{"id":3761,"date":"2026-03-20T15:09:30","date_gmt":"2026-03-20T19:09:30","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/?post_type=chapter&p=3761"},"modified":"2026-03-23T17:19:57","modified_gmt":"2026-03-23T21:19:57","slug":"quiz-solutions","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/chapter\/quiz-solutions\/","title":{"raw":"Quiz Solutions","rendered":"Quiz Solutions"},"content":{"raw":"1. In any process chemical industry, including hydrometallurgy, it is important to gain an understanding of the periodic table. This question begins to address that. Match the compound or element on the right to the definition or term on the left. There will be one unique compound\/element per term or definition. By process of elimination determine the most suitable match for each phrase.\r\n\r\n\\[\r\n\\begin{array}{l|l}\r\n\\text{Strong acid} & \\ce{H2SO4} \\\\\r\n\\text{Most electronegative metal} & \\ce{Au} \\\\\r\n\\text{Main group metal} & \\ce{Pb} \\\\\r\n\\text{Strong oxidant} & \\ce{Cl2} \\\\\r\n\\text{Most important industrial base} & \\ce{CaO} \\\\\r\n\\text{Strong base} & \\ce{NaOH} \\\\\r\n\\text{Strong reducing agent} & \\ce{Na} \\\\\r\n\\text{Transition metal} & \\ce{Ti} \\\\\r\n\\text{Strong electrolyte} & \\ce{NaCl} \\\\\r\n\\text{Strongly sulphophilic metal} & \\ce{Fe} \\\\\r\n\\text{Amphoteric ion} & \\ce{HSO3^-} \\\\\r\n\\text{Weak acid} & \\ce{H2S} \\\\\r\n\\text{Semi-metal} & \\ce{As} \\\\\r\n\\text{Incorrect chemical formula} & \\ce{NaCO3} \\\\\r\n\\end{array}\r\n\\]\r\n\r\n2. Fill in the blanks:\r\n(a) In an electron transfer reaction when an oxidizing agent reacts it gets reduced<\/span>.\r\n(b) Reduction is gain<\/span> of electrons.\r\n(c) A reducing agent loses<\/span> electrons.\r\n(d) What is the term we use for -log 10 [H + ]? pH<\/span>\r\n(e) A salt is a compound composed of cations<\/span> and anions<\/span>.\r\n\r\n3. Select the most comprehensive and correct definition for a base in aqueous solution. A base is a\r\ncompound that\r\n(a) has an ionizable OH-<\/sup> group and raises the pH.\r\n\r\n(b) has an ionizable OH-<\/sup> group and lowers the pH.\r\n\r\n(c) accepts a proton and raises the pH.<\/span>\r\n\r\n(d) accepts a proton and lowers the pH.\r\n\r\n4. Which of the following compounds\r\n(a) will act as an acid in water: CH4<\/sub> , CO2 <\/sub>, HCO3<\/sub>- <\/sup>, HNO3<\/sub><\/span> , NH3<\/sub>\r\n(b) will acts as a base in water: HCO3<\/sub>-<\/sup> , SO4<\/sub>2-<\/sup> , NaCH3<\/sub>CO2<\/sub><\/span> , N2<\/sub> , C6<\/sub>H5<\/sub>OH (phenol)\r\n\r\n5. Indicate which of the following are soluble salts in water:\r\nHCl, Ag2<\/sub>S, Na2<\/sub>S<\/span>, C12<\/sub>H22<\/sub>O11<\/sub> (sugar), S8<\/sub> , NH3<\/sub> , CuO , PbCl2<\/sub> , CaO , CO2<\/sub> , H2<\/sub> , Te\r\n\r\nOnly Na2<\/sub>S. Note that HCl, C12<\/sub>H22<\/sub>O11<\/sub> and NH3<\/sub> are all soluble, but are not salts. S8<\/sub> , H2<\/sub> , CO2<\/sub> and Te are not salts\r\nand not soluble. Ag2<\/sub>S, CuO, PbCl2<\/sub> and CaO are insoluble salts.\r\n\r\n6. Perform the required calculations for the following aqueous solutions:\r\n\r\n(a) Calculate the pH of a solution made up of 5.00 mL of 1.2 M HCl diluted to precisely 250 mL.\r\n\r\n\\[\r\n\\frac{5 \\cdot 1.2\\,\\text{M}\\,\\mathrm{H^+}}{250}\r\n= 0.024\\,\\text{M}\\,\\mathrm{H^+}\r\n\\]\r\n

(HCl is a strong acid; fully dissociates.)<\/p>\r\n\\[\r\n\\text{pH} = -\\log_{10}(0.024) = \\textcolor{blue}{1.62}\r\n\\]\r\n\r\n(b) A solution contains a total ammonia concentration of 0.8 M (the sum of ammonia and ammonium concentrations). The volume is 1.5 L and the pH is 9.6. What volume (mL) of aqueous ammonia (density = 0.90 g\/mL and 28% by mass, i.e. 280 g NH3<\/sub> \/kg of solution) and mass of ammonium sulfate (in g) were added to make up the solution? Use Kb<\/sub> for ammonia = 1.8 x 10-5<\/sup>.\r\n\r\n\\[\r\n[\\mathrm{NH_3}] + [\\mathrm{NH_4^+}] = 0.8\r\n\\]\r\n\r\n\\[\r\n\\text{pH} = \\text{p}K_a + \\log\\!\\left(\\frac{[\\mathrm{NH_3}]}{[\\mathrm{NH_4^+}]}\\right) = 9.6\r\n\\]\r\n\r\n\\[\r\nK_a = \\frac{1 \\times 10^{-14}}{1.8 \\times 10^{-5}}\r\n= 5.556 \\times 10^{-10}\r\n\\]\r\n\r\n\\[\r\n\\text{p}K_a = -\\log_{10}(K_a) = 9.2553\r\n\\]\r\n\r\n\\[\r\n9.6 = 9.2553 + \\log\\!\\left(\\frac{[\\mathrm{NH_3}]}{[\\mathrm{NH_4^+}]}\\right)\r\n\\]\r\n\r\n\\[\r\n10^{0.3447} = 2.212 = \\frac{[\\mathrm{NH_3}]}{[\\mathrm{NH_4^+}]}\r\n\\]\r\n\r\n\\[\r\n[\\mathrm{NH_3}] = 2.212\\, [\\mathrm{NH_4^+}]\r\n\\]\r\n\r\n\\[\r\n2.212[\\mathrm{NH_4^+}] + [\\mathrm{NH_4^+}] = 0.8\r\n\\]\r\n\r\n\\[\r\n3.212[\\mathrm{NH_4^+}] = 0.8\r\n\\]\r\n\r\n\\[\r\n[\\mathrm{NH_4^+}] = 0.2491\\ \\text{M}\r\n\\qquad\r\n[\\mathrm{NH_3}] = 0.5509\\ \\text{M}\r\n\\]\r\n\r\n\\[\r\n\\text{moles of } \\mathrm{NH_4^+}\r\n= 1.5\\,\\text{L} \\times 0.2491\\,\\frac{\\text{mol}}{\\text{L}}\r\n= 0.37365\\,\\text{mol}\r\n\\]\r\n\r\n\\[\r\n\\text{Since } (\\mathrm{NH_4})_2\\mathrm{SO_4} \\text{ provides 2 mol } \\mathrm{NH_4^+}\r\n\\text{ per mol salt:}\r\n\\]\r\n\r\n\\[\r\n\\text{moles } (\\mathrm{NH_4})_2\\mathrm{SO_4}\r\n= \\frac{0.37365}{2}\r\n\\]\r\n\r\n\\[\r\n\\text{mass } (\\mathrm{NH_4})_2\\mathrm{SO_4}\r\n= 132.136\\,\\frac{\\text{g}}{\\text{mol}}\r\n\\times \\frac{0.37365}{2}\r\n= \\textcolor{blue}{24.69\\,\\text{g}}\r\n\\]\r\n\r\n\\[\r\n1\\,\\text{L aqueous NH}_3 \\text{ weighs } 900\\,\\text{g}\r\n\\]\r\n\r\n\\[\r\n0.28 \\times 900\\,\\text{g} = 252\\,\\text{g NH}_3\r\n\\]\r\n\r\n\\[\r\n\\text{moles NH}_3\r\n= \\frac{252\\,\\text{g}}{17.031\\,\\text{g\/mol}}\r\n= 14.797\\,\\text{mol\/L}\r\n\\]\r\n\r\n\\[\r\nV_{\\mathrm{NH_3}}\r\n= \\frac{1.5\\,\\text{L} \\times 0.5509\\,\\text{mol\/L}}\r\n{14.797\\,\\text{mol\/L}}\r\n= 0.0558\\,\\text{L}\r\n= \\textcolor{blue}{55.8\\,\\text{mL}}\r\n\\]\r\n\r\n(c) An acetic acid solution has a pH of 2.78. What concentration of the acetic acid (in molar) was added\r\nto make up the solution? Take K a for acetic acid to be 1.75 x 10 -5 . Acetic acid has the formula CH 3 CO 2 H.\r\n\r\n\\[\r\n\\ce{HA <=> H+ + A-}\r\n\\]\r\n\r\n\\[\r\nK_a = \\frac{[\\ce{H+}][\\ce{A-}]}{[\\ce{HA}]}\r\n\\]\r\n\r\n\\[\r\n\\begin{array}{c|ccc}\r\n& \\ce{HA} & \\ce{H+} & \\ce{A-} \\\\ \\hline\r\n\\text{Initial} & C_0 & 0 & 0 \\\\\r\n\\text{Change} & -x & +x & +x \\\\\r\n\\text{Equilibrium} & C_0 - x & x & x\r\n\\end{array}\r\n\\]\r\n\r\n\\[\r\n\\text{Let } x = [\\ce{H+}] = [\\ce{A-}] \\text{ at equilibrium.}\r\n\\]\r\n\r\n\\[\r\n\\frac{x^{2}}{C_{0} - x} = K_a\r\n\\]\r\n\r\n\\[\r\n\\frac{x^{2}}{K_a} + x = C_{0}\r\n\\]\r\n\r\n\\[\r\nx = 10^{-2.78} = 0.0016596\r\n\\]\r\n\r\n\\[\r\nC_{0} = 0.159\\,\\text{M} = \\text{[acetic acid added]}\r\n\\]\r\n\r\n(d) Estimate the pH of a solution that contains 0.5 M nickel sulfate and solid NiO. (Look up the necessary\r\ndata in the Chemistry Review Part II notes.)\r\n\r\n\\[\r\n\\ce{NiO + H2O <=> Ni^{2+} + 2OH^-}\r\n\\]\r\n\r\n\\[\r\nK_\\mathrm{sp} = 1.6 \\times 10^{-16}\r\n\\]\r\n\r\n\\[\r\nK_\\mathrm{sp} = [\\ce{Ni^{2+}}][\\ce{OH^-}]^{2}\r\n= 0.5\\, [\\ce{OH^-}]^{2}\r\n= 1.6 \\times 10^{-16}\r\n\\]\r\n\r\n\\[\r\n[\\ce{OH^-}]\r\n= \\sqrt{\\frac{1.6 \\times 10^{-16}}{0.5}}\r\n= 1.7889 \\times 10^{-8}\\,\\text{M}\r\n\\]\r\n\r\n\\[\r\n[\\ce{H^+}]\r\n= \\frac{1.0 \\times 10^{-14}}{1.7889 \\times 10^{-8}}\r\n= 5.5902 \\times 10^{-7}\\,\\text{M}\r\n\\]\r\n\r\n\\[\r\n\\text{pH}\r\n= -\\log_{10}(5.5902 \\times 10^{-7})\r\n= \\textcolor{blue}{6.25}\r\n\\]\r\n\r\n7. The solubility of many metal oxides is enhanced by complexation. Ammonia is an example of a\r\ncomplexing agent that finds some use in hydrometallurgy. Nickel leaching from various minerals is\r\npracticed using ammonia as a complexing agent.\r\n\r\n(i) What is the solubility of Ni(OH) 2 in water alone? (Express your answer in g Ni\/L.)\r\n\r\n\\[\r\n\\ce{Ni(OH)2 <=> Ni^{2+} + 2OH^-}\r\n\\]\r\n\r\n\\[\r\nK_\\mathrm{sp} = [\\ce{Ni^{2+}}][\\ce{OH^-}]^{2}\r\n\\]\r\n\r\n\\[\r\n\\text{Let } x = [\\ce{Ni^{2+}}]\r\n\\quad\\Rightarrow\\quad\r\n[\\ce{OH^-}] = 2x\r\n\\]\r\n\r\n\\[\r\nK_\\mathrm{sp} = x(2x)^{2}\r\n= 6 \\times 10^{-16}\r\n\\]\r\n\r\n\\[\r\nx = 5.313 \\times 10^{-6}\\,\\text{M}\r\n\\]\r\n\r\n\\[\r\n[\\ce{Ni^{2+}}]\r\n= 5.313 \\times 10^{-6}\\,\\frac{\\text{mol}}{\\text{L}}\r\n\\times 58.6934\\,\\frac{\\text{g}}{\\text{mol}}\r\n= \\textcolor{blue}{3.12 \\times 10^{-4}\\,\\frac{\\text{g}}{\\text{L}}}\r\n\\]\r\n\r\n(ii) The reaction below depicts leaching of Ni(OH)2<\/sub> in ammonia solution to form [Ni(NH3<\/sub>)5<\/sub>]+2<\/sup> :\r\n\r\n\\[\r\n\\ce{Ni(OH)2 + 5NH3 <=> [Ni(NH3)5]^{2+} + 2OH^-}\r\n\\]\r\n\r\n\\[\r\n\\text{Determine the equilibrium constant for the reaction.}\r\n\\]\r\n\r\n\\[\r\n\\ce{Ni(OH)2 <=> Ni^{2+} + 2OH^-}\r\n\\qquad\r\nK_\\mathrm{sp}\r\n\\]\r\n\r\n\\[\r\n\\ce{Ni^{2+} + 5NH3 <=> [Ni(NH3)5]^{2+}}\r\n\\qquad\r\n\\beta_5\r\n\\]\r\n\r\n\\[\r\n\\text{Overall: } K = K_\\mathrm{sp}\\,\\beta_5\r\n\\]\r\n\r\n\\[\r\n\\beta_5 = 7.762 \\times 10^{7}\r\n\\]\r\n\r\n% if using stepwise constants: K_n = 10^{\\log K_n}\r\n\\[\r\nK = 6.762 \\times 10^{7} \\times 6 \\times 10^{-16}\r\n= \\textcolor{blue}{4.6575 \\times 10^{-8}}\r\n\\]\r\n\r\n(iii) What is the solubility of Ni(OH)2<\/sub> in 1 M aqueous ammonia solution (in g Ni\/L)? Assume that only the\r\n[Ni(NH3<\/sub>)5<\/sub>]+2<\/sup> complex forms. (This is a significant over-simplification, but it will serve to illustrate the\r\npoint*.)\r\n\r\nLet x = [Ni+2<\/sup>] at equilibrium (as the Ni +2<\/sup>-NH3<\/sub> complex); other concentrations then are as shown below:\r\n\r\n\\[\r\n\\ce{Ni(OH)2 + 5NH3 <=> [Ni(NH3)5]^{2+} + 2OH^-}\r\n\\]\r\n\r\n\\[\r\n\\text{If } x = [\\ce{[Ni(NH3)5]^{2+}}]_{\\text{eq}},\r\n\\text{ then the equilibrium concentrations are:}\r\n\\]\r\n\r\n\\[\r\n\\begin{array}{c|ccc}\r\n& \\ce{NH3} & \\ce{[Ni(NH3)5]^{2+}} & \\ce{OH^-} \\\\ \\hline\r\n\\text{Initial} & 1 & 0 & 0 \\\\\r\n\\text{Change} & -5x & +x & +2x \\\\\r\n\\text{Equilibrium} & 1 - 5x & x & 2x\r\n\\end{array}\r\n\\]\r\n\r\nSolving for x is best done with approximate methods, such as used by Solver in Excel.\r\n\r\n\\[\r\nx = 0.0022246\\,\\text{M}\r\n\\]\r\n\r\n\\[\r\n[\\ce{Ni^{2+}}]\r\n= 0.0022246\\,\\frac{\\text{mol}}{\\text{L}}\r\n\\times 58.6934\\,\\frac{\\text{g}}{\\text{mol}}\r\n= \\textcolor{blue}{0.131\\,\\text{g\\,L^{-1}\\,(as Ni^{2+})}}\r\n\\]\r\n\r\nThis is a great improvement over solubility of Ni(OH)2<\/sub> in water alone, but still quite low.\r\n\r\n(iv) Add a suitable multiple of the ammonia base dissociation reaction to the reaction in part (ii) to\r\nobtain a reaction that converts all the OH-<\/sup> to NH3<\/sub> . (Just enough NH4<\/sub>+<\/sup> to react with all the OH-<\/sup>.) Determine\r\nthe equilibrium constant for the reaction.\r\n\r\n\\[\r\n\\ce{Ni(OH)2 + 5NH3 <=> [Ni(NH3)5]^{2+} + 2OH^-}\r\n\\]\r\n\r\n\\[\r\n\\text{To consume the } \\ce{OH^-}:\r\n\\qquad\r\n\\ce{2NH4+ + 2OH^- <=> 2NH3 + 2H2O}\r\n\\]\r\n\r\n\\[\r\nK = \\frac{1}{K_b^2}\r\n= \\frac{K_a^2}{K_w^2}\r\n\\]\r\n\r\n\\[\r\n\\ce{Ni(OH)2 <=> Ni^{2+} + 2OH^-}\r\n\\qquad\r\nK_\\mathrm{sp}\r\n\\]\r\n\r\n\\[\r\n\\ce{Ni^{2+} + 5NH3 <=> [Ni(NH3)5]^{2+}}\r\n\\qquad\r\n\\beta_5\r\n\\]\r\n\r\n\\[\r\n\\text{Adding the reactions:}\r\n\\]\r\n\r\n\\[\r\n\\ce{Ni(OH)2 + 2NH4+ + 3NH3 <=> [Ni(NH3)5]^{2+} + 2H2O}\r\n\\]\r\n\r\n\\[\r\nK = K_\\mathrm{overall}\r\n= K_\\mathrm{sp}\\,\\beta_5\r\n\\left(\\frac{K_a^2}{K_w^2}\\right)\r\n\\]\r\n\r\n\\[\r\nK = 4.657 \\times 10^{-8}\r\n\\times \\left(5.75 \\times 10^{-10}\\right)^2\r\n\\div \\left(1.0 \\times 10^{-14}\\right)^2\r\n\\]\r\n\r\n\\[\r\nK = \\textcolor{blue}{154.0}\r\n\\]\r\n\r\n(v) What is the solubility of Ni(OH) 2 in an ammonia solution with a total ammonia concentration of 1 M\r\n([NH 3 ] + [NH 4 + ] = 1) where [NH 4 + ]\/[NH 3 ] = 2\/3 (as per the reaction)?\r\n\r\n\\[\r\n[\\ce{NH3}]_0 = 0.6\\,\\text{M},\r\n\\qquad\r\n[\\ce{NH4+}]_0 = 0.4\\,\\text{M}\r\n\\]\r\n\r\n\\[\r\n\\ce{Ni(OH)2 + 2NH4+ + 3NH3 <=> [Ni(NH3)5]^{2+} + 2H2O}\r\n\\]\r\n\r\n\\[\r\n\\text{Let } x = [\\ce{Ni^{2+}}]_{\\text{eq}}\r\n\\]\r\n\r\n\\[\r\n\\begin{array}{c|ccc}\r\n& \\ce{NH4+} & \\ce{NH3} & \\ce{[Ni(NH3)5]^{2+}} \\\\ \\hline\r\n\\text{Initial} & 0.4 & 0.6 & 0 \\\\\r\n\\text{Change} & -2x & -3x & +x \\\\\r\n\\text{Equilibrium} & 0.4 - 2x & 0.6 - 3x & x\r\n\\end{array}\r\n\\]\r\n\r\n\\[\r\nx = 0.10823\\,\\text{M}\r\n\\]\r\n\r\n\\[\r\n[\\ce{Ni^{2+}}]\r\n= 0.10823\\,\\frac{\\text{mol}}{\\text{L}}\r\n\\times 58.6934\\,\\frac{\\text{g}}{\\text{mol}}\r\n= \\textcolor{blue}{6.35\\,\\frac{\\text{g}}{\\text{L}}}\r\n\\]\r\n\r\nThis is greater than Ni(OH)2<\/sub> solubility when ammonia alone is used.\r\n\r\nThe ratio of 2:3 ratio of [NH4<\/sub>+<\/sup>]:[NH3<\/sub>] gives the maximum solubility of Ni(OH)2 <\/sub>in ammonia-ammonium solution. Just\r\nenough NH4<\/sub>+<\/sup> is used in the reaction to consume all the OH-<\/sup> that is released. If there is more NH4<\/sub>+<\/sup> then there is less\r\nNH3<\/sub> to complex Ni+2<\/sup> . More NH3<\/sub> and there is not enough NH4<\/sub>+<\/sup> to react with the OH-<\/sup> released. Either way the\r\nsolubility of Ni(OH)2<\/sub> declines.\r\n\r\n*There is actually much more at play here. There is also a [Ni(NH3<\/sub>)6<\/sub>]+2<\/sup> complex, though its formation <\/em>constant is relatively small. However, each [Ni(NH3<\/sub>)n<\/sub>]+2<\/sup> complex will contribute to the total nickel <\/em>dissolved. The approximation that the only complex contributing to dissolution of Ni(OH)2<\/sub> is the <\/em>[Ni(NH3<\/sub>)5]+2<\/sup> is imperfect. However, it does demonstrate what is going on, and gives a decent first-order <\/em>estimate of the effect. Although it takes a bit more effort to get a more accurate picture, the <\/em>methodology is not difficult.<\/em>","rendered":"

1. In any process chemical industry, including hydrometallurgy, it is important to gain an understanding of the periodic table. This question begins to address that. Match the compound or element on the right to the definition or term on the left. There will be one unique compound\/element per term or definition. By process of elimination determine the most suitable match for each phrase.<\/p>\n

\\[
\n\\begin{array}{l|l}
\n\\text{Strong acid} & \\ce{H2SO4} \\\\
\n\\text{Most electronegative metal} & \\ce{Au} \\\\
\n\\text{Main group metal} & \\ce{Pb} \\\\
\n\\text{Strong oxidant} & \\ce{Cl2} \\\\
\n\\text{Most important industrial base} & \\ce{CaO} \\\\
\n\\text{Strong base} & \\ce{NaOH} \\\\
\n\\text{Strong reducing agent} & \\ce{Na} \\\\
\n\\text{Transition metal} & \\ce{Ti} \\\\
\n\\text{Strong electrolyte} & \\ce{NaCl} \\\\
\n\\text{Strongly sulphophilic metal} & \\ce{Fe} \\\\
\n\\text{Amphoteric ion} & \\ce{HSO3^-} \\\\
\n\\text{Weak acid} & \\ce{H2S} \\\\
\n\\text{Semi-metal} & \\ce{As} \\\\
\n\\text{Incorrect chemical formula} & \\ce{NaCO3} \\\\
\n\\end{array}
\n\\]<\/p>\n

2. Fill in the blanks:
\n(a) In an electron transfer reaction when an oxidizing agent reacts it gets reduced<\/span>.
\n(b) Reduction is gain<\/span> of electrons.
\n(c) A reducing agent loses<\/span> electrons.
\n(d) What is the term we use for -log 10 [H + ]? pH<\/span>
\n(e) A salt is a compound composed of cations<\/span> and anions<\/span>.<\/p>\n

3. Select the most comprehensive and correct definition for a base in aqueous solution. A base is a
\ncompound that
\n(a) has an ionizable OH–<\/sup> group and raises the pH.<\/p>\n

(b) has an ionizable OH–<\/sup> group and lowers the pH.<\/p>\n

(c) accepts a proton and raises the pH.<\/span><\/p>\n

(d) accepts a proton and lowers the pH.<\/p>\n

4. Which of the following compounds
\n(a) will act as an acid in water: CH4<\/sub> , CO2 <\/sub>, HCO3<\/sub>– <\/sup>, HNO3<\/sub><\/span> , NH3<\/sub>
\n(b) will acts as a base in water: HCO3<\/sub>–<\/sup> , SO4<\/sub>2-<\/sup> , NaCH3<\/sub>CO2<\/sub><\/span> , N2<\/sub> , C6<\/sub>H5<\/sub>OH (phenol)<\/p>\n

5. Indicate which of the following are soluble salts in water:
\nHCl, Ag2<\/sub>S, Na2<\/sub>S<\/span>, C12<\/sub>H22<\/sub>O11<\/sub> (sugar), S8<\/sub> , NH3<\/sub> , CuO , PbCl2<\/sub> , CaO , CO2<\/sub> , H2<\/sub> , Te<\/p>\n

Only Na2<\/sub>S. Note that HCl, C12<\/sub>H22<\/sub>O11<\/sub> and NH3<\/sub> are all soluble, but are not salts. S8<\/sub> , H2<\/sub> , CO2<\/sub> and Te are not salts
\nand not soluble. Ag2<\/sub>S, CuO, PbCl2<\/sub> and CaO are insoluble salts.<\/p>\n

6. Perform the required calculations for the following aqueous solutions:<\/p>\n

(a) Calculate the pH of a solution made up of 5.00 mL of 1.2 M HCl diluted to precisely 250 mL.<\/p>\n

\\[
\n\\frac{5 \\cdot 1.2\\,\\text{M}\\,\\mathrm{H^+}}{250}
\n= 0.024\\,\\text{M}\\,\\mathrm{H^+}
\n\\]<\/p>\n

(HCl is a strong acid; fully dissociates.)<\/p>\n

\\[
\n\\text{pH} = -\\log_{10}(0.024) = \\textcolor{blue}{1.62}
\n\\]<\/p>\n

(b) A solution contains a total ammonia concentration of 0.8 M (the sum of ammonia and ammonium concentrations). The volume is 1.5 L and the pH is 9.6. What volume (mL) of aqueous ammonia (density = 0.90 g\/mL and 28% by mass, i.e. 280 g NH3<\/sub> \/kg of solution) and mass of ammonium sulfate (in g) were added to make up the solution? Use Kb<\/sub> for ammonia = 1.8 x 10-5<\/sup>.<\/p>\n

\\[
\n[\\mathrm{NH_3}] + [\\mathrm{NH_4^+}] = 0.8
\n\\]<\/p>\n

\\[
\n\\text{pH} = \\text{p}K_a + \\log\\!\\left(\\frac{[\\mathrm{NH_3}]}{[\\mathrm{NH_4^+}]}\\right) = 9.6
\n\\]<\/p>\n

\\[
\nK_a = \\frac{1 \\times 10^{-14}}{1.8 \\times 10^{-5}}
\n= 5.556 \\times 10^{-10}
\n\\]<\/p>\n

\\[
\n\\text{p}K_a = -\\log_{10}(K_a) = 9.2553
\n\\]<\/p>\n

\\[
\n9.6 = 9.2553 + \\log\\!\\left(\\frac{[\\mathrm{NH_3}]}{[\\mathrm{NH_4^+}]}\\right)
\n\\]<\/p>\n

\\[
\n10^{0.3447} = 2.212 = \\frac{[\\mathrm{NH_3}]}{[\\mathrm{NH_4^+}]}
\n\\]<\/p>\n

\\[
\n[\\mathrm{NH_3}] = 2.212\\, [\\mathrm{NH_4^+}]
\n\\]<\/p>\n

\\[
\n2.212[\\mathrm{NH_4^+}] + [\\mathrm{NH_4^+}] = 0.8
\n\\]<\/p>\n

\\[
\n3.212[\\mathrm{NH_4^+}] = 0.8
\n\\]<\/p>\n

\\[
\n[\\mathrm{NH_4^+}] = 0.2491\\ \\text{M}
\n\\qquad
\n[\\mathrm{NH_3}] = 0.5509\\ \\text{M}
\n\\]<\/p>\n

\\[
\n\\text{moles of } \\mathrm{NH_4^+}
\n= 1.5\\,\\text{L} \\times 0.2491\\,\\frac{\\text{mol}}{\\text{L}}
\n= 0.37365\\,\\text{mol}
\n\\]<\/p>\n

\\[
\n\\text{Since } (\\mathrm{NH_4})_2\\mathrm{SO_4} \\text{ provides 2 mol } \\mathrm{NH_4^+}
\n\\text{ per mol salt:}
\n\\]<\/p>\n

\\[
\n\\text{moles } (\\mathrm{NH_4})_2\\mathrm{SO_4}
\n= \\frac{0.37365}{2}
\n\\]<\/p>\n

\\[
\n\\text{mass } (\\mathrm{NH_4})_2\\mathrm{SO_4}
\n= 132.136\\,\\frac{\\text{g}}{\\text{mol}}
\n\\times \\frac{0.37365}{2}
\n= \\textcolor{blue}{24.69\\,\\text{g}}
\n\\]<\/p>\n

\\[
\n1\\,\\text{L aqueous NH}_3 \\text{ weighs } 900\\,\\text{g}
\n\\]<\/p>\n

\\[
\n0.28 \\times 900\\,\\text{g} = 252\\,\\text{g NH}_3
\n\\]<\/p>\n

\\[
\n\\text{moles NH}_3
\n= \\frac{252\\,\\text{g}}{17.031\\,\\text{g\/mol}}
\n= 14.797\\,\\text{mol\/L}
\n\\]<\/p>\n

\\[
\nV_{\\mathrm{NH_3}}
\n= \\frac{1.5\\,\\text{L} \\times 0.5509\\,\\text{mol\/L}}
\n{14.797\\,\\text{mol\/L}}
\n= 0.0558\\,\\text{L}
\n= \\textcolor{blue}{55.8\\,\\text{mL}}
\n\\]<\/p>\n

(c) An acetic acid solution has a pH of 2.78. What concentration of the acetic acid (in molar) was added
\nto make up the solution? Take K a for acetic acid to be 1.75 x 10 -5 . Acetic acid has the formula CH 3 CO 2 H.<\/p>\n

\\[
\n\\ce{HA <=> H+ + A-}
\n\\]<\/p>\n

\\[
\nK_a = \\frac{[\\ce{H+}][\\ce{A-}]}{[\\ce{HA}]}
\n\\]<\/p>\n

\\[
\n\\begin{array}{c|ccc}
\n& \\ce{HA} & \\ce{H+} & \\ce{A-} \\\\ \\hline
\n\\text{Initial} & C_0 & 0 & 0 \\\\
\n\\text{Change} & -x & +x & +x \\\\
\n\\text{Equilibrium} & C_0 – x & x & x
\n\\end{array}
\n\\]<\/p>\n

\\[
\n\\text{Let } x = [\\ce{H+}] = [\\ce{A-}] \\text{ at equilibrium.}
\n\\]<\/p>\n

\\[
\n\\frac{x^{2}}{C_{0} – x} = K_a
\n\\]<\/p>\n

\\[
\n\\frac{x^{2}}{K_a} + x = C_{0}
\n\\]<\/p>\n

\\[
\nx = 10^{-2.78} = 0.0016596
\n\\]<\/p>\n

\\[
\nC_{0} = 0.159\\,\\text{M} = \\text{[acetic acid added]}
\n\\]<\/p>\n

(d) Estimate the pH of a solution that contains 0.5 M nickel sulfate and solid NiO. (Look up the necessary
\ndata in the Chemistry Review Part II notes.)<\/p>\n

\\[
\n\\ce{NiO + H2O <=> Ni^{2+} + 2OH^-}
\n\\]<\/p>\n

\\[
\nK_\\mathrm{sp} = 1.6 \\times 10^{-16}
\n\\]<\/p>\n

\\[
\nK_\\mathrm{sp} = [\\ce{Ni^{2+}}][\\ce{OH^-}]^{2}
\n= 0.5\\, [\\ce{OH^-}]^{2}
\n= 1.6 \\times 10^{-16}
\n\\]<\/p>\n

\\[
\n[\\ce{OH^-}]
\n= \\sqrt{\\frac{1.6 \\times 10^{-16}}{0.5}}
\n= 1.7889 \\times 10^{-8}\\,\\text{M}
\n\\]<\/p>\n

\\[
\n[\\ce{H^+}]
\n= \\frac{1.0 \\times 10^{-14}}{1.7889 \\times 10^{-8}}
\n= 5.5902 \\times 10^{-7}\\,\\text{M}
\n\\]<\/p>\n

\\[
\n\\text{pH}
\n= -\\log_{10}(5.5902 \\times 10^{-7})
\n= \\textcolor{blue}{6.25}
\n\\]<\/p>\n

7. The solubility of many metal oxides is enhanced by complexation. Ammonia is an example of a
\ncomplexing agent that finds some use in hydrometallurgy. Nickel leaching from various minerals is
\npracticed using ammonia as a complexing agent.<\/p>\n

(i) What is the solubility of Ni(OH) 2 in water alone? (Express your answer in g Ni\/L.)<\/p>\n

\\[
\n\\ce{Ni(OH)2 <=> Ni^{2+} + 2OH^-}
\n\\]<\/p>\n

\\[
\nK_\\mathrm{sp} = [\\ce{Ni^{2+}}][\\ce{OH^-}]^{2}
\n\\]<\/p>\n

\\[
\n\\text{Let } x = [\\ce{Ni^{2+}}]
\n\\quad\\Rightarrow\\quad
\n[\\ce{OH^-}] = 2x
\n\\]<\/p>\n

\\[
\nK_\\mathrm{sp} = x(2x)^{2}
\n= 6 \\times 10^{-16}
\n\\]<\/p>\n

\\[
\nx = 5.313 \\times 10^{-6}\\,\\text{M}
\n\\]<\/p>\n

\\[
\n[\\ce{Ni^{2+}}]
\n= 5.313 \\times 10^{-6}\\,\\frac{\\text{mol}}{\\text{L}}
\n\\times 58.6934\\,\\frac{\\text{g}}{\\text{mol}}
\n= \\textcolor{blue}{3.12 \\times 10^{-4}\\,\\frac{\\text{g}}{\\text{L}}}
\n\\]<\/p>\n

(ii) The reaction below depicts leaching of Ni(OH)2<\/sub> in ammonia solution to form [Ni(NH3<\/sub>)5<\/sub>]+2<\/sup> :<\/p>\n

\\[
\n\\ce{Ni(OH)2 + 5NH3 <=> [Ni(NH3)5]^{2+} + 2OH^-}
\n\\]<\/p>\n

\\[
\n\\text{Determine the equilibrium constant for the reaction.}
\n\\]<\/p>\n

\\[
\n\\ce{Ni(OH)2 <=> Ni^{2+} + 2OH^-}
\n\\qquad
\nK_\\mathrm{sp}
\n\\]<\/p>\n

\\[
\n\\ce{Ni^{2+} + 5NH3 <=> [Ni(NH3)5]^{2+}}
\n\\qquad
\n\\beta_5
\n\\]<\/p>\n

\\[
\n\\text{Overall: } K = K_\\mathrm{sp}\\,\\beta_5
\n\\]<\/p>\n

\\[
\n\\beta_5 = 7.762 \\times 10^{7}
\n\\]<\/p>\n

% if using stepwise constants: K_n = 10^{\\log K_n}
\n\\[
\nK = 6.762 \\times 10^{7} \\times 6 \\times 10^{-16}
\n= \\textcolor{blue}{4.6575 \\times 10^{-8}}
\n\\]<\/p>\n

(iii) What is the solubility of Ni(OH)2<\/sub> in 1 M aqueous ammonia solution (in g Ni\/L)? Assume that only the
\n[Ni(NH3<\/sub>)5<\/sub>]+2<\/sup> complex forms. (This is a significant over-simplification, but it will serve to illustrate the
\npoint*.)<\/p>\n

Let x = [Ni+2<\/sup>] at equilibrium (as the Ni +2<\/sup>-NH3<\/sub> complex); other concentrations then are as shown below:<\/p>\n

\\[
\n\\ce{Ni(OH)2 + 5NH3 <=> [Ni(NH3)5]^{2+} + 2OH^-}
\n\\]<\/p>\n

\\[
\n\\text{If } x = [\\ce{[Ni(NH3)5]^{2+}}]_{\\text{eq}},
\n\\text{ then the equilibrium concentrations are:}
\n\\]<\/p>\n

\\[
\n\\begin{array}{c|ccc}
\n& \\ce{NH3} & \\ce{[Ni(NH3)5]^{2+}} & \\ce{OH^-} \\\\ \\hline
\n\\text{Initial} & 1 & 0 & 0 \\\\
\n\\text{Change} & -5x & +x & +2x \\\\
\n\\text{Equilibrium} & 1 – 5x & x & 2x
\n\\end{array}
\n\\]<\/p>\n

Solving for x is best done with approximate methods, such as used by Solver in Excel.<\/p>\n

\\[
\nx = 0.0022246\\,\\text{M}
\n\\]<\/p>\n

\\[
\n[\\ce{Ni^{2+}}]
\n= 0.0022246\\,\\frac{\\text{mol}}{\\text{L}}
\n\\times 58.6934\\,\\frac{\\text{g}}{\\text{mol}}
\n= \\textcolor{blue}{0.131\\,\\text{g\\,L^{-1}\\,(as Ni^{2+})}}
\n\\]<\/p>\n

This is a great improvement over solubility of Ni(OH)2<\/sub> in water alone, but still quite low.<\/p>\n

(iv) Add a suitable multiple of the ammonia base dissociation reaction to the reaction in part (ii) to
\nobtain a reaction that converts all the OH–<\/sup> to NH3<\/sub> . (Just enough NH4<\/sub>+<\/sup> to react with all the OH–<\/sup>.) Determine
\nthe equilibrium constant for the reaction.<\/p>\n

\\[
\n\\ce{Ni(OH)2 + 5NH3 <=> [Ni(NH3)5]^{2+} + 2OH^-}
\n\\]<\/p>\n

\\[
\n\\text{To consume the } \\ce{OH^-}:
\n\\qquad
\n\\ce{2NH4+ + 2OH^- <=> 2NH3 + 2H2O}
\n\\]<\/p>\n

\\[
\nK = \\frac{1}{K_b^2}
\n= \\frac{K_a^2}{K_w^2}
\n\\]<\/p>\n

\\[
\n\\ce{Ni(OH)2 <=> Ni^{2+} + 2OH^-}
\n\\qquad
\nK_\\mathrm{sp}
\n\\]<\/p>\n

\\[
\n\\ce{Ni^{2+} + 5NH3 <=> [Ni(NH3)5]^{2+}}
\n\\qquad
\n\\beta_5
\n\\]<\/p>\n

\\[
\n\\text{Adding the reactions:}
\n\\]<\/p>\n

\\[
\n\\ce{Ni(OH)2 + 2NH4+ + 3NH3 <=> [Ni(NH3)5]^{2+} + 2H2O}
\n\\]<\/p>\n

\\[
\nK = K_\\mathrm{overall}
\n= K_\\mathrm{sp}\\,\\beta_5
\n\\left(\\frac{K_a^2}{K_w^2}\\right)
\n\\]<\/p>\n

\\[
\nK = 4.657 \\times 10^{-8}
\n\\times \\left(5.75 \\times 10^{-10}\\right)^2
\n\\div \\left(1.0 \\times 10^{-14}\\right)^2
\n\\]<\/p>\n

\\[
\nK = \\textcolor{blue}{154.0}
\n\\]<\/p>\n

(v) What is the solubility of Ni(OH) 2 in an ammonia solution with a total ammonia concentration of 1 M
\n([NH 3 ] + [NH 4 + ] = 1) where [NH 4 + ]\/[NH 3 ] = 2\/3 (as per the reaction)?<\/p>\n

\\[
\n[\\ce{NH3}]_0 = 0.6\\,\\text{M},
\n\\qquad
\n[\\ce{NH4+}]_0 = 0.4\\,\\text{M}
\n\\]<\/p>\n

\\[
\n\\ce{Ni(OH)2 + 2NH4+ + 3NH3 <=> [Ni(NH3)5]^{2+} + 2H2O}
\n\\]<\/p>\n

\\[
\n\\text{Let } x = [\\ce{Ni^{2+}}]_{\\text{eq}}
\n\\]<\/p>\n

\\[
\n\\begin{array}{c|ccc}
\n& \\ce{NH4+} & \\ce{NH3} & \\ce{[Ni(NH3)5]^{2+}} \\\\ \\hline
\n\\text{Initial} & 0.4 & 0.6 & 0 \\\\
\n\\text{Change} & -2x & -3x & +x \\\\
\n\\text{Equilibrium} & 0.4 – 2x & 0.6 – 3x & x
\n\\end{array}
\n\\]<\/p>\n

\\[
\nx = 0.10823\\,\\text{M}
\n\\]<\/p>\n

\\[
\n[\\ce{Ni^{2+}}]
\n= 0.10823\\,\\frac{\\text{mol}}{\\text{L}}
\n\\times 58.6934\\,\\frac{\\text{g}}{\\text{mol}}
\n= \\textcolor{blue}{6.35\\,\\frac{\\text{g}}{\\text{L}}}
\n\\]<\/p>\n

This is greater than Ni(OH)2<\/sub> solubility when ammonia alone is used.<\/p>\n

The ratio of 2:3 ratio of [NH4<\/sub>+<\/sup>]:[NH3<\/sub>] gives the maximum solubility of Ni(OH)2 <\/sub>in ammonia-ammonium solution. Just
\nenough NH4<\/sub>+<\/sup> is used in the reaction to consume all the OH–<\/sup> that is released. If there is more NH4<\/sub>+<\/sup> then there is less
\nNH3<\/sub> to complex Ni+2<\/sup> . More NH3<\/sub> and there is not enough NH4<\/sub>+<\/sup> to react with the OH–<\/sup> released. Either way the
\nsolubility of Ni(OH)2<\/sub> declines.<\/p>\n

*There is actually much more at play here. There is also a [Ni(NH3<\/sub>)6<\/sub>]+2<\/sup> complex, though its formation <\/em>constant is relatively small. However, each [Ni(NH3<\/sub>)n<\/sub>]+2<\/sup> complex will contribute to the total nickel <\/em>dissolved. The approximation that the only complex contributing to dissolution of Ni(OH)2<\/sub> is the <\/em>[Ni(NH3<\/sub>)5]+2<\/sup> is imperfect. However, it does demonstrate what is going on, and gives a decent first-order <\/em>estimate of the effect. Although it takes a bit more effort to get a more accurate picture, the <\/em>methodology is not difficult.<\/em><\/p>\n","protected":false},"author":2529,"menu_order":10,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-3761","chapter","type-chapter","status-publish","hentry"],"part":1126,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters\/3761","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/users\/2529"}],"version-history":[{"count":10,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters\/3761\/revisions"}],"predecessor-version":[{"id":3886,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters\/3761\/revisions\/3886"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/parts\/1126"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapters\/3761\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/media?parent=3761"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/pressbooks\/v2\/chapter-type?post=3761"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/contributor?post=3761"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/hydrometallurgy\/wp-json\/wp\/v2\/license?post=3761"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}