"text-align: center\">[latex]\\text{Ba(OH)}_2(aq) + 2\\text{HNO}_3(aq) \\longrightarrow \\text{Ba(NO}_3)_2(aq) + 2\\text{H}_2 \\text{O}(l)[\/latex]<\/div>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm20270192\"><em><strong>Test Yourself<\/strong><\/em><br \/>\nWrite the net ionic equation representing the neutralization of any strong acid with an ionic hydroxide. Hint: Consider the ions produced when a strong acid is dissolved in water.<\/p>\n<p>&nbsp;<\/p>\n<p><em><strong>Answer<\/strong><\/em><\/p>\n<p>[latex]\\text{H}_3 \\text{O}^{+}(aq) + \\text{OH}^{-}(aq) \\longrightarrow 2\\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3 class=\"title\">Example 3<\/h3>\n<p id=\"ball-ch04_s05_p24\" class=\"para\">Oxalic acid, H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>(s), and Ca(OH)<sub class=\"subscript\">2<\/sub>(s) react very slowly. What is the net ionic equation between these two substances if the salt formed is insoluble? The anion in oxalic acid is the oxalate ion, C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong>Solution<\/strong><\/p>\n<p id=\"ball-ch04_s05_p25\" class=\"para\">The products of the neutralization reaction will be water and calcium oxalate:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>(s) +\u00a0Ca(OH)<sub class=\"subscript\">2<\/sub>(s) [latex]\\longrightarrow[\/latex] 2 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0CaC<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub>(s)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_p26\" class=\"para\">Because nothing is dissolved, there are no substances to separate into ions, so the net ionic equation is the equation of the three solids and one liquid.<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis bolditalic\">Test Yourself<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s05_p27\" class=\"para\">What is the net ionic equation between HNO<sub class=\"subscript\">3<\/sub>(aq) and Ti(OH)<sub class=\"subscript\">4<\/sub>(s)?<\/p>\n<p>&nbsp;<\/p>\n<p class=\"simpara\"><strong><em class=\"emphasis\">Answer<\/em><\/strong><\/p>\n<p id=\"ball-ch04_s05_p28\" class=\"para\">4 H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Ti(OH)<sub class=\"subscript\">4<\/sub>(s) [latex]\\longrightarrow[\/latex] 4 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0Ti<sup class=\"superscript\">4+<\/sup>(aq)<\/p>\n<\/div>\n<div id=\"fs-idp164722352\" class=\"textbox shaded\">\n<p><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/chem1114langaracollege\/wp-content\/uploads\/sites\/387\/2018\/04\/CNX_Interactive_200DPI-4-3.png\" alt=\"\u00a0\" width=\"132\" height=\"82\" class=\"alignleft\" \/><\/p>\n<p>&nbsp;<\/p>\n<p id=\"fs-idm40317104\">Explore the microscopic <a href=\"http:\/\/openstaxcollege.org\/l\/16AcidsBases\">view<\/a> of strong and weak acids and bases.<\/p>\n<p>&nbsp;<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idm51820592\" class=\"summary\">\n<h2>Gas-forming Acid-Base reactions<\/h2>\n<p>A driving force for certain acid-base reactions is the formation of a gas. Common gases formed are\u00a0 H<sub>2<\/sub>, O<sub>2<\/sub>, and CO<sub>2<\/sub>.<\/p>\n<p>For example:<\/p>\n<p style=\"text-align: center\">2HCl(aq) + Na<sub>2<\/sub>CO<sub>3<\/sub>(aq) [latex]\\longrightarrow[\/latex] H<sub>2<\/sub>CO<sub>3<\/sub>(aq) + 2NaCl(aq) [latex]\\longrightarrow[\/latex] CO<sub>2<\/sub>(g) + H<sub>2<\/sub>O(l) + 2NaCl(aq)<\/p>\n<p>The above example can be viewed as an acid-base reaction followed by a decomposition. The driving force in this case is the gas formation. \u00a0The decomposition of H<sub>2<\/sub>CO<sub>3\u00a0<\/sub>into CO<sub>2\u00a0<\/sub>and H<sub>2<\/sub>O is a very common reaction. Both Na<sub>2<\/sub>CO<sub>3<\/sub> and NaHCO<sub>3<\/sub> mixed with acid result in a gas-forming acid-base reaction.<\/p>\n<p style=\"text-align: center\">HCl(aq) + NaHCO<sub>3<\/sub>(aq) [latex]\\longrightarrow[\/latex] H<sub>2<\/sub>CO<sub>3<\/sub>(aq) + NaCl(aq) [latex]\\longrightarrow[\/latex] CO<sub>2<\/sub>(g) + H<sub>2<\/sub>O(l) + NaCl(aq)<\/p>\n<div class=\"textbox shaded\">\n<div class=\"callout block\" id=\"ball-ch04_s06_n04\">\n<h3 class=\"title\">Food and Drink App: Acids in Foods<\/h3>\n<p id=\"ball-ch04_s06_p67\" class=\"para\">Many foods and beverages contain acids. Acids impart a sour note to the taste of foods, which may add some pleasantness to the food. For example, orange juice contains citric acid, H<sub class=\"subscript\">3<\/sub>C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">7<\/sub>. Note how this formula shows hydrogen atoms in two places; the first hydrogen atoms written are the hydrogen atoms that can form H<sup class=\"superscript\">+<\/sup> ions, while the second hydrogen atoms written are part of the citrate ion, C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">7<\/sub><sup class=\"superscript\">3\u2212<\/sup>. Lemons and limes contain much more citric acid\u2014about 60 times as much\u2014which accounts for these citrus fruits being more sour than most oranges. Vinegar is essentially a ~5% solution of acetic acid (HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub>) in water. Apples contain malic acid (H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">5<\/sub>; the name <em class=\"emphasis\">malic acid<\/em> comes from the apple\u2019s botanical genus name, <em class=\"emphasis\">malus<\/em>), while lactic acid (HC<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">3<\/sub>) is found in wine and sour milk products, such as yogurt and some cottage cheeses.<\/p>\n<p id=\"ball-ch04_s06_p68\" class=\"para\"><a class=\"xref\" href=\"#ball-ch04_s06_t01\">Table 3 &#8220;Various Acids Found in Food and Beverages&#8221;<\/a> lists some acids found in foods, either naturally or as an additive. Frequently, the salts of acid anions are used as additives, such as monosodium glutamate (MSG), which is the sodium salt derived from glutamic acid. As you read the list, you should come to the inescapable conclusion that it is impossible to avoid acids in food and beverages.<\/p>\n<div class=\"table\" id=\"ball-ch04_s06_t01\">\n<table style=\"border-spacing: 0px\" cellpadding=\"0\">\n<thead>\n<tr style=\"height: 38px\">\n<th style=\"height: 38px\">Acid Name<\/th>\n<th style=\"height: 38px\">Acid Formula<\/th>\n<th style=\"height: 38px\">Use and Appearance<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">acetic acid<\/td>\n<td style=\"height: 24px\">HC<sub class=\"subscript\">2<\/sub>H<sub class=\"subscript\">3<\/sub>O<sub class=\"subscript\">2<\/sub><\/td>\n<td style=\"height: 24px\">flavouring; found in vinegar<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">adipic acid<\/td>\n<td style=\"height: 24px\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">8<\/sub>O<sub class=\"subscript\">4<\/sub><\/td>\n<td style=\"height: 24px\">flavouring; found in processed foods and some antacids<\/td>\n<\/tr>\n<tr style=\"height: 19px\">\n<td style=\"height: 19px\">alginic acid<\/td>\n<td style=\"height: 19px\">various<\/td>\n<td style=\"height: 19px\">thickener; found in drinks, ice cream, and weight loss products<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">ascorbic acid<\/td>\n<td style=\"height: 24px\">HC<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">7<\/sub>O<sub class=\"subscript\">6<\/sub><\/td>\n<td style=\"height: 24px\">antioxidant, also known as vitamin C; found in fruits and vegetables<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">benzoic acid<\/td>\n<td style=\"height: 24px\">HC<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>CO<sub class=\"subscript\">2<\/sub><\/td>\n<td style=\"height: 24px\">preservative; found in processed foods<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">citric acid<\/td>\n<td style=\"height: 24px\">H<sub class=\"subscript\">3<\/sub>C<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">7<\/sub><\/td>\n<td style=\"height: 24px\">flavouring; found in citrus fruits<\/td>\n<\/tr>\n<tr style=\"height: 38px\">\n<td style=\"height: 38px\">dehydroacetic acid<\/td>\n<td style=\"height: 38px\">HC<sub class=\"subscript\">8<\/sub>H<sub class=\"subscript\">7<\/sub>O<sub class=\"subscript\">4<\/sub><\/td>\n<td style=\"height: 38px\">preservative, especially for strawberries and squash<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">erythrobic acid<\/td>\n<td style=\"height: 24px\">HC<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">7<\/sub>O<sub class=\"subscript\">6<\/sub><\/td>\n<td style=\"height: 24px\">antioxidant; found in processed foods<\/td>\n<\/tr>\n<tr style=\"height: 19px\">\n<td style=\"height: 19px\">fatty acids<\/td>\n<td style=\"height: 19px\">various<\/td>\n<td style=\"height: 19px\">thickener and emulsifier; found in processed foods<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">fumaric acid<\/td>\n<td style=\"height: 24px\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">2<\/sub>O<sub class=\"subscript\">4<\/sub><\/td>\n<td style=\"height: 24px\">flavouring; acid reactant in some baking powders<\/td>\n<\/tr>\n<tr style=\"height: 38px\">\n<td style=\"height: 38px\">glutamic acid<\/td>\n<td style=\"height: 38px\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">5<\/sub>H<sub class=\"subscript\">7<\/sub>NO<sub class=\"subscript\">4<\/sub><\/td>\n<td style=\"height: 38px\">flavouring; found in processed foods and in tomatoes, some cheeses, and soy products<\/td>\n<\/tr>\n<tr style=\"height: 38px\">\n<td style=\"height: 38px\">lactic acid<\/td>\n<td style=\"height: 38px\">HC<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">3<\/sub><\/td>\n<td style=\"height: 38px\">flavouring; found in wine, yogurt, cottage cheese, and other sour milk products<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">malic acid<\/td>\n<td style=\"height: 24px\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">5<\/sub><\/td>\n<td style=\"height: 24px\">flavouring; found in apples and unripe fruit<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">phosphoric acid<\/td>\n<td style=\"height: 24px\">H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub><\/td>\n<td style=\"height: 24px\">flavouring; found in some colas<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">propionic acid<\/td>\n<td style=\"height: 24px\">HC<sub class=\"subscript\">3<\/sub>H<sub class=\"subscript\">5<\/sub>O<sub class=\"subscript\">2<\/sub><\/td>\n<td style=\"height: 24px\">preservative; found in baked goods<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">sorbic acid<\/td>\n<td style=\"height: 24px\">HC<sub class=\"subscript\">6<\/sub>H<sub class=\"subscript\">7<\/sub>O<sub class=\"subscript\">2<\/sub><\/td>\n<td style=\"height: 24px\">preservative; found in processed foods<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">stearic acid<\/td>\n<td style=\"height: 24px\">HC<sub class=\"subscript\">18<\/sub>H<sub class=\"subscript\">35<\/sub>O<sub class=\"subscript\">2<\/sub><\/td>\n<td style=\"height: 24px\">anticaking agent; found in hard candies<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">succinic acid<\/td>\n<td style=\"height: 24px\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">4<\/sub><\/td>\n<td style=\"height: 24px\">flavouring; found in wine and beer<\/td>\n<\/tr>\n<tr style=\"height: 24px\">\n<td style=\"height: 24px\">tartaric acid<\/td>\n<td style=\"height: 24px\">H<sub class=\"subscript\">2<\/sub>C<sub class=\"subscript\">4<\/sub>H<sub class=\"subscript\">4<\/sub>O<sub class=\"subscript\">6<\/sub><\/td>\n<td style=\"height: 24px\">flavouring; found in grapes, bananas, and tamarinds<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p><strong><span class=\"title-prefix\">Table 3.<\/span><\/strong>\u00a0Various Acids Found in Food and Beverages<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Key Concepts and Summary<\/h2>\n<p>Chemical reactions are classified according to similar patterns of behaviour. Acid-base reactions involve the transfer of hydrogen ions between reactants.<\/p>\n<p>General acid-base reactions, also called neutralization reactions can be summarized with the following reaction equation:<\/p>\n<p style=\"text-align: center\">ACID(aq) + BASE(aq)\u00a0[latex]\\longrightarrow[\/latex] H<sub>2<\/sub>O(l) + SALT(aq) or (s)<\/p>\n<p>The DRIVING FORCE for a general acid-base reaction is the formation of water.<\/p>\n<p>Gas-forming acid-base reactions can be summarized with the following reaction equation:<\/p>\n<p style=\"text-align: center\">ACID(aq) + NaHCO<sub>3<\/sub> or Na<sub>2<\/sub>CO<sub>3<\/sub>(aq)\u00a0[latex]\\longrightarrow[\/latex] H<sub>2<\/sub>O(l) + CO<sub>2<\/sub>(g) + SALT(aq) or (s)<\/p>\n<p>The DRIVING FORCE for a gas-forming acid-base reaction is the formation of gas.\u00a0There are three ways of<\/p>\n<p>There are three ways of representing a neutralization reaction, using a molecular equation, complete ionic equation or net ionic equation, as described in section 6.1.<\/p>\n<div class=\"key_takeaways editable block\" id=\"ball-ch04_s05_n04\">\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<div class=\"qandaset block\" id=\"ball-ch04_s05_qs01\">\n<div class=\"question\">\n<p id=\"ball-ch04_s05_qs01_p1\" class=\"para\">1. What is the Arrhenius definition of an acid?<\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">2. What is the Arrhenius definition of a base?<\/span><\/p>\n<p class=\"para\"><span style=\"font-size: 1em\">3. Predict the products of each acid-base combination listed. Assume that a neutralization reaction occurs.<\/span><\/p>\n<\/div>\n<p>a) \u00a0HCl and KOH<\/p>\n<p>b) \u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> and KOH<\/p>\n<p>c) \u00a0H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> and Ni(OH)<sub class=\"subscript\">2<\/sub><\/p>\n<div class=\"question\">\n<p id=\"ball-ch04_s05_qs01_p7\" class=\"para\">4. \u00a0Write a balanced chemical equation for each neutralization reaction in Exercise 3.<\/p>\n<\/div>\n<p><span style=\"font-size: 1em\">5. \u00a0Write a balanced chemical equation for the neutralization reaction between each given acid and base. Include the proper phase labels.<\/span><\/p>\n<div class=\"question\">\n<p>a) \u00a0HI(aq) +\u00a0KOH(aq) [latex]\\longrightarrow[\/latex] ?<\/p>\n<p>b) \u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex] ?<\/p>\n<\/div>\n<div class=\"question\">\n<p id=\"ball-ch04_s05_qs01_p10\" class=\"para\"><span style=\"font-size: 1em\">6. \u00a0Write the net ionic equation for each neutralization reaction in Exercise 7.<\/span><\/p>\n<\/div>\n<p><span style=\"font-size: 1em\">7. \u00a0Write the complete and net ionic equations for the neutralization reaction between HClO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">(aq) and Zn(OH)<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">(s). Assume the salt is soluble.<\/span><\/p>\n<p><span style=\"font-size: 1em\">8. \u00a0Explain why the net ionic equation for the neutralization reaction between HCl(aq) and KOH(aq) is the same as the net ionic equation for the neutralization reaction between HNO<\/span><sub class=\"subscript\">3<\/sub><span style=\"font-size: 1em\">(aq) and RbOH.<\/span><\/p>\n<p><span style=\"font-size: 1em\">9. \u00a0Write the complete and net ionic equations for the neutralization reaction between HCl(aq) and KOH(aq) using the hydronium ion in place of H<\/span><sup class=\"superscript\">+<\/sup><span style=\"font-size: 1em\">. What difference does it make when using the hydronium ion?<\/span><\/p>\n<p>10.\u00a0Complete and balance the following acid-base equations:<\/p>\n<p id=\"fs-idp63939984\">a) HCl gas reacts with solid Ca(OH)<sub>2<\/sub>(<em>s<\/em>).<\/p>\n<p id=\"fs-idm54386864\">b) A solution of Sr(OH)<sub>2<\/sub> is added to a solution of HNO<sub>3<\/sub>.<\/p>\n<p>11. Complete and balance the equations for the following acid-base neutralization reactions. If water is used as a solvent, write the reactants and products as aqueous ions. In some cases, there may be more than one correct answer, depending on the amounts of reactants used.<\/p>\n<p id=\"fs-idm49795408\">a) [latex]\\text{Mg(OH)}_2(s) + \\text{HClO}_4(aq) \\longrightarrow[\/latex]<\/p>\n<p id=\"fs-idp205785072\">b) [latex]\\text{SrO}(s) + \\text{H}_2 \\text{SO}_4(l) \\longrightarrow[\/latex]<\/p>\n<p>12. Complete and balance the equations of the following reactions, each of which could be used to remove hydrogen sulfide from natural gas:<\/p>\n<p id=\"fs-idp89463616\">a) [latex]\\text{Ca(OH)}_2(s) + \\text{H}_2 \\text{S}(g) \\longrightarrow[\/latex]<\/p>\n<p id=\"fs-idp2916112\">b) [latex]\\text{Na}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{S}(g) \\longrightarrow[\/latex]<\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<p><b>Answers<\/b><\/p>\n<p><span style=\"font-size: 1em\">1. An Arrhenius acid increases the amount of H<\/span><sup class=\"superscript\">+<\/sup><span style=\"font-size: 1em\"> ions in an aqueous solution.<\/span><\/p>\n<div class=\"answer\">\n<p class=\"para\">2.\u00a0<span style=\"font-size: 1em\">An Arrhenius base increases the amount of OH<\/span><sup class=\"superscript\">&#8211;<\/sup><span style=\"font-size: 1em\"> ions in an aqueous solution.<\/span><\/p>\n<\/div>\n<p>3.\u00a0a) \u00a0KCl and H<sub class=\"subscript\">2<\/sub>O<\/p>\n<div class=\"answer\">\n<p>b) \u00a0K<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> and H<sub class=\"subscript\">2<\/sub>O<\/p>\n<p>c) \u00a0Ni<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub> and H<sub class=\"subscript\">2<\/sub>O<\/p>\n<\/div>\n<p>4.\u00a0a) \u00a0 \u00a0HCl +\u00a0KOH [latex]\\longrightarrow[\/latex] KCl +\u00a0H<sub class=\"subscript\">2<\/sub>O<\/p>\n<p>b) \u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> +\u00a02 KOH [latex]\\longrightarrow[\/latex] K<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub> +\u00a02 H<sub class=\"subscript\">2<\/sub>O<\/p>\n<p>c) \u00a02 H<sub class=\"subscript\">3<\/sub>PO<sub class=\"subscript\">4<\/sub> +\u00a03 Ni(OH)<sub class=\"subscript\">2<\/sub> [latex]\\longrightarrow[\/latex] Ni<sub class=\"subscript\">3<\/sub>(PO<sub class=\"subscript\">4<\/sub>)<sub class=\"subscript\">2<\/sub> +\u00a06 H<sub class=\"subscript\">2<\/sub>O<\/p>\n<p><span style=\"font-size: 1em\">5.\u00a0<\/span><span style=\"font-size: 1em\">a) \u00a0HI(aq) +\u00a0KOH(aq) [latex]\\longrightarrow[\/latex] KCl(aq) +\u00a0H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O(\u2113)<\/span><\/p>\n<div class=\"answer\">\n<p>b) \u00a0H<sub class=\"subscript\">2<\/sub>SO<sub class=\"subscript\">4<\/sub>(aq) +\u00a0Ba(OH)<sub class=\"subscript\">2<\/sub>(aq) [latex]\\longrightarrow[\/latex] BaSO<sub class=\"subscript\">4<\/sub>(s) +\u00a02 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/p>\n<\/div>\n<p><span style=\"font-size: 1em\">6.\u00a0<\/span><span style=\"font-size: 1em\">a) \u00a0H<\/span><sup class=\"superscript\">+<\/sup><span style=\"font-size: 1em\">(aq) +\u00a0OH<\/span><sup class=\"superscript\">\u2212<\/sup><span style=\"font-size: 1em\">(aq) [latex]\\longrightarrow[\/latex] H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O(\u2113)<\/span><\/p>\n<div class=\"answer\">\n<p>b) \u00a02 H<sup class=\"superscript\">+<\/sup>(aq) +\u00a0SO<sub class=\"subscript\">4<\/sub><sup class=\"superscript\">2\u2212<\/sup>(aq) +\u00a0Ba<sup class=\"superscript\">2+<\/sup>(aq) +\u00a02 OH<sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] BaSO<sub class=\"subscript\">4<\/sub>(s) +\u00a02 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/p>\n<p>7. \u00a0Complete ionic equation:<\/p>\n<\/div>\n<p><span style=\"font-size: 1em\">2 H<\/span><sup class=\"superscript\">+<\/sup><span style=\"font-size: 1em\">(aq) +\u00a02 ClO<\/span><sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup><span style=\"font-size: 1em\">(aq) +\u00a0Zn<\/span><sup class=\"superscript\">2+<\/sup><span style=\"font-size: 1em\">(aq) +\u00a02 OH<\/span><sup class=\"superscript\">\u2212<\/sup><span style=\"font-size: 1em\">(aq) [latex]\\longrightarrow[\/latex] Zn<\/span><sup class=\"superscript\">2+<\/sup><span style=\"font-size: 1em\">(aq) +\u00a02 ClO<\/span><sub class=\"subscript\">3<\/sub><sup class=\"superscript\">\u2212<\/sup><span style=\"font-size: 1em\">(aq) +\u00a02 H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O(\u2113)<\/span><\/p>\n<div class=\"answer\">\n<p id=\"ball-ch04_s05_qs01_p15_ans\" class=\"para\">Net ionic equation:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">2 H<sup class=\"superscript\">+<\/sup>(aq) +\u00a02 OH<sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] 2 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<\/div>\n<p><span style=\"font-size: 1em\">8. \u00a0Because the salts are soluble in both cases, the net ionic reaction is just H<\/span><sup class=\"superscript\">+<\/sup><span style=\"font-size: 1em\">(aq) +\u00a0OH<\/span><sup class=\"superscript\">\u2212<\/sup><span style=\"font-size: 1em\">(aq) [latex]\\longrightarrow[\/latex] H<\/span><sub class=\"subscript\">2<\/sub><span style=\"font-size: 1em\">O(\u2113).<\/span><\/p>\n<p><span style=\"font-size: 1em\">9. \u00a0Complete ionic equation:<\/span><\/p>\n<div class=\"answer\">\n<p><span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq) +\u00a0K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] 2 H<sub class=\"subscript\">2<\/sub>O(\u2113) +\u00a0K<sup class=\"superscript\">+<\/sup>(aq) +\u00a0Cl<sup class=\"superscript\">\u2212<\/sup>(aq)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_qs01_p25_ans\" class=\"para\">Net ionic equation:<\/p>\n<p><span class=\"informalequation\"><span class=\"mathphrase\">H<sub class=\"subscript\">3<\/sub>O<sup class=\"superscript\">+<\/sup>(aq) +\u00a0OH<sup class=\"superscript\">\u2212<\/sup>(aq) [latex]\\longrightarrow[\/latex] 2 H<sub class=\"subscript\">2<\/sub>O(\u2113)<\/span><\/span><\/p>\n<p id=\"ball-ch04_s05_qs01_p26_ans\" class=\"para\">The difference is simply the presence of an extra water molecule as a product.<\/p>\n<p>10.\u00a0<span style=\"font-size: 1em\">a) [latex]2\\text{HCl}(g) + \\text{Ca(OH)}_2(s) \\longrightarrow \\text{CaCl}_2(s) + 2\\text{H}_2 \\text{O}(l)[\/latex];<\/span><\/p>\n<p id=\"fs-idm141927648\">b) [latex]\\text{Sr(OH)}_2(aq) + 2\\text{HNO}_3(aq) \\longrightarrow \\text{Sr(NO}_3)_2(aq) + 2\\text{H}_2 \\text{O}(l)[\/latex];<\/p>\n<p id=\"fs-idp32868368\">11.\u00a0a) [latex]\\text{Mg(OH)}_2(s) + 2\\text{HClO}_4(aq) \\longrightarrow \\text{Mg}^{2+}(aq) + 2{\\text{ClO}_4}^{-}(aq) + 2\\text{H}_2 \\text{O}(l);[\/latex]<br \/>\nb) [latex]\\text{SrO}(s) + \\text{H}_2 \\text{SO}_4(l) \\longrightarrow \\text{SrSO}_4(s) + \\text{H}_2 \\text{O}[\/latex]<\/p>\n<p id=\"fs-idp19195056\">12.\u00a0a) [latex]\\text{Ca(OH)}_2(s) + \\text{H}_2 \\text{S}(g) \\longrightarrow \\text{CaS}(s) + 2\\text{H}_2\\text{O}(l);[\/latex]<br \/>\nb) [latex]\\text{Na}_2 \\text{CO}_3(aq) + \\text{H}_2 \\text{S}(g) \\longrightarrow \\text{Na}_2 \\text{S}(aq) + \\text{CO}_2(g) + \\text{H}_2 \\text{O}(l)[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/div>\n<h2>Glossary<\/h2>\n<p><strong>acid:\u00a0<\/strong>substance that produces H<sub>3<\/sub>O<sup>+<\/sup> when dissolved in water<\/p>\n<p><strong>acid-base reaction:\u00a0<\/strong>reaction involving the transfer of a hydrogen ion between reactant species<\/p>\n<p><strong>base:\u00a0<\/strong>substance that produces OH<sup>\u2212<\/sup> when dissolved in water<\/p>\n<p><strong>neutralization reaction:\u00a0<\/strong>reaction between an acid and a base to produce salt and water<\/p>\n<p><strong>salt:\u00a0<\/strong>ionic compound that can be formed by the reaction of an acid with a base that contains a cation and an anion other than hydroxide or oxide<\/p>\n<p><strong>strong acid:\u00a0<\/strong>acid that reacts completely when dissolved in water to yield hydronium ions<\/p>\n<p><strong>strong base:\u00a0<\/strong>base that reacts completely when dissolved in water to yield hydroxide ions<\/p>\n<p><strong>weak acid:\u00a0<\/strong>acid that reacts only to a slight extent when dissolved in water to yield hydronium ions<\/p>\n<p><strong>weak base:\u00a0<\/strong>base that reacts only to a slight extent when dissolved in water to yield hydroxide ions<\/p>\n<\/section>\n<\/div>\n<\/div>\n","protected":false},"author":330,"menu_order":4,"template":"","meta":{"pb_show_title":"on","pb_short_title":"6.3 Acid-Base 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