{"id":1892,"date":"2020-04-30T17:55:15","date_gmt":"2020-04-30T21:55:15","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/inorganicchemistrychem250\/chapter\/bronsted-lowry-acids-and-bases\/"},"modified":"2021-11-08T17:09:35","modified_gmt":"2021-11-08T22:09:35","slug":"bronsted-lowry-acids-and-bases","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/inorganicchemistrychem250\/chapter\/bronsted-lowry-acids-and-bases\/","title":{"raw":"8.2 Br\u00f8nsted-Lowry Acids and Bases","rendered":"8.2 Br\u00f8nsted-Lowry Acids and Bases"},"content":{"raw":"[latexpage]\r\n
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Learning Objectives<\/p>\r\n\r\n<\/header>\r\n

\r\n\r\nBy the end of this section, you will be able to:\r\n
    \r\n \t
  • Identify acids, bases, and conjugate acid-base pairs according to the Br\u00f8nsted-Lowry definition<\/li>\r\n \t
  • Write equations for acid and base ionization reactions<\/li>\r\n \t
  • Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations<\/li>\r\n \t
  • Describe the acid-base behavior of amphiprotic substances<\/li>\r\n<\/ul>\r\n<\/div>\r\n<\/div>\r\n

    The acid-base reaction class has been studied for quite some time. In 1680, Robert Boyle<\/span> reported traits of acid solutions that included their ability to dissolve many substances, to change the colors of certain natural dyes, and to lose these traits after coming in contact with alkali (base) solutions. In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO2<\/sub>), and interact with alkalis to form neutral substances. In 1815, Humphry Davy<\/span> contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius<\/span> defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.<\/p>\r\n

    Johannes Br\u00f8nsted and Thomas Lowry proposed a more general description in 1923 in which acids and bases were defined in terms of the transfer of hydrogen ions, H+<\/sup>. (Note that these hydrogen ions are often referred to simply as protons<\/em>, since that subatomic particle is the only component of cations derived from the most abundant hydrogen isotope, 1<\/sup>H.) A compound that donates a proton to another compound is called a Br\u00f8nsted-Lowry acid<\/span>, and a compound that accepts a proton is called a Br\u00f8nsted-Lowry base<\/span>. An acid-base reaction is, thus, the transfer of a proton from a donor (acid) to an acceptor (base).<\/p>\r\n

    The concept of conjugate pairs<\/em> is useful in describing Br\u00f8nsted-Lowry acid-base reactions (and other reversible reactions, as well). When an acid donates H+<\/sup>, the species that remains is called the conjugate base<\/strong> of the acid because it reacts as a proton acceptor in the reverse reaction. Likewise, when a base accepts H+<\/sup>, it is converted to its conjugate acid<\/strong>. The reaction between water and ammonia illustrates this idea. In the forward direction, water acts as an acid by donating a proton to ammonia and subsequently becoming a hydroxide ion, OH\u2212<\/sup>, the conjugate base of water. The ammonia acts as a base in accepting this proton, becoming an ammonium ion, \\({{\\text{NH}}_{4}}^{\\text{+}}\\text{,}\\) the conjugate acid of ammonia. In the reverse direction, a hydroxide ion acts as a base in accepting a proton from ammonium ion, which acts as an acid.<\/p>\r\n\u00a0<\/em>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"This Scheme 8.2.1 - A Br\u00f8nsted-Lowry acid reacts with a base to form the conjugate base and acid, as shown with H2O (acid) reacting with NH3 (base) to form their respective conjugates.<\/strong>[\/caption]\r\n

    The reaction between a Br\u00f8nsted-Lowry acid and water is called acid ionization<\/span>. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:<\/p>\r\n\u00a0<\/em>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"782\"]\"This Scheme 8.2.2 - The acid ionization reaction of an acid (HF) with water, resulting in a hydronium ionsand the conjugate base of the acid.<\/strong>[\/caption]\r\n

    Base ionization<\/span> of a species occurs when it accepts protons from water molecules. In the example below, pyridine molecules, C5<\/sub>NH5<\/sub>, undergo base ionization when dissolved in water, yielding hydroxide and pyridinium ions:<\/p>\r\n\u00a0<\/em>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"1029\"]\"This Scheme 8.2.3 - Pyridine reacts with water to form the pyridinium ion and hydroxide ion.<\/strong>[\/caption]\r\n

    The preceding ionization reactions suggest that water may function as both a base (as in its reaction with hydrogen fluoride) and an acid (as in its reaction with ammonia). Species capable of either donating or accepting protons are called amphiprotric<\/span>, or more generally, amphoteric<\/span>, a term that may be used for acids and bases per definitions other than the Br\u00f8nsted-Lowry one. The equations below show the two possible acid-base reactions for two amphiprotic species, bicarbonate ion and water:<\/p>\r\n\u00a0<\/em>\r\n

    \\({\\text{HCO}}_{3}{}^{\u2013}\\left(\\text{aq}\\right)+{\\text{H}}_{2}\\text{O}\\left(\\text{l}\\right)\\rightleftharpoons\\phantom{\\rule{4em}{0ex}}{\\text{CO}}_{3}{}^{2\u2013}\\left(\\text{aq}\\right)+{\\text{H}}_{3}{\\text{O}}^{+}\\left(\\text{aq}\\right)\\)<\/div>\r\n
    \u00a0<\/em><\/div>\r\n
    \\({\\text{HCO}}_{3}{}^{\u2013}\\left(\\text{aq}\\right)+{\\text{H}}_{2}\\text{O}\\left(\\text{l}\\right)\\rightleftharpoons\\phantom{\\rule{4em}{0ex}}{\\text{H}}_{2}{\\text{CO}}_{3}\\left(\\text{aq}\\right)+{\\text{OH}}^{\u2013}\\left(\\text{aq}\\right)\\)<\/div>\r\n
    \u00a0<\/em><\/div>\r\n
    <\/div>\r\n

    The first equation represents the reaction of bicarbonate as an acid with water as a base, whereas the second represents reaction of bicarbonate as a base with water as an acid. When bicarbonate is added to water, both these equilibria are established simultaneously and the composition of the resulting solution may be determined through appropriate equilibrium calculations, as described later in this chapter.<\/p>\r\n

    In the liquid state, molecules of an amphiprotic substance can react with one another as illustrated for water in the equations below:<\/p>\r\n\u00a0<\/em>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"975\"]\"This Scheme 8.2.4 - Water is amphiprotic and can react with itself to from hydronium and hydroxide ions.<\/strong>[\/caption]\r\n

    The process in which like molecules react to yield ions is called autoionization<\/span>. Liquid water undergoes autoionization to a very slight extent; at 25 \u00b0C, approximately two out of every billion water molecules are ionized. The extent of the water autoionization process is reflected in the value of its equilibrium constant, the ion-product constant for water, K<\/em>w<\/sub><\/span>:<\/p>\r\n\u00a0<\/em>\r\n

    \\({\\text{H}}_{2}\\text{O}\\left(l\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\phantom{\\rule{0.2em}{0ex}}\\rightleftharpoons\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{OH}}^{\\text{-}}\\left(aq\\right)\\phantom{\\rule{5em}{0ex}}{K}_{\\text{w}}=\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{OH}^-}^{\\text{}}\\right]\\)<\/div>\r\n
    \u00a0<\/em><\/div>\r\n

    The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 \u00b0C, K<\/em>w<\/sub> has a value of 1.0 \\(\\times\\) 10\u221214<\/sup>. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 \u00b0C, the value for K<\/em>w<\/sub> is about 5.6 \\(\\times\\) 10\u221213<\/sup>, roughly 50 times larger than the value at 25 \u00b0C.<\/p>\r\n\r\n

    \r\n
    \r\n

    Activity 8.2.1 - Ion Concentrations in Pure Water<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWhat are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 \u00b0C?\r\n

    Solution<\/h2>\r\nThe autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, [H3<\/sub>O+<\/sup>] = [OH\u2212<\/sup>] = x<\/em>. At 25 \u00b0C:\r\n\r\n\u00a0<\/em>\r\n
    \\({K}_{\\text{w}}=\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{OH}^-}^{\\text{}}\\right]=x=1.0\\phantom{\\rule{0.2em}{0ex}}\\times\\phantom{\\rule{0.2em}{0ex}}{10}^{-14}\\)<\/div>\r\n
    \u00a0<\/em><\/div>\r\n

    So:<\/p>\r\n\u00a0<\/em>\r\n

    \\(x=\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]=\\left[{\\text{OH}^-}^{\\text{}}\\right]=\\sqrt{1.0\\phantom{\\rule{0.2em}{0ex}}\\times\\phantom{\\rule{0.2em}{0ex}}{10}^{-14}}=1.0\\phantom{\\rule{0.2em}{0ex}}\\times\\phantom{\\rule{0.2em}{0ex}}{10}^{-7}\\phantom{\\rule{0.2em}{0ex}}M\\)<\/div>\r\n
    \u00a0<\/em><\/div>\r\n

    The hydronium ion concentration and the hydroxide ion concentration are the same, 1.0 \\(\\times\\) 10\u22127<\/sup>M<\/em>.<\/p>\r\n\r\n\r\n

    \r\n\r\n

    Check Your Learning<\/span><\/h2>\r\nThe ion product of water at 80 \u00b0C is 2.4 \\(\\times\\) 10\u221213<\/sup>. What are the concentrations of hydronium and hydroxide ions in pure water at 80 \u00b0C?\r\n
    \r\n

    Answer<\/h3>\r\n

    [H3<\/sub>O+<\/sup>] = [OH\u2212<\/sup>] = 4.9 \\(\\times\\) 10\u22127<\/sup>M<\/em><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    Activity 8.2.2 - The Inverse Relation between [H3<\/sub>O+<\/sup>] and [OH\u2212<\/sup>]<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nA solution of an acid in water has a hydronium ion concentration of 2.0 \\(\\times\\) 10\u22126<\/sup>M<\/em>. What is the concentration of hydroxide ion at 25 \u00b0C?\r\n

    Solution<\/h2>\r\nUse the value of the ion-product constant for water at 25 \u00b0C:\r\n\r\n\u00a0<\/em>\r\n
    \\(2{\\text{H}}_{2}\\text{O}\\left(l\\right)\\phantom{\\rule{0.2em}{0ex}}\\rightleftharpoons\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{OH}^-}^{\\text{}}\\left(aq\\right)\\phantom{\\rule{5em}{0ex}}{K}_{\\text{w}}=\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{OH}^-}^{\\text{}}\\right]=1.0\\phantom{\\rule{0.2em}{0ex}}\\times\\phantom{\\rule{0.2em}{0ex}}{10}^{-14}\\)<\/div>\r\n
    \u00a0<\/em><\/div>\r\n
    <\/div>\r\n

    to calculate the missing equilibrium concentration.<\/p>\r\n

    Rearrangement of the K<\/em>w<\/sub> expression shows that [OH\u2212<\/sup>] is inversely proportional to [H3<\/sub>O+<\/sup>]:<\/p>\r\n\u00a0<\/em>\r\n

    \\(\\left[{\\text{OH}^-}^{\\text{}}\\right]=\\phantom{\\rule{0.2em}{0ex}}\\frac{{K}_{\\text{w}}}{\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]}\\phantom{\\rule{0.2em}{0ex}}=\\phantom{\\rule{0.2em}{0ex}}\\frac{1.0\\phantom{\\rule{0.2em}{0ex}}\\times\\phantom{\\rule{0.2em}{0ex}}{10}^{-14}}{2.0\\phantom{\\rule{0.2em}{0ex}}\\times\\phantom{\\rule{0.2em}{0ex}}{10}^{-6}}\\phantom{\\rule{0.2em}{0ex}}=5.0\\phantom{\\rule{0.2em}{0ex}}\\times\\phantom{\\rule{0.2em}{0ex}}{10}^{-9}\\)<\/div>\r\n
    \u00a0<\/em><\/div>\r\n
    <\/div>\r\n

    Compared with pure water, a solution of acid exhibits a higher concentration of hydronium ions (due to ionization of the acid) and a proportionally lower concentration of hydroxide ions. This may be explained via Le Ch\u00e2telier\u2019s principle as a left shift in the water autoionization equilibrium resulting from the stress of increased hydronium ion concentration.<\/p>\r\n

    Substituting the ion concentrations into the K<\/em>w<\/sub> expression confirms this calculation, resulting in the expected value:<\/p>\r\n\u00a0<\/em>\r\n

    \\({K}_{\\text{w}}=\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{OH}^-}^{\\text{}}\\right]=\\left(2.0\\phantom{\\rule{0.2em}{0ex}}\\times\\phantom{\\rule{0.2em}{0ex}}{10}^{-6}\\right)\\left(5.0\\phantom{\\rule{0.2em}{0ex}}\\times\\phantom{\\rule{0.2em}{0ex}}{10}^{-9}\\right)=1.0\\phantom{\\rule{0.2em}{0ex}}\\times\\phantom{\\rule{0.2em}{0ex}}{10}^{-14}\\)<\/div>\r\n
    \r\n\r\n
    \r\n\r\n<\/div>\r\n

    Check Your Learning<\/span><\/h2>\r\nWhat is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M<\/em> at 25 \u00b0C?\r\n
    \r\n

    Answer<\/h3>\r\n

    [H3<\/sub>O+<\/sup>] = 1 \\(\\times\\) 10\u221211<\/sup>M<\/em><\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    \r\n
    \r\n

    Activity 8.2.3 - Representing the Acid-Base Behavior of an Amphoteric Substance<\/p>\r\n\r\n<\/header>\r\n

    \r\n\r\nWrite separate equations representing the reaction of \\({\\text{HSO}}_{3}{^-}^{\\text{}}\\)\r\n

    (a) as an acid with OH\u2212<\/sup><\/p>\r\n

    (b) as a base with HI<\/p>\r\n\r\n

    Solution<\/h2>\r\n(a) \\({\\text{HSO}}_{3}{}^{\\text{-}}\\left(aq\\right)+{\\text{OH}^-}^{\\text{}}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}\\rightleftharpoons\\phantom{\\rule{0.2em}{0ex}}{\\text{SO}}_{3}{^{2-}}^{\\text{}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\)\r\n

    (b) \\({\\text{HSO}}_{3}{^-}^{\\text{}}\\left(aq\\right)+\\text{HI}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}\\rightleftharpoons\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{2}{\\text{SO}}_{3}\\left(aq\\right)+{\\text{I}^-}^{\\text{}}\\left(aq\\right)\\)<\/p>\r\n\r\n\r\n

    \r\n\r\n

    Check Your Learning<\/span><\/h2>\r\nWrite separate equations representing the reaction of \\({\\text{H}}_{2}{\\text{PO}}_{4}{^-}^{\\text{}}\\)\r\n

    (a) as a base with HBr<\/p>\r\n

    (b) as an acid with OH\u2212<\/sup><\/p>\r\n\r\n

    \r\n

    Answer<\/h3>\r\n

    (a) \\({\\text{H}}_{2}{\\text{PO}}_{4}{^-}^{\\text{}}\\left(aq\\right)+\\text{HBr}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}\\rightleftharpoons\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{3}{\\text{PO}}_{4}\\left(aq\\right)+{\\text{Br}^-}^{\\text{}}\\left(aq\\right);\\)<\/p>\r\n

    (b) \\({\\text{H}}_{2}{\\text{PO}}_{4}{}^{\\text{\u2212}}\\left(aq\\right)+{\\text{OH}^-}^{\\text{}}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}\\rightleftharpoons\\phantom{\\rule{0.2em}{0ex}}{\\text{HPO}}_{4}{^{2-}}^{\\text{}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

    Key Concepts and Summary<\/h1>\r\n

    A compound that can donate a proton (a hydrogen ion) to another compound is called a Br\u00f8nsted-Lowry acid. The compound that accepts the proton is called a Br\u00f8nsted-Lowry base. The species remaining after a Br\u00f8nsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Br\u00f8nsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, H3<\/sub>O+<\/sup>, and the hydroxide ion, OH\u2212<\/sup> when it undergoes autoionization:<\/p>\r\n\u00a0<\/em>\r\n\r\n\\({\\text{2H}}_{2}\\text{O(}l\\right)\\phantom{\\rule{0.2em}{0ex}}\\rightleftharpoons\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{3}{\\text{O}}^{+}\\left(aq\\right)+{\\text{OH}^-}^{\\text{}}\\left(aq\\right)\\)\r\n\r\n\u00a0<\/em>\r\n

    The ion product of water, K<\/em>w<\/sub> is the equilibrium constant for the autoionization reaction:<\/p>\r\n\u00a0<\/em>\r\n\r\n\\({K}_{\\text{w}}=\\left[{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\right]\\left[{\\text{OH}^-}^{\\text{}}\\right]=1.0\\phantom{\\rule{0.2em}{0ex}}\\times\\phantom{\\rule{0.2em}{0ex}}1{0}^{-14}\\phantom{\\rule{0.2em}{0ex}}\\text{at}\\phantom{\\rule{0.2em}{0ex}}25\\phantom{\\rule{0.2em}{0ex}}{^\\circ}C}\\)\r\n

    Key Equations<\/h2>\r\n
      \r\n \t
    • K<\/em>w<\/sub> = [H3<\/sub>O+<\/sup>][OH\u2212<\/sup>] = 1.0 \\(\\times\\) 10\u221214<\/sup> (at 25 \u00b0C)<\/li>\r\n<\/ul>\r\n
      \r\n

      End of Chapter Exercises<\/p>\r\n\r\n<\/header>\r\n

      \r\n
      \r\n
      \r\n

      (1) Write equations that show NH3<\/sub> as both a conjugate acid and a conjugate base.<\/p>\r\n\u00a0<\/em>\r\n

      Solution<\/em><\/p>\r\n\r\n<\/div>\r\n

      \r\n

      One example for NH3<\/sub> as a conjugate acid:<\/p>\r\n\\({\\text{NH}}_{2}{^-}^{\\text{}}+{\\text{H}}^{\\text{+}}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{NH}}_{3};\\) as a conjugate base: \\({\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}^-}^{\\text{}}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right)\\)\r\n\r\n<\/div>\r\n\u00a0\u00a0<\/em>\r\n\r\n<\/div>\r\n

      \r\n
      \r\n

      (2) Write equations that show \\({\\text{H}}_{2}{\\text{PO}}_{4}{^-}^{\\text{}}\\) acting both as an acid and as a base.<\/p>\r\n\u00a0<\/em>\r\n\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      \r\n

      (3) Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry acid:<\/p>\r\n

      (3a) \\({\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\)<\/p>\r\n

      (3b) HCl<\/p>\r\n

      (3c) NH3<\/sub><\/p>\r\n

      (3d) CH3<\/sub>CO2<\/sub>H<\/p>\r\n

      (3e) \\({\\text{NH}}_{4}{}^{\\text{+}}\\)<\/p>\r\n

      (3f) \\({\\text{HSO}}_{4}{^-}^{\\text{}}\\)<\/p>\r\n \r\n

      Solution<\/em><\/p>\r\n\r\n<\/div>\r\n

      \r\n

      (a) \\({\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{H}}_{2}\\text{O}\\left(l\\right);\\)<\/p>\r\n(b) \\(\\text{HCl}\\left(\\mathrm{aq}\\right)\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}^-}^{\\text{}}\\left(aq\\right);\\)\r\n\r\n(c) \\({\\text{NH}}_{3}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{NH}}_{2}{^-}^{\\text{}}\\left(aq\\right);\\)\r\n\r\n(d) \\({\\text{CH}}_{3}{\\text{CO}}_{2}\\text{H}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{CH}}_{3}{\\text{CO}}_{2}{^-}^{\\text{}}\\left(aq\\right);\\)\r\n\r\n(e) \\({\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}^{\\text{+}}\\left(aq\\right)+{\\text{NH}}_{3}\\left(aq\\right);\\)\r\n\r\n(f) \\({\\text{HSO}}_{4}{^-}^{\\text{}}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}^{+}\\left(aq\\right)+{\\text{SO}}_{4}{^{2-}}^{\\text{}}\\left(aq\\right)\\)\r\n\r\n<\/div>\r\n\u00a0\u00a0<\/em>\r\n\r\n<\/div>\r\n

      \r\n
      \r\n

      (4) Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry acid:<\/p>\r\n

      (4a) HNO3<\/sub><\/p>\r\n

      (4b) \\({\\text{PH}}_{4}{}^{\\text{+}}\\)<\/p>\r\n

      (4c) H2<\/sub>S<\/p>\r\n

      (4d) CH3<\/sub>CH2<\/sub>COOH<\/p>\r\n

      (4e) \\({\\text{H}}_{2}{\\text{PO}}_{4}{^-}^{\\text{}}\\)<\/p>\r\n

      (4f) HS\u2212<\/sup><\/p>\r\n\u00a0<\/em>\r\n\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      \r\n

      (5) Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry base:<\/p>\r\n

      (5a) H2<\/sub>O<\/p>\r\n

      (5b) OH\u2212<\/sup><\/p>\r\n

      (5c) NH3<\/sub><\/p>\r\n

      (5d) CN\u2212<\/sup><\/p>\r\n

      (5e) S2\u2212<\/sup><\/p>\r\n

      (5f) \\({\\text{H}}_{2}{\\text{PO}}_{4}{^-}^{\\text{}}\\)<\/p>\r\n \r\n

      Solution<\/em><\/p>\r\n\r\n<\/div>\r\n

      \r\n

      (a) \\({\\text{H}}_{2}\\text{O}\\left(l\\right)+{\\text{H}}^{\\text{+}}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right);\\)<\/p>\r\n(b) \\({\\text{OH}}^{\\text{\u2212}}\\left(aq\\right)+{\\text{H}}^{\\text{+}}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{2}\\text{O}\\left(l\\right);\\)\r\n\r\n(c) \\({\\text{NH}}_{3}\\left(aq\\right)+{\\text{H}}^{\\text{+}}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right);\\)\r\n\r\n(d) \\({\\text{CN}^-}^{\\text{}}\\left(aq\\right)+{\\text{H}}^{\\text{+}}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}\\text{HCN}\\left(aq\\right);\\)\r\n\r\n(e) \\({\\text{S}}^{2-}\\left(aq\\right)+{\\text{H}}^{\\text{+}}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{HS}^-}^{\\text{}}\\left(aq\\right);\\)\r\n\r\n(f) \\({\\text{H}}_{2}{\\text{PO}}_{4}{^-}^{\\text{}}\\left(aq\\right)+{\\text{H}}^{\\text{+}}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{3}{\\text{PO}}_{4}\\left(aq\\right)\\)\r\n\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      \r\n\r\n \r\n

      (6) Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry base:<\/p>\r\n

      (6a) HS\u2212<\/sup><\/p>\r\n

      (6b) \\({\\text{PO}}_{4}{^{3-}}^{\\text{}}\\)<\/p>\r\n

      (6c) \\({\\text{NH}}_{2}{^-}^{\\text{}}\\)<\/p>\r\n

      (6d) C2<\/sub>H5<\/sub>OH<\/p>\r\n

      (6e) O2\u2212<\/sup><\/p>\r\n

      (6f) \\({\\text{H}}_{2}{\\text{PO}}_{4}{^-}^{\\text{}}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      \r\n\r\n \r\n

      (7) What is the conjugate acid of each of the following? What is the conjugate base of each?<\/p>\r\n

      (7a) OH\u2212<\/sup><\/p>\r\n

      (7b) H2<\/sub>O<\/p>\r\n

      (7c) \\({\\text{HCO}}_{3}{^-}^{\\text{}}\\)<\/p>\r\n

      (7d) NH3<\/sub><\/p>\r\n

      (7e) \\({\\text{HSO}}_{4}{^-}^{\\text{}}\\)<\/p>\r\n

      (7f) H2<\/sub>O2<\/sub><\/p>\r\n

      (7g) HS\u2212<\/sup><\/p>\r\n

      (7h) \\({\\text{H}}_{5}{\\text{N}}_{2}{}^{\\text{+}}\\)<\/p>\r\n \r\n

      Solution<\/em><\/p>\r\n\r\n<\/div>\r\n

      \r\n

      (a) H2<\/sub>O, O2\u2212<\/sup>;<\/p>\r\n(b) H3<\/sub>O+<\/sup>, OH\u2212<\/sup>;\r\n\r\n(c) H2<\/sub>CO3<\/sub>, \\({\\text{CO}}_{3}{^{2-}}^{\\text{}};\\)\r\n\r\n(d) \\({\\text{NH}}_{4}{}^{\\text{+}},\\) \\({\\text{NH}}_{2}{^-}^{\\text{}};\\)\r\n\r\n(e) H2<\/sub>SO4<\/sub>, \\({\\text{SO}}_{4}{^{2-}}^{\\text{}};\\)\r\n\r\n(f) \\({\\text{H}}_{3}{\\text{O}}_{2}{}^{\\text{+}},\\) \\({\\text{HO}}_{2}{^-}^{\\text{}};\\)\r\n\r\n(g) H2<\/sub>S; S2\u2212<\/sup>;\r\n\r\n(h) \\({\\text{H}}_{6}{\\text{N}}_{2}{}^{2+},\\) H4<\/sub>N2<\/sub>\r\n\r\n<\/div>\r\n\u00a0\u00a0<\/em>\r\n\r\n<\/div>\r\n

      \r\n
      \r\n

      (8) What is the conjugate acid of each of the following? What is the conjugate base of each?<\/p>\r\n

      (8a) H2<\/sub>S<\/p>\r\n

      (8b) \\({\\text{H}}_{2}{\\text{PO}}_{4}{^-}^{\\text{}}\\)<\/p>\r\n

      (8c) PH3<\/sub><\/p>\r\n

      (8d) HS\u2212<\/sup><\/p>\r\n

      (8e) \\({\\text{HSO}}_{3}{^-}^{\\text{}}\\)<\/p>\r\n

      (8f) \\({\\text{H}}_{3}{\\text{O}}_{2}{}^{\\text{+}}\\)<\/p>\r\n

      (8g) H4<\/sub>N2<\/sub><\/p>\r\n

      (8h) CH3<\/sub>OH<\/p>\r\n\u00a0\u00a0<\/em>\r\n\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      \r\n

      (9) Identify and label the Br\u00f8nsted-Lowry acid, its conjugate base, the Br\u00f8nsted-Lowry base, and its conjugate acid in each of the following equations:<\/p>\r\n

      (9a) \\({\\text{HNO}}_{3}+{\\text{H}}_{2}\\text{O}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{3}{\\text{O}}^{\\text{+}}+{\\text{NO}}_{3}{^-}^{\\text{}}\\)<\/p>\r\n

      (9b) \\({\\text{CN}^-}^{\\text{}}+{\\text{H}}_{2}\\text{O}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}\\text{HCN}+{\\text{OH}^-}^{\\text{}}\\)<\/p>\r\n

      (9c) \\({\\text{H}}_{2}{\\text{SO}}_{4}+{\\text{Cl}^-}^{\\text{}}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}\\text{HCl}+{\\text{HSO}}_{4}{^-}^{\\text{}}\\)<\/p>\r\n

      (9d) \\({\\text{HSO}}_{4}{^-}^{\\text{}}+{\\text{OH}^-}^{\\text{}}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{SO}}_{4}{^{2-}}^{\\text{}}+{\\text{H}}_{2}\\text{O}\\)<\/p>\r\n

      (9e) \\({\\text{O}}^{2-}+{\\text{H}}_{2}\\text{O}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}2{\\mathrm{OH}^-}^{\\text{}}\\)<\/p>\r\n

      (9f) \\({\\left[\\text{Cu}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{3}\\left(\\text{OH}\\right)\\right]}^{\\text{+}}+{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}\\right]}^{3+}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\left[\\text{Cu}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{4}\\right]}^{2+}+{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}\\left(\\text{OH}\\right)\\right]}^{2+}\\)<\/p>\r\n

      (9g) \\({\\text{H}}_{2}\\text{S}+{\\text{NH}}_{2}{^-}^{\\text{}}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{HS}^-}^{\\text{}}+{\\text{NH}}_{3}\\)<\/p>\r\n \r\n

      Solution<\/em><\/p>\r\n\r\n<\/div>\r\n

      \r\n

      The labels are Br\u00f8nsted-Lowry acid = BA; its conjugate base = CB; Br\u00f8nsted-Lowry base = BB; its conjugate acid = CA.<\/p>\r\n(a) HNO3<\/sub>(BA), H2<\/sub>O(BB), H3<\/sub>O+<\/sup>(CA), \\({\\text{NO}}_{3}{^-}^{\\text{}}\\left(\\text{CB}\\right)\\)\r\n\r\n(b) CN\u2212<\/sup>(BB), H2<\/sub>O(BA), HCN(CA), OH\u2212<\/sup>(CB);\r\n\r\n(c) H2<\/sub>SO4<\/sub>(BA), Cl\u2212<\/sup>(BB), HCl(CA), \\({\\text{HSO}}_{4}{^-}^{\\text{}}\\left(\\text{CB}\\right)\\)\r\n\r\n(d) \\({\\text{HSO}}_{4}{^-}^{\\text{}}\\left(\\text{BA}\\right),\\) OH\u2212<\/sup>(BB), \\({\\text{SO}}_{4}{^{2-}}^{\\text{}}\\)(CB), H2<\/sub>O(CA)\r\n\r\n(e) O2\u2212<\/sup>(BB), H2<\/sub>O(BA) OH\u2212<\/sup>(CB and CA)\r\n\r\n(f) [Cu(H2<\/sub>O)3<\/sub>(OH)]+<\/sup>(BB), [Al(H2<\/sub>O)6<\/sub>]3+<\/sup>(BA), [Cu(H2<\/sub>O)4<\/sub>]2+<\/sup>(CA), [Al(H2<\/sub>O)5<\/sub>(OH)]2+<\/sup>(CB)\r\n\r\n(g) H2<\/sub>S(BA), \\({\\text{NH}}_{2}{^-}^{\\text{}}\\left(\\text{BB}\\right),\\) HS\u2212<\/sup>(CB), NH3<\/sub>(CA)\r\n\r\n<\/div>\r\n\u00a0 \u00a0<\/em>\r\n\r\n<\/div>\r\n

      \r\n
      \r\n

      (10) Identify and label the Br\u00f8nsted-Lowry acid, its conjugate base, the Br\u00f8nsted-Lowry base, and its conjugate acid in each of the following equations:<\/p>\r\n

      (10a) \\({\\text{NO}}_{2}{^-}^{\\text{}}+{\\text{H}}_{2}\\text{O}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{HNO}}_{2}+{\\text{OH}^-}^{\\text{}}\\)<\/p>\r\n

      (10b) \\(\\text{HBr}+{\\text{H}}_{2}\\text{O}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{3}{\\text{O}}^{\\text{+}}+{\\text{Br}^-}^{\\text{}}\\)<\/p>\r\n

      (10c) \\({\\text{HS}}^{\\text{\u2212}}+{\\text{H}}_{2}\\text{O}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{2}\\text{S}+{\\text{OH}^-}^{\\text{}}\\)<\/p>\r\n

      (10d) \\({\\text{H}}_{2}{\\text{PO}}_{4}{^-}^{\\text{}}+{\\text{OH}^-}^{\\text{}}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{HPO}}_{4}{^{2-}}^{\\text{}}+{\\text{H}}_{2}\\text{O}\\)<\/p>\r\n

      (10e) \\({\\text{H}}_{2}{\\text{PO}}_{4}{^-}^{\\text{}}+\\text{HCl}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{3}{\\text{PO}}_{4}+{\\text{Cl}^-}^{\\text{}}\\)<\/p>\r\n

      (10f) \\({\\left[\\text{Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}\\left(\\text{OH}\\right)\\right]}^{2+}+{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}\\right]}^{3+}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{[Fe}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{6}\\right]}^{3+}+{\\left[\\text{Al}{\\left({\\text{H}}_{2}\\text{O}\\right)}_{5}\\left(\\text{OH}\\right)\\right]}^{2+}\\)<\/p>\r\n

      (10g) \\({\\text{CH}}_{3}\\text{OH}+{\\text{H}^-}^{\\text{}}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{CH}}_{3}{\\text{O}^-}^{\\text{}}+{\\text{H}}_{2}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      \r\n\r\n \r\n

      (11) What are amphiprotic species? Illustrate with suitable equations.<\/p>\r\n \r\n

      Solution<\/em><\/p>\r\n\r\n<\/div>\r\n

      \r\n

      Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is H2<\/sub>O.<\/p>\r\nAs an acid: \\({\\text{H}}_{2}\\text{O}\\left(aq\\right)+{\\text{NH}}_{3}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}\\rightleftharpoons\\phantom{\\rule{0.2em}{0ex}}{\\text{NH}}_{4}{}^{\\text{+}}\\left(aq\\right)+{\\text{OH}^-}^{\\text{}}\\left(aq\\right).\\)\r\n\r\nAs a base: \\({\\text{H}}_{2}\\text{O}\\left(aq\\right)+\\text{HCl}\\left(aq\\right)\\phantom{\\rule{0.2em}{0ex}}\\rightleftharpoons\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\left(aq\\right)+{\\text{Cl}^-}^{\\text{}}\\left(aq\\right)\\)\r\n\r\n<\/div>\r\n\u00a0\u00a0<\/em>\r\n\r\n<\/div>\r\n

      \r\n
      \r\n

      (12) State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species:<\/p>\r\n

      (12a) H2<\/sub>O<\/p>\r\n

      (12b) \\({\\text{H}}_{2}{\\text{PO}}_{4}{^-}^{\\text{}}\\)<\/p>\r\n

      (12c) S2\u2212<\/sup><\/p>\r\n

      (12d) \\({\\text{CO}}_{3}{^{2-}}^{\\text{}}\\)<\/p>\r\n

      (12e) \\({\\text{HSO}}_{4}{^-}^{\\text{}}\\)<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n

      \r\n
      \r\n\r\n \r\n

      (13) State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.<\/p>\r\n

      (13a) NH3<\/sub><\/p>\r\n

      (13b) \\({\\text{HPO}}_{4}{^-}^{\\text{}}\\)<\/p>\r\n

      (13c) Br\u2212<\/sup><\/p>\r\n

      (13d) \\({\\text{NH}}_{4}{}^{\\text{+}}\\)<\/p>\r\n

      (13e) \\({\\text{ASO}}_{4}{^{3-}}^{\\text{}}\\)<\/p>\r\n \r\n

      Solution<\/em><\/p>\r\n\r\n<\/div>\r\n

      \r\n

      amphiprotic:<\/p>\r\n(a) \\({\\text{NH}}_{3}+{\\text{H}}_{3}{\\text{O}}^{\\text{+}}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{NH}}_{4}\\text{OH}+{\\text{H}}_{2}\\text{O},\\) \\({\\text{NH}}_{3}+{\\text{OCH}}_{3}{^-}^{\\text{}}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{NH}}_{2}{^-}^{\\text{}}+{\\text{CH}}_{3}\\text{OH};\\)\r\n\r\n(b) \\({\\text{HPO}}_{4}{^{2-}}^{\\text{}}+{\\text{OH}^-}^{\\text{}}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{PO}}_{4}{^{3-}}^{\\text{}}+{\\text{H}}_{2}\\text{O},\\) \\({\\text{HPO}}_{4}{^{2-}}^{\\text{}}+{\\text{HClO}}_{4}\\phantom{\\rule{0.2em}{0ex}}$\\rightarrow$\\phantom{\\rule{0.2em}{0ex}}{\\text{H}}_{2}{\\text{PO}}_{4}{^-}^{\\text{}}+{\\text{ClO}}_{4}{^-}^{\\text{}};\\) not\r\n\r\namphiprotic:\r\n\r\n(c) Br\u2212<\/sup>;\r\n\r\n(d) \\({\\text{NH}}_{4}{}^{\\text{+}};\\)\r\n\r\n(e) \\({\\text{AsO}}_{4}{^{3-}}^{\\text{}}\\)\r\n\r\n<\/div>\r\n\u00a0\u00a0<\/em>\r\n\r\n<\/div>\r\n

      \r\n
      \r\n

      (14) Is the self-ionization of water endothermic or exothermic? The ionization constant for water (K<\/em>w<\/sub>) is 2.9 \\(\\times\\) 10\u221214<\/sup> at 40 \u00b0C and 9.3 \\(\\times\\) 10\u221214<\/sup> at 60 \u00b0C.<\/p>\r\n\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n<\/div>\r\n

      \r\n

      Glossary<\/h2>\r\n
      \r\n \t
      [pb_glossary id=\"3225\"]acid ionization[\/pb_glossary]<\/dt>\r\n \t
      reaction involving the transfer of a proton from an acid to water, yielding hydronium ions and the conjugate base of the acid<\/dd>\r\n<\/dl>\r\n
      \r\n \t
      [pb_glossary id=\"3226\"]amphiprotic[\/pb_glossary]<\/dt>\r\n \t
      species that may either donate or accept a proton in a Bronsted-Lowry acid-base reaction<\/dd>\r\n<\/dl>\r\n
      \r\n \t
      [pb_glossary id=\"3227\"]amphoteric[\/pb_glossary]<\/dt>\r\n \t
      species that can act as either an acid or a base<\/dd>\r\n<\/dl>\r\n
      \r\n \t
      [pb_glossary id=\"3228\"]autoionization[\/pb_glossary]<\/dt>\r\n \t
      reaction between identical species yielding ionic products; for water, this reaction involves transfer of protons to yield hydronium and hydroxide ions<\/dd>\r\n<\/dl>\r\n
      \r\n \t
      [pb_glossary id=\"3229\"]base ionization[\/pb_glossary]<\/dt>\r\n \t
      reaction involving the transfer of a proton from water to a base, yielding hydroxide ions and the conjugate acid of the base<\/dd>\r\n<\/dl>\r\n
      \r\n \t
      [pb_glossary id=\"3230\"]Br\u00f8nsted-Lowry acid[\/pb_glossary]<\/dt>\r\n \t
      proton donor<\/dd>\r\n<\/dl>\r\n
      \r\n \t
      [pb_glossary id=\"3231\"]Br\u00f8nsted-Lowry base[\/pb_glossary]<\/dt>\r\n \t
      proton acceptor<\/dd>\r\n<\/dl>\r\n
      \r\n \t
      [pb_glossary id=\"3232\"]conjugate acid[\/pb_glossary]<\/dt>\r\n \t
      substance formed when a base gains a proton<\/dd>\r\n<\/dl>\r\n
      \r\n \t
      [pb_glossary id=\"3233\"]conjugate base[\/pb_glossary]<\/dt>\r\n \t
      substance formed when an acid loses a proton<\/dd>\r\n<\/dl>\r\n
      \r\n \t
      [pb_glossary id=\"3234\"]ion-product constant for water (K<\/em>w<\/sub>)[\/pb_glossary]<\/dt>\r\n \t
      equilibrium constant for the autoionization of water<\/dd>\r\n<\/dl>\r\n<\/div>","rendered":"
      \n
      \n
      \n

      Learning Objectives<\/p>\n<\/header>\n

      \n

      By the end of this section, you will be able to:<\/p>\n

        \n
      • Identify acids, bases, and conjugate acid-base pairs according to the Br\u00f8nsted-Lowry definition<\/li>\n
      • Write equations for acid and base ionization reactions<\/li>\n
      • Use the ion-product constant for water to calculate hydronium and hydroxide ion concentrations<\/li>\n
      • Describe the acid-base behavior of amphiprotic substances<\/li>\n<\/ul>\n<\/div>\n<\/div>\n

        The acid-base reaction class has been studied for quite some time. In 1680, Robert Boyle<\/span> reported traits of acid solutions that included their ability to dissolve many substances, to change the colors of certain natural dyes, and to lose these traits after coming in contact with alkali (base) solutions. In the eighteenth century, it was recognized that acids have a sour taste, react with limestone to liberate a gaseous substance (now known to be CO2<\/sub>), and interact with alkalis to form neutral substances. In 1815, Humphry Davy<\/span> contributed greatly to the development of the modern acid-base concept by demonstrating that hydrogen is the essential constituent of acids. Around that same time, Joseph Louis Gay-Lussac concluded that acids are substances that can neutralize bases and that these two classes of substances can be defined only in terms of each other. The significance of hydrogen was reemphasized in 1884 when Svante Arrhenius<\/span> defined an acid as a compound that dissolves in water to yield hydrogen cations (now recognized to be hydronium ions) and a base as a compound that dissolves in water to yield hydroxide anions.<\/p>\n

        Johannes Br\u00f8nsted and Thomas Lowry proposed a more general description in 1923 in which acids and bases were defined in terms of the transfer of hydrogen ions, H+<\/sup>. (Note that these hydrogen ions are often referred to simply as protons<\/em>, since that subatomic particle is the only component of cations derived from the most abundant hydrogen isotope, 1<\/sup>H.) A compound that donates a proton to another compound is called a Br\u00f8nsted-Lowry acid<\/span>, and a compound that accepts a proton is called a Br\u00f8nsted-Lowry base<\/span>. An acid-base reaction is, thus, the transfer of a proton from a donor (acid) to an acceptor (base).<\/p>\n

        The concept of conjugate pairs<\/em> is useful in describing Br\u00f8nsted-Lowry acid-base reactions (and other reversible reactions, as well). When an acid donates H+<\/sup>, the species that remains is called the conjugate base<\/strong> of the acid because it reacts as a proton acceptor in the reverse reaction. Likewise, when a base accepts H+<\/sup>, it is converted to its conjugate acid<\/strong>. The reaction between water and ammonia illustrates this idea. In the forward direction, water acts as an acid by donating a proton to ammonia and subsequently becoming a hydroxide ion, OH\u2212<\/sup>, the conjugate base of water. The ammonia acts as a base in accepting this proton, becoming an ammonium ion, \"{{\text{NH}}_{4}}^{\text{+}}\text{,}\" the conjugate acid of ammonia. In the reverse direction, a hydroxide ion acts as a base in accepting a proton from ammonium ion, which acts as an acid.<\/p>\n

        \u00a0<\/em><\/p>\n

        \"This
        Scheme 8.2.1 – A Br\u00f8nsted-Lowry acid reacts with a base to form the conjugate base and acid, as shown with H2O (acid) reacting with NH3 (base) to form their respective conjugates.<\/strong><\/figcaption><\/figure>\n

        The reaction between a Br\u00f8nsted-Lowry acid and water is called acid ionization<\/span>. For example, when hydrogen fluoride dissolves in water and ionizes, protons are transferred from hydrogen fluoride molecules to water molecules, yielding hydronium ions and fluoride ions:<\/p>\n

        \u00a0<\/em><\/p>\n

        \"This
        Scheme 8.2.2 – The acid ionization reaction of an acid (HF) with water, resulting in a hydronium ionsand the conjugate base of the acid.<\/strong><\/figcaption><\/figure>\n

        Base ionization<\/span> of a species occurs when it accepts protons from water molecules. In the example below, pyridine molecules, C5<\/sub>NH5<\/sub>, undergo base ionization when dissolved in water, yielding hydroxide and pyridinium ions:<\/p>\n

        \u00a0<\/em><\/p>\n

        \"This
        Scheme 8.2.3 – Pyridine reacts with water to form the pyridinium ion and hydroxide ion.<\/strong><\/figcaption><\/figure>\n

        The preceding ionization reactions suggest that water may function as both a base (as in its reaction with hydrogen fluoride) and an acid (as in its reaction with ammonia). Species capable of either donating or accepting protons are called amphiprotric<\/span>, or more generally, amphoteric<\/span>, a term that may be used for acids and bases per definitions other than the Br\u00f8nsted-Lowry one. The equations below show the two possible acid-base reactions for two amphiprotic species, bicarbonate ion and water:<\/p>\n

        \u00a0<\/em><\/p>\n

        \"{\text{HCO}}_{3}{}^{-}\left(\text{aq}\right)+{\text{H}}_{2}\text{O}\left(\text{l}\right)\rightleftharpoons\phantom{\rule{4em}{0ex}}{\text{CO}}_{3}{}^{2-}\left(\text{aq}\right)+{\text{H}}_{3}{\text{O}}^{+}\left(\text{aq}\right)\"<\/div>\n
        \u00a0<\/em><\/div>\n
        \"{\text{HCO}}_{3}{}^{-}\left(\text{aq}\right)+{\text{H}}_{2}\text{O}\left(\text{l}\right)\rightleftharpoons\phantom{\rule{4em}{0ex}}{\text{H}}_{2}{\text{CO}}_{3}\left(\text{aq}\right)+{\text{OH}}^{-}\left(\text{aq}\right)\"<\/div>\n
        \u00a0<\/em><\/div>\n
        <\/div>\n

        The first equation represents the reaction of bicarbonate as an acid with water as a base, whereas the second represents reaction of bicarbonate as a base with water as an acid. When bicarbonate is added to water, both these equilibria are established simultaneously and the composition of the resulting solution may be determined through appropriate equilibrium calculations, as described later in this chapter.<\/p>\n

        In the liquid state, molecules of an amphiprotic substance can react with one another as illustrated for water in the equations below:<\/p>\n

        \u00a0<\/em><\/p>\n

        \"This
        Scheme 8.2.4 – Water is amphiprotic and can react with itself to from hydronium and hydroxide ions.<\/strong><\/figcaption><\/figure>\n

        The process in which like molecules react to yield ions is called autoionization<\/span>. Liquid water undergoes autoionization to a very slight extent; at 25 \u00b0C, approximately two out of every billion water molecules are ionized. The extent of the water autoionization process is reflected in the value of its equilibrium constant, the ion-product constant for water, K<\/em>w<\/sub><\/span>:<\/p>\n

        \u00a0<\/em><\/p>\n

        \"{\text{H}}_{2}\text{O}\left(l\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{OH}}^{\text{-}}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{\text{w}}=\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{OH}^-}^{\text{}}\right]\"<\/div>\n
        \u00a0<\/em><\/div>\n

        The slight ionization of pure water is reflected in the small value of the equilibrium constant; at 25 \u00b0C, K<\/em>w<\/sub> has a value of 1.0 \"\times\" 10\u221214<\/sup>. The process is endothermic, and so the extent of ionization and the resulting concentrations of hydronium ion and hydroxide ion increase with temperature. For example, at 100 \u00b0C, the value for K<\/em>w<\/sub> is about 5.6 \"\times\" 10\u221213<\/sup>, roughly 50 times larger than the value at 25 \u00b0C.<\/p>\n

        \n
        \n
        \n

        Activity 8.2.1 – Ion Concentrations in Pure Water<\/p>\n<\/header>\n

        \n

        What are the hydronium ion concentration and the hydroxide ion concentration in pure water at 25 \u00b0C?<\/p>\n

        Solution<\/h2>\n

        The autoionization of water yields the same number of hydronium and hydroxide ions. Therefore, in pure water, [H3<\/sub>O+<\/sup>] = [OH\u2212<\/sup>] = x<\/em>. At 25 \u00b0C:<\/p>\n

        \u00a0<\/em><\/p>\n

        \"{K}_{\text{w}}=\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{OH}^-}^{\text{}}\right]=x=1.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-14}\"<\/div>\n
        \u00a0<\/em><\/div>\n

        So:<\/p>\n

        \u00a0<\/em><\/p>\n

        \"x=\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]=\left[{\text{OH}^-}^{\text{}}\right]=\sqrt{1.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-14}}=1.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-7}\phantom{\rule{0.2em}{0ex}}M\"<\/div>\n
        \u00a0<\/em><\/div>\n

        The hydronium ion concentration and the hydroxide ion concentration are the same, 1.0 \"\times\" 10\u22127<\/sup>M<\/em>.<\/p>\n

        \n

        Check Your Learning<\/span><\/h2>\n

        The ion product of water at 80 \u00b0C is 2.4 \"\times\" 10\u221213<\/sup>. What are the concentrations of hydronium and hydroxide ions in pure water at 80 \u00b0C?<\/p>\n

        \n

        Answer<\/h3>\n

        [H3<\/sub>O+<\/sup>] = [OH\u2212<\/sup>] = 4.9 \"\times\" 10\u22127<\/sup>M<\/em><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n
        \n

        Activity 8.2.2 – The Inverse Relation between [H3<\/sub>O+<\/sup>] and [OH\u2212<\/sup>]<\/p>\n<\/header>\n

        \n

        A solution of an acid in water has a hydronium ion concentration of 2.0 \"\times\" 10\u22126<\/sup>M<\/em>. What is the concentration of hydroxide ion at 25 \u00b0C?<\/p>\n

        Solution<\/h2>\n

        Use the value of the ion-product constant for water at 25 \u00b0C:<\/p>\n

        \u00a0<\/em><\/p>\n

        \"2{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{OH}^-}^{\text{}}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{\text{w}}=\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{OH}^-}^{\text{}}\right]=1.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-14}\"<\/div>\n
        \u00a0<\/em><\/div>\n
        <\/div>\n

        to calculate the missing equilibrium concentration.<\/p>\n

        Rearrangement of the K<\/em>w<\/sub> expression shows that [OH\u2212<\/sup>] is inversely proportional to [H3<\/sub>O+<\/sup>]:<\/p>\n

        \u00a0<\/em><\/p>\n

        \"\left[{\text{OH}^-}^{\text{}}\right]=\phantom{\rule{0.2em}{0ex}}\frac{{K}_{\text{w}}}{\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]}\phantom{\rule{0.2em}{0ex}}=\phantom{\rule{0.2em}{0ex}}\frac{1.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-14}}{2.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-6}}\phantom{\rule{0.2em}{0ex}}=5.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-9}\"<\/div>\n
        \u00a0<\/em><\/div>\n
        <\/div>\n

        Compared with pure water, a solution of acid exhibits a higher concentration of hydronium ions (due to ionization of the acid) and a proportionally lower concentration of hydroxide ions. This may be explained via Le Ch\u00e2telier\u2019s principle as a left shift in the water autoionization equilibrium resulting from the stress of increased hydronium ion concentration.<\/p>\n

        Substituting the ion concentrations into the K<\/em>w<\/sub> expression confirms this calculation, resulting in the expected value:<\/p>\n

        \u00a0<\/em><\/p>\n

        \"{K}_{\text{w}}=\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{OH}^-}^{\text{}}\right]=\left(2.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-6}\right)\left(5.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-9}\right)=1.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}{10}^{-14}\"<\/div>\n
        \n
        \n<\/div>\n

        Check Your Learning<\/span><\/h2>\n

        What is the hydronium ion concentration in an aqueous solution with a hydroxide ion concentration of 0.001 M<\/em> at 25 \u00b0C?<\/p>\n

        \n

        Answer<\/h3>\n

        [H3<\/sub>O+<\/sup>] = 1 \"\times\" 10\u221211<\/sup>M<\/em><\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        \n
        \n
        \n

        Activity 8.2.3 – Representing the Acid-Base Behavior of an Amphoteric Substance<\/p>\n<\/header>\n

        \n

        Write separate equations representing the reaction of \"{\text{HSO}}_{3}{^-}^{\text{}}\"<\/p>\n

        (a) as an acid with OH\u2212<\/sup><\/p>\n

        (b) as a base with HI<\/p>\n

        Solution<\/h2>\n

        (a) \"{\text{HSO}}_{3}{}^{\text{-}}\left(aq\right)+{\text{OH}^-}^{\text{}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{SO}}_{3}{^{2-}}^{\text{}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\"<\/p>\n

        (b) \"{\text{HSO}}_{3}{^-}^{\text{}}\left(aq\right)+\text{HI}\left(aq\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}{\text{SO}}_{3}\left(aq\right)+{\text{I}^-}^{\text{}}\left(aq\right)\"<\/p>\n

        \n

        Check Your Learning<\/span><\/h2>\n

        Write separate equations representing the reaction of \"{\text{H}}_{2}{\text{PO}}_{4}{^-}^{\text{}}\"<\/p>\n

        (a) as a base with HBr<\/p>\n

        (b) as an acid with OH\u2212<\/sup><\/p>\n

        \n

        Answer<\/h3>\n

        (a) \"{\text{H}}_{2}{\text{PO}}_{4}{^-}^{\text{}}\left(aq\right)+\text{HBr}\left(aq\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{PO}}_{4}\left(aq\right)+{\text{Br}^-}^{\text{}}\left(aq\right);\"<\/p>\n

        (b) \"{\text{H}}_{2}{\text{PO}}_{4}{}^{\text{−}}\left(aq\right)+{\text{OH}^-}^{\text{}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{HPO}}_{4}{^{2-}}^{\text{}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\"<\/p>\n<\/div>\n<\/div>\n<\/div>\n<\/div>\n

        Key Concepts and Summary<\/h1>\n

        A compound that can donate a proton (a hydrogen ion) to another compound is called a Br\u00f8nsted-Lowry acid. The compound that accepts the proton is called a Br\u00f8nsted-Lowry base. The species remaining after a Br\u00f8nsted-Lowry acid has lost a proton is the conjugate base of the acid. The species formed when a Br\u00f8nsted-Lowry base gains a proton is the conjugate acid of the base. Thus, an acid-base reaction occurs when a proton is transferred from an acid to a base, with formation of the conjugate base of the reactant acid and formation of the conjugate acid of the reactant base. Amphiprotic species can act as both proton donors and proton acceptors. Water is the most important amphiprotic species. It can form both the hydronium ion, H3<\/sub>O+<\/sup>, and the hydroxide ion, OH\u2212<\/sup> when it undergoes autoionization:<\/p>\n

        \u00a0<\/em><\/p>\n

        \"{\text{2H}}_{2}\text{O(}l\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{+}\left(aq\right)+{\text{OH}^-}^{\text{}}\left(aq\right)\"<\/p>\n

        \u00a0<\/em><\/p>\n

        The ion product of water, K<\/em>w<\/sub> is the equilibrium constant for the autoionization reaction:<\/p>\n

        \u00a0<\/em><\/p>\n

        \"{K}_{\text{w}}=\left[{\text{H}}_{3}{\text{O}}^{\text{+}}\right]\left[{\text{OH}^-}^{\text{}}\right]=1.0\phantom{\rule{0.2em}{0ex}}\times\phantom{\rule{0.2em}{0ex}}1{0}^{-14}\phantom{\rule{0.2em}{0ex}}\text{at}\phantom{\rule{0.2em}{0ex}}25\phantom{\rule{0.2em}{0ex}}{^\circ}C}\"<\/p>\n

        Key Equations<\/h2>\n
          \n
        • K<\/em>w<\/sub> = [H3<\/sub>O+<\/sup>][OH\u2212<\/sup>] = 1.0 \"\times\" 10\u221214<\/sup> (at 25 \u00b0C)<\/li>\n<\/ul>\n
          \n
          \n

          End of Chapter Exercises<\/p>\n<\/header>\n

          \n
          \n
          \n

          (1) Write equations that show NH3<\/sub> as both a conjugate acid and a conjugate base.<\/p>\n

          \u00a0<\/em><\/p>\n

          Solution<\/em><\/p>\n<\/div>\n

          \n

          One example for NH3<\/sub> as a conjugate acid:<\/p>\n

          \"{\text{NH}}_{2}{^-}^{\text{}}+{\text{H}}^{\text{+}}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{3};\" as a conjugate base: \"{\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{OH}^-}^{\text{}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{3}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\"<\/p>\n<\/div>\n

          \u00a0\u00a0<\/em><\/p>\n<\/div>\n

          \n
          \n

          (2) Write equations that show \"{\text{H}}_{2}{\text{PO}}_{4}{^-}^{\text{}}\" acting both as an acid and as a base.<\/p>\n

          \u00a0<\/em><\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          (3) Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry acid:<\/p>\n

          (3a) \"{\text{H}}_{3}{\text{O}}^{\text{+}}\"<\/p>\n

          (3b) HCl<\/p>\n

          (3c) NH3<\/sub><\/p>\n

          (3d) CH3<\/sub>CO2<\/sub>H<\/p>\n

          (3e) \"{\text{NH}}_{4}{}^{\text{+}}\"<\/p>\n

          (3f) \"{\text{HSO}}_{4}{^-}^{\text{}}\"<\/p>\n

           <\/p>\n

          Solution<\/em><\/p>\n<\/div>\n

          \n

          (a) \"{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}^{\text{+}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right);\"<\/p>\n

          (b) \"\text{HCl}\left(\mathrm{aq}\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}^{\text{+}}\left(aq\right)+{\text{Cl}^-}^{\text{}}\left(aq\right);\"<\/p>\n

          (c) \"{\text{NH}}_{3}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}^{\text{+}}\left(aq\right)+{\text{NH}}_{2}{^-}^{\text{}}\left(aq\right);\"<\/p>\n

          (d) \"{\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}^{\text{+}}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{^-}^{\text{}}\left(aq\right);\"<\/p>\n

          (e) \"{\text{NH}}_{4}{}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}^{\text{+}}\left(aq\right)+{\text{NH}}_{3}\left(aq\right);\"<\/p>\n

          (f) \"{\text{HSO}}_{4}{^-}^{\text{}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}^{+}\left(aq\right)+{\text{SO}}_{4}{^{2-}}^{\text{}}\left(aq\right)\"<\/p>\n<\/div>\n

          \u00a0\u00a0<\/em><\/p>\n<\/div>\n

          \n
          \n

          (4) Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry acid:<\/p>\n

          (4a) HNO3<\/sub><\/p>\n

          (4b) \"{\text{PH}}_{4}{}^{\text{+}}\"<\/p>\n

          (4c) H2<\/sub>S<\/p>\n

          (4d) CH3<\/sub>CH2<\/sub>COOH<\/p>\n

          (4e) \"{\text{H}}_{2}{\text{PO}}_{4}{^-}^{\text{}}\"<\/p>\n

          (4f) HS\u2212<\/sup><\/p>\n

          \u00a0<\/em><\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          (5) Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry base:<\/p>\n

          (5a) H2<\/sub>O<\/p>\n

          (5b) OH\u2212<\/sup><\/p>\n

          (5c) NH3<\/sub><\/p>\n

          (5d) CN\u2212<\/sup><\/p>\n

          (5e) S2\u2212<\/sup><\/p>\n

          (5f) \"{\text{H}}_{2}{\text{PO}}_{4}{^-}^{\text{}}\"<\/p>\n

           <\/p>\n

          Solution<\/em><\/p>\n<\/div>\n

          \n

          (a) \"{\text{H}}_{2}\text{O}\left(l\right)+{\text{H}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right);\"<\/p>\n

          (b) \"{\text{OH}}^{\text{−}}\left(aq\right)+{\text{H}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{O}\left(l\right);\"<\/p>\n

          (c) \"{\text{NH}}_{3}\left(aq\right)+{\text{H}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{4}{}^{\text{+}}\left(aq\right);\"<\/p>\n

          (d) \"{\text{CN}^-}^{\text{}}\left(aq\right)+{\text{H}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}\text{HCN}\left(aq\right);\"<\/p>\n

          (e) \"{\text{S}}^{2-}\left(aq\right)+{\text{H}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{HS}^-}^{\text{}}\left(aq\right);\"<\/p>\n

          (f) \"{\text{H}}_{2}{\text{PO}}_{4}{^-}^{\text{}}\left(aq\right)+{\text{H}}^{\text{+}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{PO}}_{4}\left(aq\right)\"<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

           <\/p>\n

          (6) Show by suitable net ionic equations that each of the following species can act as a Br\u00f8nsted-Lowry base:<\/p>\n

          (6a) HS\u2212<\/sup><\/p>\n

          (6b) \"{\text{PO}}_{4}{^{3-}}^{\text{}}\"<\/p>\n

          (6c) \"{\text{NH}}_{2}{^-}^{\text{}}\"<\/p>\n

          (6d) C2<\/sub>H5<\/sub>OH<\/p>\n

          (6e) O2\u2212<\/sup><\/p>\n

          (6f) \"{\text{H}}_{2}{\text{PO}}_{4}{^-}^{\text{}}\"<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

           <\/p>\n

          (7) What is the conjugate acid of each of the following? What is the conjugate base of each?<\/p>\n

          (7a) OH\u2212<\/sup><\/p>\n

          (7b) H2<\/sub>O<\/p>\n

          (7c) \"{\text{HCO}}_{3}{^-}^{\text{}}\"<\/p>\n

          (7d) NH3<\/sub><\/p>\n

          (7e) \"{\text{HSO}}_{4}{^-}^{\text{}}\"<\/p>\n

          (7f) H2<\/sub>O2<\/sub><\/p>\n

          (7g) HS\u2212<\/sup><\/p>\n

          (7h) \"{\text{H}}_{5}{\text{N}}_{2}{}^{\text{+}}\"<\/p>\n

           <\/p>\n

          Solution<\/em><\/p>\n<\/div>\n

          \n

          (a) H2<\/sub>O, O2\u2212<\/sup>;<\/p>\n

          (b) H3<\/sub>O+<\/sup>, OH\u2212<\/sup>;<\/p>\n

          (c) H2<\/sub>CO3<\/sub>, \"{\text{CO}}_{3}{^{2-}}^{\text{}};\"<\/p>\n

          (d) \"{\text{NH}}_{4}{}^{\text{+}},\" \"{\text{NH}}_{2}{^-}^{\text{}};\"<\/p>\n

          (e) H2<\/sub>SO4<\/sub>, \"{\text{SO}}_{4}{^{2-}}^{\text{}};\"<\/p>\n

          (f) \"{\text{H}}_{3}{\text{O}}_{2}{}^{\text{+}},\" \"{\text{HO}}_{2}{^-}^{\text{}};\"<\/p>\n

          (g) H2<\/sub>S; S2\u2212<\/sup>;<\/p>\n

          (h) \"{\text{H}}_{6}{\text{N}}_{2}{}^{2+},\" H4<\/sub>N2<\/sub><\/p>\n<\/div>\n

          \u00a0\u00a0<\/em><\/p>\n<\/div>\n

          \n
          \n

          (8) What is the conjugate acid of each of the following? What is the conjugate base of each?<\/p>\n

          (8a) H2<\/sub>S<\/p>\n

          (8b) \"{\text{H}}_{2}{\text{PO}}_{4}{^-}^{\text{}}\"<\/p>\n

          (8c) PH3<\/sub><\/p>\n

          (8d) HS\u2212<\/sup><\/p>\n

          (8e) \"{\text{HSO}}_{3}{^-}^{\text{}}\"<\/p>\n

          (8f) \"{\text{H}}_{3}{\text{O}}_{2}{}^{\text{+}}\"<\/p>\n

          (8g) H4<\/sub>N2<\/sub><\/p>\n

          (8h) CH3<\/sub>OH<\/p>\n

          \u00a0\u00a0<\/em><\/p>\n<\/div>\n<\/div>\n

          \n
          \n

          (9) Identify and label the Br\u00f8nsted-Lowry acid, its conjugate base, the Br\u00f8nsted-Lowry base, and its conjugate acid in each of the following equations:<\/p>\n

          (9a) \"{\text{HNO}}_{3}+{\text{H}}_{2}\text{O}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}+{\text{NO}}_{3}{^-}^{\text{}}\"<\/p>\n

          (9b) \"{\text{CN}^-}^{\text{}}+{\text{H}}_{2}\text{O}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}\text{HCN}+{\text{OH}^-}^{\text{}}\"<\/p>\n

          (9c) \"{\text{H}}_{2}{\text{SO}}_{4}+{\text{Cl}^-}^{\text{}}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}\text{HCl}+{\text{HSO}}_{4}{^-}^{\text{}}\"<\/p>\n

          (9d) \"{\text{HSO}}_{4}{^-}^{\text{}}+{\text{OH}^-}^{\text{}}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{SO}}_{4}{^{2-}}^{\text{}}+{\text{H}}_{2}\text{O}\"<\/p>\n

          (9e) \"{\text{O}}^{2-}+{\text{H}}_{2}\text{O}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}2{\mathrm{OH}^-}^{\text{}}\"<\/p>\n

          (9f) \"{\left[\text{Cu}{\left({\text{H}}_{2}\text{O}\right)}_{3}\left(\text{OH}\right)\right]}^{\text{+}}+{\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}\right]}^{3+}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\left[\text{Cu}{\left({\text{H}}_{2}\text{O}\right)}_{4}\right]}^{2+}+{\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left(\text{OH}\right)\right]}^{2+}\"<\/p>\n

          (9g) \"{\text{H}}_{2}\text{S}+{\text{NH}}_{2}{^-}^{\text{}}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{HS}^-}^{\text{}}+{\text{NH}}_{3}\"<\/p>\n

           <\/p>\n

          Solution<\/em><\/p>\n<\/div>\n

          \n

          The labels are Br\u00f8nsted-Lowry acid = BA; its conjugate base = CB; Br\u00f8nsted-Lowry base = BB; its conjugate acid = CA.<\/p>\n

          (a) HNO3<\/sub>(BA), H2<\/sub>O(BB), H3<\/sub>O+<\/sup>(CA), \"{\text{NO}}_{3}{^-}^{\text{}}\left(\text{CB}\right)\"<\/p>\n

          (b) CN\u2212<\/sup>(BB), H2<\/sub>O(BA), HCN(CA), OH\u2212<\/sup>(CB);<\/p>\n

          (c) H2<\/sub>SO4<\/sub>(BA), Cl\u2212<\/sup>(BB), HCl(CA), \"{\text{HSO}}_{4}{^-}^{\text{}}\left(\text{CB}\right)\"<\/p>\n

          (d) \"{\text{HSO}}_{4}{^-}^{\text{}}\left(\text{BA}\right),\" OH\u2212<\/sup>(BB), \"{\text{SO}}_{4}{^{2-}}^{\text{}}\"(CB), H2<\/sub>O(CA)<\/p>\n

          (e) O2\u2212<\/sup>(BB), H2<\/sub>O(BA) OH\u2212<\/sup>(CB and CA)<\/p>\n

          (f) [Cu(H2<\/sub>O)3<\/sub>(OH)]+<\/sup>(BB), [Al(H2<\/sub>O)6<\/sub>]3+<\/sup>(BA), [Cu(H2<\/sub>O)4<\/sub>]2+<\/sup>(CA), [Al(H2<\/sub>O)5<\/sub>(OH)]2+<\/sup>(CB)<\/p>\n

          (g) H2<\/sub>S(BA), \"{\text{NH}}_{2}{^-}^{\text{}}\left(\text{BB}\right),\" HS\u2212<\/sup>(CB), NH3<\/sub>(CA)<\/p>\n<\/div>\n

          \u00a0 \u00a0<\/em><\/p>\n<\/div>\n

          \n
          \n

          (10) Identify and label the Br\u00f8nsted-Lowry acid, its conjugate base, the Br\u00f8nsted-Lowry base, and its conjugate acid in each of the following equations:<\/p>\n

          (10a) \"{\text{NO}}_{2}{^-}^{\text{}}+{\text{H}}_{2}\text{O}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{HNO}}_{2}+{\text{OH}^-}^{\text{}}\"<\/p>\n

          (10b) \"\text{HBr}+{\text{H}}_{2}\text{O}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}+{\text{Br}^-}^{\text{}}\"<\/p>\n

          (10c) \"{\text{HS}}^{\text{−}}+{\text{H}}_{2}\text{O}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}\text{S}+{\text{OH}^-}^{\text{}}\"<\/p>\n

          (10d) \"{\text{H}}_{2}{\text{PO}}_{4}{^-}^{\text{}}+{\text{OH}^-}^{\text{}}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{HPO}}_{4}{^{2-}}^{\text{}}+{\text{H}}_{2}\text{O}\"<\/p>\n

          (10e) \"{\text{H}}_{2}{\text{PO}}_{4}{^-}^{\text{}}+\text{HCl}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{PO}}_{4}+{\text{Cl}^-}^{\text{}}\"<\/p>\n

          (10f) \"{\left[\text{Fe}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left(\text{OH}\right)\right]}^{2+}+{\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{6}\right]}^{3+}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{[Fe}{\left({\text{H}}_{2}\text{O}\right)}_{6}\right]}^{3+}+{\left[\text{Al}{\left({\text{H}}_{2}\text{O}\right)}_{5}\left(\text{OH}\right)\right]}^{2+}\"<\/p>\n

          (10g) \"{\text{CH}}_{3}\text{OH}+{\text{H}^-}^{\text{}}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{CH}}_{3}{\text{O}^-}^{\text{}}+{\text{H}}_{2}\"<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

           <\/p>\n

          (11) What are amphiprotic species? Illustrate with suitable equations.<\/p>\n

           <\/p>\n

          Solution<\/em><\/p>\n<\/div>\n

          \n

          Amphiprotic species may either gain or lose a proton in a chemical reaction, thus acting as a base or an acid. An example is H2<\/sub>O.<\/p>\n

          As an acid: \"{\text{H}}_{2}\text{O}\left(aq\right)+{\text{NH}}_{3}\left(aq\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{4}{}^{\text{+}}\left(aq\right)+{\text{OH}^-}^{\text{}}\left(aq\right).\"<\/p>\n

          As a base: \"{\text{H}}_{2}\text{O}\left(aq\right)+\text{HCl}\left(aq\right)\phantom{\rule{0.2em}{0ex}}\rightleftharpoons\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{Cl}^-}^{\text{}}\left(aq\right)\"<\/p>\n<\/div>\n

          \u00a0\u00a0<\/em><\/p>\n<\/div>\n

          \n
          \n

          (12) State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species:<\/p>\n

          (12a) H2<\/sub>O<\/p>\n

          (12b) \"{\text{H}}_{2}{\text{PO}}_{4}{^-}^{\text{}}\"<\/p>\n

          (12c) S2\u2212<\/sup><\/p>\n

          (12d) \"{\text{CO}}_{3}{^{2-}}^{\text{}}\"<\/p>\n

          (12e) \"{\text{HSO}}_{4}{^-}^{\text{}}\"<\/p>\n<\/div>\n<\/div>\n

          \n
          \n

           <\/p>\n

          (13) State which of the following species are amphiprotic and write chemical equations illustrating the amphiprotic character of these species.<\/p>\n

          (13a) NH3<\/sub><\/p>\n

          (13b) \"{\text{HPO}}_{4}{^-}^{\text{}}\"<\/p>\n

          (13c) Br\u2212<\/sup><\/p>\n

          (13d) \"{\text{NH}}_{4}{}^{\text{+}}\"<\/p>\n

          (13e) \"{\text{ASO}}_{4}{^{3-}}^{\text{}}\"<\/p>\n

           <\/p>\n

          Solution<\/em><\/p>\n<\/div>\n

          \n

          amphiprotic:<\/p>\n

          (a) \"{\text{NH}}_{3}+{\text{H}}_{3}{\text{O}}^{\text{+}}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{4}\text{OH}+{\text{H}}_{2}\text{O},\" \"{\text{NH}}_{3}+{\text{OCH}}_{3}{^-}^{\text{}}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{NH}}_{2}{^-}^{\text{}}+{\text{CH}}_{3}\text{OH};\"<\/p>\n

          (b) \"{\text{HPO}}_{4}{^{2-}}^{\text{}}+{\text{OH}^-}^{\text{}}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{PO}}_{4}{^{3-}}^{\text{}}+{\text{H}}_{2}\text{O},\" \"{\text{HPO}}_{4}{^{2-}}^{\text{}}+{\text{HClO}}_{4}\phantom{\rule{0.2em}{0ex}}$\rightarrow$\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}{\text{PO}}_{4}{^-}^{\text{}}+{\text{ClO}}_{4}{^-}^{\text{}};\" not<\/p>\n

          amphiprotic:<\/p>\n

          (c) Br\u2212<\/sup>;<\/p>\n

          (d) \"{\text{NH}}_{4}{}^{\text{+}};\"<\/p>\n

          (e) \"{\text{AsO}}_{4}{^{3-}}^{\text{}}\"<\/p>\n<\/div>\n

          \u00a0\u00a0<\/em><\/p>\n<\/div>\n