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# Download 1 Ag PO 7.5 10 1.79 10 418.57 mol x gL x M g

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Chapter 18 Practice Exercises 18.1 Ba3(PO4)2(s) 3Ba2+(aq) + 2PO43+(aq) Ksp = [Ba2+]3[PO43+]2 18.2 (a) 18.3 TlI(s) Tl+(aq) + I–(aq) + Ksp = [Tl ][I–] 1 mol TlI –5 mol TlI = 5.9 × 10–3 g = 1.78 × 10 mol TlI 331.3 g TlI Ksp = [Ca2+][C2O42–] (b) Ksp = [Ag+]2[SO42–] 1.78 ×10−5 mol = 1.78 × 10–5 M 1L Ksp = (1.8 × 10–5)(1.8 × 10–5) = 3.2 × 10–10 [Tl+] = [I–] = 18.4 ( K sp = 2.15 × 10−3 18.5 Ksp = Pb 2+ F− Pb2+(aq) + 2F–(aq) PbF2(s) ) ( 2 ( 2.15 × 10 )) −3 2 2 = 3.98 × 10–8 3Ag+(aq) + PO42–(aq) Ag3PO4(s) Ksp = [Ag+]3[PO42–] The molar solubility of the salt is: ( 7.5x10 −3 gL−1 ) 1mol Ag3PO4 = 1.79 x10 −5 M 418.57 g Ksp (3 x 1.79 x 10-5)3(1.79 x 10-5) = 2.78 x 10-18 18.6 (a) AgBr(s) Ag+(aq) + Br-(aq) Ksp = [Ag+][Br-] = 5.4 x 10-13 The molar solubility for AgBr, x, is equal to the molar concentration of either Ag+ or Br- in the solution since the stoichiometry is 1:1. x2 = 5.4 x 10-13 x = 7.3 x 10-7 M (b) Ag2CO3(s) 2Ag+(aq) + CO32-(aq) The molar solubility of Ag2CO3, x, is equal to the molar concentration of CO32- in solution. Ksp = [Ag+]2[CO32-] = (2x)2(x) = 8.5 x 10-12 x = 1.3 x 10-4 M 181 Chapter 18 18.7 3Ca2+(aq) + 2PO43-(aq) Ca3(PO4)2(s) Let the molar solubility equal x. Ksp = [Ca2+]3[PO43-]2 = (3x)3(2x)2 = 2.0 x 10-20 x = 4.5 x 10-5 M 18.8 Ag+(aq) + I–(aq) AgI(s) Ksp = [Ag+][I–] = 8.5 × 10–17 The initial concentration of I– is 2 × 0.20 M from the CaI2. [Ag+] [I–] I – 0.40 C +x +x E +x 0.40 + x Substituting the above values for equilibrium concentrations into the expression for Ksp gives: Ksp = 8.5 × 10–17 = [Ag+][I–] = (x)(0.40 + x) We know that the value of Ksp is very small, and it suggests the simplifying assumption that (0.40 + x) ≈ 0.40: Hence, 8.5 × 10–17 = (0.40)x, and x = 2.125 × 10–16. The assumption that (0.40 + x) ≈ 0.40 is seen to be valid indeed. Thus the molar concentration of AgI in a 0.20 M CaI2 solution will be 2.1 × 10–16 M . In pure water, Ksp = 8.5 × 10–17 = [Ag+][I–] = (x)(x) x = [AgI(aq)] = 9.2 × 10–9 M (much more soluble) 18.9 Fe(OH)3(s) I C E Fe3+(aq) + 3OH–(aq) [Fe3+] – +x +x Ksp = [Fe3+][OH–]3 = 2.8 × 10–39 [OH–] 0.050 + 3x 0.050 + 3x Substituting the above values for equilibrium concentrations into the expression for Ksp gives: Ksp = 2.8 × 10–39 = [Fe3+][OH–]3 = (x)[0.050 + 3x]3 We try to simplify by making the approximation that (0.050 + 3x) ≈ 0.050: 2.8 × 10–39 = (x)(0.050)3 or x = 2.24 × 10–35 Clearly the assumption that (0.050 + 3x) ≈ 0.050 is justified. Thus 2.24 × 10–35 M of Fe(OH)3 will dissolve in a 0.050 M sodium hydroxide solution. 18.10 The expression for Ksp is Ksp = [Ca2+][SO42–] = 2.4 × 10–5 and the ion product for this solution would be: [Ca2+][SO42–] = (2.5 × 10–3)(3.0 × 10–2) = 7.5 × 10–5 Since the ion product is larger than the value of Ksp, a precipitate will form. 182 Chapter 18 18.11 The solubility product constant is Ksp = [Ag+]2[CrO42–] = 1.2 × 10–12 and the ion product for this solution would be: [Ag+]2[CrO42–] = (4.8 × 10–5)2(3.4 × 10–4) = 7.8 × 10–13 Since the ion product is smaller than the value of Ksp, no precipitate will form. 18.12 We expect PbSO4(s) since nitrates are soluble. Because two solutions are to be mixed together, there will be a dilution of the concentrations of the various ions, and the diluted ion concentrations must be used. In general, on dilution, the following relationship is found for the concentrations of the initial solution (Mi) and the concentration of the final solution (Mf): MiVi = MfVf Thus the final or diluted concentrations are: Pb 2+ = 1.0 × 10−3 M 100.0 mL = 5.0 × 10−4 M 200.0 mL ( ) SO 4 2− = 2.0 × 10 −3 M 100.0 mL = 1.0 × 10−3 M 200.0 mL ( ) The value of the ion product for the final (diluted) solution is: [Pb2+][SO42–] = (5.0 × 10–4)(1.0 × 10–3) = 5.0 × 10–7 Since this is larger than the value of Ksp (2.53 × 10–8), a precipitate of PbSO4 is expected. 18.13 We expect a precipitate of PbCl2 since nitrates are soluble. We proceed as in Practice Exercise 12 MiVi = MfVf Pb 2+ = ( 0.10 M ) 50.0 mL = 0.071 M 70.0 mL Cl− = ( 0.040 M ) 20.0 mL = 0.011 M 70.0 mL The value of the ion product for such a solution would be: [Pb2+][Cl–]2 = (7.1 × 10–2)(1.1 × 10–2)2 = 8.6 × 10–6 Since the ion product is smaller than Ksp (1.7 × 10–5), a precipitate of PbCl2 is not expected. 18.14 Follow the exact procedure outlined in Example 18.8. Ksp = [Ca2+][SO42–]= 4.9 × 10–5 Ksp = [Ba2+][SO42–] = 1.1 × 10–10 CaSO4 is more soluble and will precipitate when: [SO42–] = K sp = 4.9 × 10 0.25 5 [Ca 2 + ] BaSO4 will precipitate when: K sp 1.1 × 10 [SO42 − ] = = 2+ 0.05 [Ba ] = 2.0 × 10 4 10 = 2.2 × 10 9 BaSO4 will precipitate and CaSO4 will not precipitate if [SO42–] > 2.2 × 10–9 and [SO42–] < 2.0 × 10–4. 183 Chapter 18 18.15 Follow the exact procedure outlined in Example 18.8. Ksp = [Ca2+][OH–]2 = 5.0 × 10–6 Ksp = [Mg2+][OH–]2 = 5.6 × 10–12 Ca(OH)2 is more soluble and will precipitate when: 1/2 K sp [OH ] = [Ca 2 + ] – 5.0 × 10 = 0.20 6 1/2 = 5.0 × 10 3 Mg(OH)2 will precipitate when: 1/2 K sp [OH ] = [Ba 2+ ] − 5.6 × 10 = 0.1 12 1/2 = 7.4 × 10 6 Mg(OH)2 will precipitate and Ca(OH)2 will not precipitate if 7.4 x 10-6 < [OH–] < 5.0 x 10-3. 18.16 CoS will precipitate if the H+ concentration is too low. Solving for Q and then comparing Q to Kspa, we can determine whether or not CoS will precipitate. K spa = Q= [Co 2+ ][H 2S] [H + ]2 [Co 2 + ][H 2S] + 2 = = 5 × 10−1 [0.005][0.10] −4 2 = 5 × 103 [H ] [3.16 × 10 ] Q > Kspa Since Q is greater than Kspa, then the reaction will move to reactants and CoS solid will form. 18.17 Consulting Table 18.2, we find that Fe2+ is much more soluble in acid than Hg2+. We want to make the H+ concentration large enough to prevent FeS from precipitating, but small enough that HgS does precipitate. First, we calculate the highest pH at which FeS will remain soluble, by using Kspa for FeS. (Recall that a saturated solution of H2S = 0.10 M.) K spa = [Fe 2+ ][H 2S] + 2 = [H ] + [H ] = 0.0013 M pH = –log [H+] = 2.9 [0.010][0.10] [H + ]2 = 6 × 102 Since Fe2+ is much more soluble in acid than Hg2+ we already know that this pH will precipitate HgS, but we can check it by using Kspa for HgS: K spa = [Hg 2+ ][H 2S] + 2 [H ] [H+] = 2.2 × 1014 M = [0.010][0.10] [H + ]2 = 2 × 10−32 (This concentration is impossibly high, but it tells us that this much acid would be required to dissolve HgS at these concentrations.) 184 Chapter 18 18.18 BaC2O4(s) Ba2+(aq) + C2O42–(aq) 2+ [Ba ] = 0.050 M 1.2 × 10–7 = (0.050)[C2O42–] [C2O42–] = 2.4 × 10–6 M Ksp = 1.2 × 10–7 = [Ba2+][C2O42–] H2C2O4 + H2O H3O+ + HC2O4– – H3O+ + C2O42– HC2O4 + H2O H2C2O4 + H2O 2H3O+ + C2O42– [H2C2O4] = 0.10 [C2O42–] = 2.4 × 10–6 M Ka1 = 6.0 × 10–2 Ka2 = 6.1 × 10–5 Ka = (6.1 × 10–5) × (6.0 × 10–2) = 3.7 × 10–6 2 Ka = 3.7 × 10–6 2 H O+ C O 2− H O+ 2.4 × 10−6 3 3 2 4 = = ( 0.10 ) [ H 2 C2 O 4 ] ( ) Since the amount of oxalate formed is so small, the concentration of oxalic acid is essentially unchanged. [H+] = 1.54 × 10–1 This is the minimum concentration of H+ that will prevent the formation of BaC2O4 precipitate. 18.19 Follow the exact procedure outlined in Example 18.10. Ksp = [Ca2+][CO32–]= 3.4 × 10–9 Ksp = [Ni2+][CO32–] = 1.4 × 10–7 NiCO3 is more soluble and will precipitate when: K sp 1.4 × 10 7 [CO32–] = = = 1.4 × 10 6 2+ 0.10 [Ni ] CaCO3 will precipitate when: K sp 3.4 × 10 9 [CO32− ] = = = 3.4 × 10 8 2+ 0.10 [Ca ] CaCO3 will precipitate and NiCO3 will not precipitate if [CO32–] > 3.4 × 10–8 and [CO32–] < 1.4 × 10–6. Now, using the equation in example 18.10 we get: 0.030 [H + ]2 = (2.4 × 10−17 ) [CO 2 − ] 3 NiCO3 will precipitate if: 0.030 −13 [H + ]2 = (2.4 × 10 −17 ) = 5.14 × 10 1.4 × 10 −6 [H+] = 7.17 × 10–7 pH = 6.14 CaCO3 will precipitate: 0.030 [H + ]2 = (2.4 × 10−17 ) = 2.12 × 10−11 −8 3.4 × 10 [H + ] = 4.6 × 10−6 pH = 5.34 So CaCO3 will precipitate and NiCO3 will not if the pH is maintained between pH = 5.34 and pH = 6.14 185 Chapter 18 18.20 Ag(NH3)2+(aq) + Cl–(aq) The overall equilibrium is AgCl(s) + 2NH3(aq) Ag(NH ) + Cl− 3 2 Kc = 2 NH3 In order to obtain a value for Kc for this reaction, we need to use the expressions for Ksp of AgCl(s) and the Kform of Ag(NH3)2+: K sp = Ag + Cl− = 1.8 × 10−10 Ag(NH ) + 3 2 K form = = 1.6 × 107 + Ag NH 2 3 Ag(NH ) + Cl− 3 2 = 2.9 × 10−3 K c = K sp × K form = 2 NH3 Now we may use an equilibrium table for the reaction in question: I C E [Ag(NH3)2+] – +x x [NH3] 0.10 – 2x 0.10 – 2x [Cl–] – +x x Substituting these values into the mass action expression gives: K c = 2.9 × 10−3 = (x)(x) (0.10 − 2x)2 (x) (0.10 − 2x) Solving for x we get, x = 4.9 × 10–3 M. The molar solubility of AgCl in 0.10 M NH3 is therefore 4.9 × 10–3 M. Take the square root of both sides to get 0.054 = In order to determine the solubility in pure water, we simply look at Ksp AgCl(s) Ag+ (aq) + Cl–(aq) Ksp = [Ag+][Cl–] = 1.8 × 10–10 At equilibrium; [Ag+] = [Cl–] = 1.3 × 10–5. Hence the molar solubility of AgCl in 0.10 M NH3 is about 380 times greater than in pure water. 186 Chapter 18 18.21 We will use the information gathered for the last problem. Specifically, AgCl(s) + 2NH3(aq) Ag(NH3)2+(aq) + Cl–(aq) Ag(NH ) + Cl− 3 2 = 2.9 × 10−3 Kc = 2 NH3 If we completely dissolve 0.20 mol of AgCl, the equilibrium [Cl–] and [Ag(NH3)2+] will be 0.20 M in a one liter container. The question asks, therefore, what amount of NH3 must be initially present so that the equilibrium concentration of Cl– is 0.20 M? [NH3] Z – 2x Z – 2x I C E K c = 2.9 × 10−3 = [Ag(NH3)2+] – +x x [Cl–] – +x x (x)(x) (Z − 2x)2 Take the square root of both sides to get; 0.054 = x 0.20 = Z − 2x Z − 0.40 We have substituted the known value of x. Solving for Z we get, Z = 4.1 M Consequently, we would need to add 4.1 moles of NH3 to a one liter container of 0.20 M AgCl in order to completely dissolve the AgCl. Review Questions 18.1 The ion product is the quantity which results from the mass action expression and it is a product of the ion concentrations. The ion product constant, also called the solubility product constant, is equal to the product of the ion concentrations for a saturated solution of a sparingly soluble substance. 3 18.2 Ba 2+ PO 3− 4 Kc = Ba 3 ( PO 4 ) 2 2 The denominator of the mass action expression is a constant since the concentration of substance within a pure solid is constant. Consequently, we define a new constant which is the product of Kc and the concentration of the pure solid. 3 K sp = Ba 2+ PO43− 2 18.3 The addition of a common ion to a saturated solution lowers the solubility of a sparingly soluble ionic salt. According to Le Châtelier’s principle, addition of an ion to a saturated solution will shift the equilibrium so as to absorb as much of the added ion as possible. This results in the precipitation of the sparingly soluble salt. In the case of AgCl, Ksp = 1.8 × 10–10, addition of NaCl to a saturated solution containing Ag+ and Cl– will result in the precipitation of AgCl(s). The common ion in this case is Cl–. It is present in the solution due to the NaCl as well as the AgCl. 18.4 When the value of the ion product is greater than Ksp, a precipitate will form. 187 Chapter 18 18.5 Ksp values do not give highly accurate compound solubilities for a number of reasons including: (a) Ksp values are determined at a specific temperature and unless you use an accurately controlled temperature bath, the measured solubility will not be precisely what was calculated. (b) Ksp values assume 100% dissociation into ions. This condition is a reasonable assumption for singly charged species but the assumption does not work well when working with charges greater than +1 or −1. Ion paring becomes a significant factor when ions charges are large and when concentrations are high. 18.6 Salts such as NaCl are strong electrolytes. We assume they dissociate 100%, just like strong acids and bases. Therefore, just like we do not need a Ka or Kb value from strong acids and bases, there is no need for an equilibrium constant for highly soluble salts. 18.7 The oxide ion reacts with the water to produce hydroxide ion: O2– + H2O 2OH– 18.8 Na2S(s) 18.9 (a) (b) 2Na+(aq) + HS–(aq) + OH–(aq) CoS(s) Co2+(aq) + HS–(aq) + OH–(aq) + CoS(s) + 2H (aq) 2+ Co (aq) + H2S(aq) Ksp = [CO2+][HS–][OH–] K spa Co2+ [ H S] 2 = 2 H+ 18.10 As the pH of an oxalic acid solution is decreased, the hydrogen ion concentration is increased, this shifts the equilibrium to the undissociated acid, thus increasing the H2C2O4 concentration. The opposite is true as the pH of the solution is increased. 18.11 By combining the three dissociation expressions for phosphoric acid, we come up with a single expression that relates the hydrogen ion concentration and the phosphate ion concentration: H3PO4(aq) 3H+(aq) + PO43–(aq) K = K a1 × K a2 × K a3 18.12 Halides of Ag+ are insoluble while those of Co2+ and Cu2+ are soluble. Therefore, Ag+ can be separated from the mixture using 1 M HCl. Sulfides of Co2+ are insoluble in base but soluble in acid while sulfides of Cu2+ are insoluble in acid. Therefore, the addition of S2− in acidic solution will precipitate the Cu2+ leaving the Co2+ in solution. To precipitate the Co2+ increase the pH to a value > 9. 18.13 MgBr2(aq) + 2 AgCl(s) MgCl2(aq) + 2AgBr(s) AgCl and AgBr are both insoluble compounds. However, AgBr is less soluble than AgCl. When solid AgCl is added to an aqueous solution of MgBr2, some of the AgCl dissociates. The Ag+ ion reacts with the aqueous Br– to form insoluble AgBr. This disrupts the AgCl/Ag+ equilibrium and additional AgCl dissociates. With sufficient stirring, and perhaps a little heat, all of the AgCl will dissolve and AgBr will precipitate. The result is a solution of MgCl2 and solid AgBr. 18.14 NH3(aq) + H2O NH4+(aq) + OH–(aq) The addition of NH4Cl to the mixture of Mg(OH)2 and water causes the Mg(OH)2 to dissolve because the NH4+ reacts with the OH– and shifts the equilibrium towards the dissolution of the solid. 18.15 AgCl(s) Ag+(aq) + Cl–(aq) 188 Chapter 18 Ag+(aq) + 2NH3(aq) Ag(NH3)2+(aq) Silver chloride is an insoluble solid. However, any Ag+ ions present react with added NH3 to form the Ag(NH3)2+ complex ion. According to Le Châtelier’s Principle, as NH3 is added to a solution containing Ag+ ions, the complex ion forms using up the Ag+ ions. This disrupts the equilibrium and forces AgCl to dissolve. Upon addition of HNO3, a strong acid, H+ reacts with NH3 to form NH4+. This disrupts the complex equilibria and causes the Ag(NH3)2+ to dissociate and form free Ag+ ions. Once again, the [Ag+] reaches a value sufficient in the presence of Cl– to precipitate AgCl in accordance with Le Châtelier’s Principle. 18.16 PbCl2(s) Pb2+(aq) + 2Cl–(aq) 2+ PbCl3–(aq) Pb (aq) + 3Cl–(aq) PbCl − 3 K form = Pb2+ Cl− 3 = mol PbCl − 3 volume 3 2+ − mol Pb volume mol Cl volume ( mol PbCl ) ( mol Pb )( mol Cl ) − 3 K form = ( volume ) × 3 2+ − 3 Notice that the above expression is the product of a ratio of mole amounts and a volume3 term. The constant Kform does not change on dilution, but the volume term is changed by dilution. This means that the ratio of moles term in the above expression must change on dilution, in order to hold the product constant. If the volume is doubled, the ratio of moles would have to become smaller by a factor of 8 (= 23) in order for the entire argument to have a constant value, i.e, in order for Kform to remain constant on dilution. This means that the concentrations of Pb2+ and Cl– must increase as the solution containing the complex ion is diluted. Eventually the ion product for PbCl2 will exceed the value of Ksp for PbCl2, resulting in its precipitation. 18.17 Any of the species that contains the conjugate base of a weak acid will be more soluble in acid. Thus, ZnS, Ca(OH)2, MgCO3 and PbF2 will increase in solubility in acid. ZnS(s) + 2H+(aq) Ca(OH)2(s) + 2H+(aq) MgCO3(s) + 2H+(aq) PbF2(s) + 2H+(aq) → → → → Zn2+(aq) + H2S(aq) 2+ Ca (aq) + 2H2O Mg2+(aq) + CO2(g) + H2O Zn2+(aq) + 2HF AgCl will increase in solubility if the base is NH3. AgCl is less soluble than Ag2O so it will not dissolve in other bases. AgOH is not favored in the reaction of silver salts with strong base. Rather, the reaction favors the formation of the oxide. AgCl(s) + 2NH3(aq) → Ag(NH3)2+(aq) + Cl-(aq) None of the compound’s solubility is independent of pH. Review Problems 18.18 (a) (b) CaF2(s) Ca2+ + 2F– Ag2CO3(s) 2Ag+ + CO32– Ksp = [Ca2+][F–]2 Ksp = [Ag+]2[CO32–] 189 Chapter 18 (c) (d) (e) (f) PbSO4(s) Pb2+ + SO42– Fe(OH)3(s) Fe3+ + 3OH– 2+ Pb + 2I– PbI2(s) Cu2+ + 2OH– Cu(OH)2(s) Ksp = [Pb2+][SO42–] Ksp = [Fe3+][OH–]3 Ksp = [Pb2+][I–]2 Ksp = [Cu2+][OH–]2 18.19 (a) (b) (c) (d) (e) (f) AgI(s) Ag+ + I– 3Ag+ + PO43– Ag3PO4(s) Pb2+ + CrO42– PbCrO4(s) Al(OH)3(s) Al3+ + 3OH– ZnCO3(s) Zn2+ + CO32– Zn2+ + 2OH– Zn(OH)2(s) Ksp = [Ag+][I–] Ksp = [Ag+]3[PO43–] Ksp = [Pb2+][CrO42–] Ksp = [Al3+][OH–]3 Ksp = [Zn2+][CO32–] Ksp = [Zn2+][OH–]2 18.20 Pb2+ + 2Cl– Ksp = [Pb2+][Cl–]2 PbCl2 2+ – At equilibrium [Pb ] = 0.016 M and [Cl ] = 0.032 M so Ksp = (0.016)(0.032)2 = 1.6 × 10–5 18.21 [Zn2+] = [C2O4] = 7.9 × 10–3 M Ksp = [Zn2+][C2O4] = (7.9 × 10–3)2 = 6.2× 10–5 18.22 1 mole BaSO 4 –5 mol BaSO 4 = ( 0.00245 g BaSO 4 ) = 1.05 × 10 mol 233.3906 g BaSO 4 [Ba2+] = [SO42–] = 1.05 × 10–5 M This is the molar solubility of BaSO4 Ksp = [Ba2+][SO42–] = (1.05 × 10–5)2 = 1.10 × 10–10 18.23 1 mol AgC2 H3O2 −3 mol AgC2 H3O 2 = ( 0.800 g AgC2 H3O 2 ) = 4.79 × 10 mol AgC2 H3O 2 166.9 g AgC H O 2 3 2 4.79 × 10−3 mol AgC2 H3O 2 = 4.79 × 10–2 M 0.100 L solution One mole of Ag+ and one mole of C2H3O2 will be produced for every mole of AgC2H3O2. Therefore, Ksp = [Ag+][C2H3O2–] = (4.79 × 10–2)2 = 2.29 × 10–3. [AgC2H3O2] = 18.24 BaF2 Ba2+(aq) + 2F–(aq) Ksp = [Ba2+][F–]-2 First find the concentration of the Ba2+ and F– that was in solution and then find the value for Ksp. Using the amount of BaF2 recovered; determine the number of moles of each ion, then find the concentration of each ion. 1 mol BaF2 –4 mole BaF2 = 0.132 g BaF2 = 7.53 × 10 mol BaF2 175.32 g BaF 2 −4 7.53 × 10 mol BaF2 1 mol Ba 2+ 1000 mL [Ba2+] = = 7.53 × 10–3 M Ba2+ 100 mL solution 1 mol BaF2 1 L 7.53 × 10−4 mol BaF 2 mol F− 1000 mL 2 [F–] = = 1.51 × 10–2 M F– 100 mL solution 1 mol BaF2 1 L Ksp = [Ba2+][F–] = (7.53 × 10–3)(1.51 × 10–2)2 = 1.7 × 10–6 190 Chapter 18 18.25 To solve this problem we need to realize that the concentration of the solution is equal to the number of moles of solid recovered divided by the volume of the solution, i.e., 0.649 g CaCrO 1 mole CaCrO 1000 mL [ CaCrO4 ] = 156 mL 4 156.1 g CaCrO4 1 L = 0.0267 M 4 2+ The equilibrium for this problem is CaCrO4(s) Ca + CrO42– 2+ 2– 2 –4 Ksp = [Ca ][CrO4 ] = (0.0267) = 7.13 × 10 18.26 Ag3PO4(s) 3Ag+ + PO43– Ksp = [Ag+]3[PO43–] –5 3 –5 Ksp = [3(1.8 × 10 )] [1.8 × 10 ] = 2.8 × 10–18 18.27 Ba3(PO4)2(s) 3Ba2+ + 2PO43– Ksp = [Ba2+]3[PO43–]2 –8 3 Ksp = [3(1.4 × 10 )] [2(1.4 × 10–8)]2 = 5.8 × 10–38 18.28 Compare the values of Ksp. The larger the value the greater the molar solubility. Ksp = 6.8 x 10-8 for MgCO3 Ksp = 1.5 x 10-10 for ZnCO3 Therefore, MgCO3 is more soluble in water. 18.29 Compare the values of Ksp. The larger the value the greater the molar solubility. Ksp = 1.1 x 10-10 for BaSO4 Ksp = 5.6 x 10-10 for SrCO3 Therefore, SrCO3 is more soluble in water. 18.30 Pb2+ + 2Br– PbBr2(s) I C E [Pb2+] – +x +x Ksp = [Pb2+][Br–]2 [Br–] – + 2x + 2x Ksp = (x)(2x)2 = 4x3 = 6.6 × 10–6, x = 18.31 Ag2CrO4(s) I C E [Ag+] – + 2x + 2x Zn(CN)2(s) 6.6 × 10−6 = 1.2 × 10−2 M 4 2Ag+ + CrO42– [CrO42–] – +x +x Ksp = (2x)2(x) = 4x3 = 1.2 × 10–12, x = 18.32 3 3 1.2 × 10−12 = 6.7 × 10−5 M 4 Zn2+(aq) + 2CN-(aq) For every mole of Zn2+ produced, 2 moles of CN- will be produced. Let x = [Zn2+] at equilibrium and [CN-] = 2x at equilibrium. Ksp = [Zn+2][CN–]2 = 3.0 × 10–16 = (x)(2x)2 = 4x3. Solving we find x = 4.2 × 10–6. Thus, the molar solubility of Zn(CN)2 is 4.2 × 10–6 moles/L. 18.33 For every mole of Pb2+ produced, 2 moles of F– will be produced. 191 Chapter 18 Let x = [Pb2+] at equilibrium and [F–] = 2x at equilibrium. Ksp = [ Pb2+][F–]2 = 3.3 × 10–8 = (x)(2x)2 = 4x3. Solving we find x = 2.0 × 10–3 M. Thus, the molar solubility of PbF2 is 2.1 × 10–3 moles/L. 18.34 To solve this problem, determine the molar solubility for each compound. Ksp = [Li+][F–] = x2 = 1.7 × 10–3 LiF: let x = [Li+] = [F–] –2 x = 4.1 × 10 moles/L = molar solubility of LiF. BaF2: let x = [Ba2+]; [F–]2 = 2x Ksp = [Ba2+][F–]2 = (x)(2x)2 = 1.7 × 10–6 3 –6 –3 4x = 1.7 × 10 , and x = 7.5 × 10 M = molar solubility of BaF2. Because the molar solubility of LiF is greater than the molar solubility of BaF2, LiF is more soluble. 18.35 To solve this problem, first determine the molar solubility for each compound. AgCN: let x = [Ag+] = [CN–] Ksp = [Ag+][CN–] = x2 = 6.0 × 10–17 –8 x = 7.7 × 10 moles/L = molar solubility of AgCN. The number of grams of AgCN that will dissolve in 100 mL is; 7.7 × 10−9 moles 133.9 g g = (100 mL) = 1.0 × 10−7 g 1 mole 1000 mL Zn(CN)2: let x = [Zn2+], [CN–] = 2x; Ksp = [Zn2+][CN–]2 = (x)(2x)2 = 3 × 10–16 3 –16 –6 4x = 3 × 10 , and x = 4.2 × 10 M = molar solubility of Zn(CN)2. The number of grams of Zn(CN)2 that will dissolve in 100 mL is; 4.2 × 10−6 moles 117.4 g g = (100 mL) = 4.9 × 10−5 g 1 mole 1000 mL More Zn(CN)2 will dissolve in 100 mL of water so it has the larger solubility. 18.36 First determine the molar solubility of the MX salt. Let x = [M+] = [X–], Ksp = [M+][X–] = (x)(x) = 3.2 × 10–10 –5 x = 1.8 × 10 M. This is the equilibrium concentration of the two ions. For the MX3 salt, let x = equilibrium concentration of M3+, [X–] = 3x. Ksp = [M+][X–]3 = (x)(3x)3 = 27x4. The value of x in this expression is the value determined in the first part of this problem. So, Ksp = (27)(1.8 × 10–5)4 = 2.8 × 10–18 18.37 First determine the molar solubility of the M2X3 salt. M2X3 (s) 2M3+ (aq) + 3X2–(aq) Ksp = [M3+]2[X2–]3 I C E [M3+] – + 2x + 2x [X2–] – + 3x + 3x Ksp = (2x)2(3x)3= 2.2 × 10–20 = 108x5 x = 4.6 × 10–5 M The molar solubility of this compound is 4.6 × 10–5 moles/L 192 Chapter 18 We want the molar solubility of the M2X compound to be twice the value just calculated or 9.2 × 10–5 moles/L. We need to solve the equilibrium expression: 2M+ (aq) + X2–(aq) M2X (s) I C E [M+] – + 2x + 2x Ksp = [M+]2[X2–] [X2–] – +x +x Ksp = (2x)2(x) = 4x3 So, Ksp = 4(9.2 × 10–5)3 = 3.1 × 10–12 where x = 9.2 × 10–5 M 18.38 CaSO4(s) Ca2+(aq) + SO42–(aq) Ksp = [Ca2+][SO42–] = 4.9 × 10–5 2+ 2– 2 Ksp = x = 4.9 × 10–5 and x = 7.0 × 10–3 M. let x = [Ca ] = [SO4 ] –3 The molar solubility of CaSO4 is 7.0 × 10 moles/L. 18.39 CaCO3(s) Ca2+ + CO32– Ksp = [Ca2+][CO32–] = 3.4 × 10–9 2– 2+ let x = [Ca ] = [CO3 ] x2 = 3.4 × 10–9; x = 5.8 × 10–5 M The solubility of CaCO3 is 5.8 × 10–5 M. So, 5.8 × 10–5 moles of CaCO3 dissolves in 1 L of H2O. If 100 mL of water are available, 5.8 × 10–6 moles of CaCO3 will dissolve. Converting this to g/100 mL we multiply by the molar mass of 100 g/mole and determine that 5.8 × 10–4 g will dissolve in 100 mL of water. 18.40 BaSO3(s) Ba2+ + SO32– Ksp = [Ba2+][SO32–] –6 –7 Ksp = (0.10)(8.0 × 10 ) = 8.0 × 10 In this problem, all of the Ba2+ comes from the BaCl2. 18.41 Ag2CrO4(s) 2Ag+ + CrO42– Ksp = [Ag+]2[CrO42–] –6 2 –13 Ksp = (0.10)(1.7 × 10 ) = 2.9 × 10 In this problem, all of the Ba2+ comes from the BaCl2. 18.42 (a) Cu+(aq) + Cl–(aq) CuCl(s) I C E [Cu+] – +x +x Ksp = x2 = 1.7 × 10–7 (b) [Cl–] – +x +x ∴ x = molar solubility = 4.1 × 10–4 M Cu+(aq) + Cl–(aq) CuCl(s) I C E [Cu+] – +x +x Ksp = [Cu+][Cl–] = 1.7 × 10–7 Ksp = [Cu+][Cl–] = 1.7 × 10–7 [Cl–] 0.0200 +x 0.0200 + x Ksp = (x)(0.0200+x) = 1.7 × 10–7 Assume that x << 0.0200 ∴ x = molar solubility = 8.5 × 10–6 M 193 Chapter 18 (c) Cu+(aq) + Cl–(aq) CuCl(s) I C E [Cu+] – +x +x Ksp = [Cu+][Cl–] = 1.7 × 10–7 [Cl–] 0.200 +x 0.200 + x Ksp = (x)(0.200+x) = 1.7 × 10–7 Assume that x << 0.200 ∴ x = molar solubility = 8.5 × 10–7 M (d) CuCl(s) Cu+(aq) + Cl–(aq) Ksp = [Cu+][Cl–] = 1.7 × 10–7 – Note that the Cl concentration equals (2)(0.150 M) since two moles of Cl– are produced for every mole of CaCl2. I C E [Cu+] – +x +x [Cl–] 0.300 +x 0.300 + x Ksp = (x)(0.300+x) = 1.7 × 10–7 Assume that x << 0.300 ∴ x = molar solubility = 5.7 × 10–7 M 18.43 Au3+ + 3Cl– AuCl3(s) (a) Ksp = [Au3+][Cl–]3 = 3.2 × 10–25 let x = [Au3+]; then [Cl–] = 3x Ksp = (x)(3x)3 = 27x4 3.2 × 10 −25 = 3.3 × 10–7 M 27 The molar solubility of AuCl3 is 3.3 × 10–7 M in H2O. x= 4 (b) [Au3+] = x; [Cl–] = 0.010 + 3x Ksp = (x)(0.010 + 3x)3: Assume 3x << 0.010 Ksp = (x)(0.010)3 x = 3.2 × 10–19 M The molar solubility of AuCl3 is 3.2 × 10–19 M in 0.010 M HCl. (c) [Au3+] = x; [Cl–] = 0.020 + 3x Ksp = (x)(0.020 + 3x)3: Assume 3x << 0.020 Ksp = (x)(0.020)3 x = 4.0 × 10–20 M The molar solubility of AuCl3 is 4.0 × 10–20 M in 0.010 M MgCl2. (d) [Au3+] = 0.010 + x; [Cl–] = 3x Ksp = (0.010 + x)(3x)3: Assume x << 0.010 Ksp = (0.010)(3x)3 3.2 × 10−25 = 1.1 × 10–8 M 0.27 The molar solubility of AuCl3 is 1.1 × 10–8 M in 0.010 M Au(NO3)3. x= 18.44 3 Mg(OH)2 Mg2+ + 2OH– Ksp = [Mg2+][OH–]2 = 5.6 × 10–12 The concentration of OH– is determined from the pH: pOH = 14 – 12.50 = 1.50 194 Chapter 18 [OH–] = 0.0316 M [Mg ] = x [OH–] = 0.0316 M 2 Ksp = x(0.0316) = 5.6 × 10–12 x = 5.6 × 10–9 M The molar solubility of Mg(OH)2 is 5.6 × 10–9 M in a solution with a pH of 12.50. 2+ 18.45 Al(OH)3(s) Al3+ + 3OH– Ksp = [Al3+][OH–]3 = 3 × 10–34 – The concentration of OH is determined from the pH: pOH = 14 – 9.50 = 4.50 [OH–] = 3.2 × 10–5 M 3+ [Al ] = x [OH–] = 3.2 × 10–5 –5 3 Ksp = x(3.2 × 10 ) = 3 × 10–34 x = 9.2 × 10–21 The molar solubility of Al(OH)3 is 9.2 × 10–21 M in a solution with a pH of 9.50. 18.46 PbCl2(s) Pb2+ + 2Cl– – [Cl ] = 0.10 M [Pb2+][Cl–]2 =[Pb2+][0.10]2 = 1.7 × 10–5 [Pb2+] = 1.7 × 10–3 M 18.47 PbBr2 will be less soluble in a solution of 0.10 M NaBr because the values of Q and Ksp for lead bromide are dependent upon the [Br–]2. 18.48 Ag2CrO4 (s) (a) I C E 2Ag+ (aq) + CrO42–(aq) [Ag+] 0.200 + 2x 0.200 + 2x Ksp = [Pb2+][Cl–]2 = 1.7 × 10–5 Ksp = [Ag+]2[CrO42–] = 1.1 × 10–12 [CrO42–] – +x +x Ksp = (0.200+2x)2(x) Assume that x << 0.200 1.1 × 10–12 = (0.200)2(x) x = 2.8 × 10–11 The molar solubility is 3.0 × 10–11 moles/L (b) I C E [Ag+] – +2x +2x [CrO42–] 0.200 +x 0.200 + x Ksp = (2x)2(0.200+x) Assume that x << 0.200 1.1 × 10–12 = (2x)2(0.200) x = 1.2 × 10–6 The molar solubility is 1.2 × 10–6 moles/L. 18.49 Mg(OH)2 (s) (a) I C E Mg2+(aq) + 2OH–(aq) [Mg2+] – +x +x Ksp = [Mg2+][OH–]2 = 5.6 × 10–12 [OH–] 0.20 0.20 + 2x 0.20 + 2x Ksp = (x)(0.20 + 2x)2 Assume 2x << 0.20 Ksp = (x)(0.20)2 = 5.6 × 10–12 x = 1.4 × 10–10 M 195 Chapter 18 The assumption is valid and the molar solubility of Mg(OH)2 in 0.20 M NaOH is 1.4 × 10–10 moles/L (b) I C E [Mg2+] 0.20 0.20 + x 0.20 + x [OH–] – + 2x + 2x Ksp = (0.20 + x)(2x)2 Assume x << 0.20 Ksp = (0.20)(2x)2 = 5.6 × 10–12 x = 2.6 × 10–6 M The assumption is valid and the molar solubility of Mg(OH)2 in 0.20 M MgSO4 is 2.6 × 10–6 moles/L 18.50 Fe(OH)2(s) Fe2+(aq) + 2 OH–(aq) Ksp = Fe 2+ OH − 2 mol OH– = 2.20 g NaOH(1 mol/40.01 g NaOH) = 0.0550 mol NaOH [OH–] = mol OH–/L solution = 0.0550 mol/0.250 L = 0.22 M I C E [Fe2+] – +x x [OH–] 0.22 + 2x 0.22 + 2x We assume that x << 0.22, so that 0.22 + 2x ≈ 0.22, then we enter the equilibrium values of the above table into the Ksp expression: 2 Ksp = Fe 2+ OH − 4.9 × 10–17 = x(0.22)2 x = molar solubility = 1.0 × 10–15 M Next, we must determine how many moles of Fe(OH)2 are formed in the reaction. This is a limiting reactant problem. The number of moles of OH– is 0.0550 (see above). The number of moles of Fe2+ is (0.250 L)(0.10 mol/L) = 0.025 mol From the balanced equation at the top, we need two OH– for every one Fe2+. This would be 2(0.025 mol) = 0.050 mol OH–. Looking at the molar quantities above, we have more than enough OH– so, Fe2+ is our limiting reactant: 0.025 mol Fe(OH)2 will form in 0.25 L solution. If dissolved, this would be a concentration of 0.025 mol/0.25 L = 0.10 M. But from above, the maximum molar solubility of is 1.0 × 10–15 M. This means that remainder of Fe(OH)2 in excess of this value precipitates: 0.10 – 1.0 × 10–15 ≈ 0.10 M. This works out to 0.25 L(0.10 mol/L) = 0.025 mol Fe(OH)2(89.8 g/mol) = 2.2 g solid Fe(OH)2 (essentially all of it). The remaining OH–, 0.005 mol, gives a concentration of OH– of 0.005 mol OH − = 0.02 M OH– 0.250 L 196 Chapter 18 4.9 × 10–17 = [Fe2+][0.02]2 [Fe2+] = 1.2 × 10–13 M 18.51 Ni(OH)2(s) Ni2+(aq) + 2OH–(aq) mol OH– = 1.75 g NaOH(1 mol/40.01 g NaOH) = 0.0437 mol NaOH First, assume that all of the ions are in solution and no precipitate has formed. At that point the concentrations are: 0.0437 mol OH − = 0.175 M OH– 0.250 L [Ni2+] = 0.10 M Next, determine the limiting reactant: 0.10 mol Ni 2+ 2 mol OH − = 0.20 M OH– 1 L soln 1 mol Ni 2+ Only 0.175 M OH– is available, therefore OH– is the limiting reagent. Using the limiting reagent, assume that all of the OH– has precipitated [OH–] = 0.0 M 0.175 mol OH − 1 mol Ni 2+ [Ni2+] = 0.10 M Ni2+ – 2 mol OH − 1 L soln = 0.10 M Ni2+ – 0.088 M Ni2+ = 0.012 M Ni2+ From here, calculate the equilibrium concentrations: [Ni2+] [OH–] I 0.012 – C +x + 2x E 0.012 + x 2x We assume that x << 0.012, so that 0.012 + 2x ≈ 0.012, then we enter the equilibrium values of the above table into the Ksp expression: [OH–] = 2 Ksp = Ni 2+ OH − 5.5 × 10–16 = 0.012(x)2 x = 2.1 × 10–7 M The amount of Ni(OH)2 formed can be calculated from the amount of OH– added to the solution since essentially all of the OH– was precipitated: 1 mol Ni(OH)2 92.71 g Ni(OH)2 = 2.03 g Ni(OH)2 g Ni(OH)2 = 0.0437 mol OH– 2 mol OH − 1 mol Ni(OH)2 To determine the pH of the final solution, the [OH–] needs to be found: From the ICE table, the equilibrium concentration of OH– = 2x x = 2.1 × 10–7 M 2x = 4.2 × 10–7 M = [OH–] We need to add in the concentration of OH–,1 × 10–7 M, from the dissociation of water: [OH–] = 4.2 × 10–7 M + 1 × 10–7 = 5.2 × 10–7 M pOH = –log[OH–] = –log[5.2 × 10–7] = 6.28 pH = 14 – pOH = 14 – 6.28 = 7.72 18.52 Fe(OH)2(s) Fe2+(aq) + 2 OH–(aq) 2 Ksp = Fe 2+ OH − = 4.9 × 10–17 pH = 9.50 pOH = 14.00 – pH = 4.50 [OH–] = 10–4.50 = 3.16 × 10–5 M 197 Chapter 18 I C E [Fe2+] – +x x [OH–] 3.16 × 10–5 + 2x (3.16 × 10–5) + 2x Since Ksp for iron(II) hydroxide is so small, we can safely assume that 2x << 3.16 × 10–5, so that (3.16 × 10–5) + 2x ≈ 3.16 × 10–5, then we enter the equilibrium values of the above table into the Ksp expression: 2 Ksp = Fe 2+ OH − 4.9 × 10–17 = x(3.16 × 10–5)2 x = molar solubility = 4.9 × 10–8 M 18.53 (a) Ca(OH)2(s) Ca2+(aq) + 2OH–(aq) Ksp = [Ca2+][OH–]2 = 5.0 × 10–5 – 2+ let 2x = [OH ], [Ca ] = x + 0.10 Ksp = (0.10 + x)(2x)2 = 5.0 × 10–5 By successive approximations, x = 0.0036 The molar solubility of Ca(OH)2 in 0.10 M CaCl2 is 0.0036 moles/L. (b) Ca(OH)2(s) Ca2+(aq) + 2OH–(aq) Ksp = [Ca2+][OH–]2 = 5.0 × 10–5 let x = [Ca2+], [OH–] = 2x + 0.10 Ksp = (x)(2x + 0.10)2 = 5.0 × 10–5 By successive approximations, x = 4.9 x 10− 4 The molar solubility of Ca(OH)2 in 0.10 M NaOH is 4.9 x 10− 4 moles/L. 18.54 In order for a precipitate to form, the value of the reaction quotient, Q, must be greater than the value of Ksp. For PbCl2, Ksp = 1.7 × 10–5 (see Table 18.1). 2 Q = Pb2+ Cl− = (0.0150)(0.0120)2 = 2.16 × 10–6. Since Q < Ksp, no precipitate will form. 18.55 In order for a precipitate to form, the value of the reaction quotient, Q, must be greater than the value of Ksp. For AgC2H3O2, Ksp = 2.3 × 10–3. Q = Ag + C2 H3O2− = (0.015)(0.50) = 7.5 × 10–3. Since Q > Ksp, a precipitate will form. (Note: The concentration of C2H3O2– is twice the concentration of Ca(C2H3O2)2 since one mole of Ca(C2H3O2)2 produces two moles of C2H3O2–). 18.56 To solve this problem, determine the value for Q and apply LeChâtelier’s Principle. (a) Pb 2+ = (50.0 mL)(0.0100 moles/L)/(100.0 mL) = 5.00 × 10–3 Br − = (50.0 mL)(0.0100 moles/L)/(100.0 mL) = 5.00 × 10–3 2 Q = Pb2+ Br − = (5.00 × 10–3)(5.00 × 10–3)2 = 1.25 × 10–7 For PbBr2, Ksp = 6.6 × 10–6 Since Q < Ksp, no precipitate will form. (b) Pb 2+ = (50.0 mL)(0.0100 moles/L)/(100.0 mL) = 5.00 × 10–3 − Br = (50.0 mL)(0.100 moles/L)/(100.0 mL) = 5.00 × 10–2 198 Chapter 18 2 Q = Pb2+ Br − = (5.00 × 10–3)(5.00 × 10–2)2 = 1.25 × 10–5 For PbBr2, Ksp = 6.6 × 10–6 Since Q > Ksp, a precipitate will form. 18.57 In order for a precipitate to form, the value of the reaction quotient, Q, must be greater than the value of Ksp. For AgC2H3O2, Ksp = 2.3 × 10–3. Ag + = (22.0 mL)(0.100 M)/(67.0 mL) = 3.28 × 10–2 M C2 H3O2− = (45.0 mL)(0.0260 M)/(67.0 mL) = 1.75 × 10–2 M + Q = Ag C2 H3O2− = (3.28 × 10–2)(1.75 × 10–2) = 5.74 × 10–4. Since Q < Ksp, no precipitate will form. 18.58 The precipitate that may form is PbBr2(s). To determine if a precipitate will form, a value for the reaction quotient, Q, must be calculated: Q = [Pb2+][Br–]2. In performing this calculation, the dilution of the ions must be considered: [Pb2+] = [Br–] = 0.00500 M. Q = [0.00500][0.00500]2 = 1.3 × 10–7 If Q > Ksp, a precipitate will form. Ksp for PbBr2(s) is 6.6 × 10–6. Therefore, a precipitate will not form. Hence, the concentrations calculated are the diluted concentrations. Since no precipitate forms, the concentrations are not equilibrium values. 18.59 Ksp = [Mn2+][OH–]2 = 1.6 × 10–13 Ksp = [Cd2+][OH–]2 = 7.2 × 10–15 Mn(OH)2 is more soluble, so we need to determine the hydroxide ion concentration when it begins to precipitate. Ksp = [Mn2+][OH–]2 = 1.6 × 10–13 = (0.10)(x)2 x = 1.3 × 10–6 M OH– – Use this value for the [OH ] to solve for the concentration of the [Cd2+] left in solution: Ksp = [Cd2+][OH–]2 = 7.2 × 10–15 = (x)[1.3 × 10–6]2 x = 4.5 × 10–3 M Cd2+ 18.60 AgCl(s) AgI(s) Ag+ + Cl– Ag+ + I– K sp = Ag + Cl− = 1.8 × 10–10 + − K sp = Ag I = 8.5 × 10–17 When AgNO3 is added to the solution, AgI will precipitate before any AgCl does due to the lower solubility of AgI. In order to answer the question, i.e., what is the [I–] when AgCl first precipitates, we need to find the minimum concentration of Ag+ that must be added to precipitate AgCl. Let x = [Ag+]; Ksp = (x)(0.050) = 1.8 × 10–10; x = 3.6 × 10–9 M When the AgCl starts to precipitate, the solution will have a [Ag+] of 3.6 × 10–9 M. Now we ask, what is the [I–] if [Ag+] = 3.6 × 10–9 M? So, K sp = Ag + I − = (3.6 × 10–9)(x) = 8.5 × 10–17; x = 2.3 × 10–8 M = [I–] 18.61 This problem is similar to 18.60 except that the Ksp constants are closer in value. We first determine the minimum amount of SO42– that must be added to initiate the precipitation of CaSO4. CaSO4 will precipitate after SrSO4 due to its larger value for Ksp: Ksp(CaSO4) = 4.9 × 10–5 and Ksp(SrSO4) = 3.4 × 10–7 (see Table 18.1) (a) Let x = Ca 2+ ; K sp = Ca 2+ SO42− = (0.15)(x) = 4.9 × 10–5 199 Chapter 18 x = SO42− = 3.3 × 10–4 M When the SO42− = 3.3 × 10–4 the CaSO4 will start to precipitate. Now we ask, what is the Sr 2+ if SO42− = 3.3 × 10–4 M? Sr2+ (aq) + SO42–(aq) SrSO4 (s) [Sr2+] – +x +x I C E K sp = Sr 2+ SO42− [SO42–] 3.3 × 10–4 3.3 × 10–4 + x 3.3 × 10–4 + x K sp = Sr 2+ SO42− = (x)(1.6 × 10–4 + x) = 3.2 × 10–7 For this problem, we must solve the quadratic equation and we determine that x = 4.4 × 10–4 M. Thus, the [Sr2+] = 4.4 × 10–4 M when the CaSO4 starts to precipitate. Initially the solution had a concentration of 0.15 M. The solution now has a [Sr2+] = 4.4 × 10–4 M. So, the percentage of Sr2+ precipitated is; (b) 0.15 − 4.4 × 10-4 × 100% = 99.7 % 0.15 18.62 The less soluble substance is PbS. We need to determine the minimum [H+] at which CoS will precipitate. K spa Co 2+ [ H 2S] (0.010)(0.1) = = = 0.5 2 [H + ]2 H+ (from Table 18.2) (0.010)(0.1) = 0.045 0.5 pH = –log[H+] = 1.35. At a pH lower than 1.35, PbS will precipitate and CoS will not. At larger values of pH, both PbS and CoS will precipitate. [H + ] = 18.63 The less soluble substance is SnS so we will determine the maximum amount of H+ that is permitted before MnS starts to precipitate. Mn 2+ [ H 2S] (0.010)(0.1) K spa = = = 3 × 107 (from Table 18.2) 2 + 2 [H ] H+ [H + ] = (0.010)(0.1) = 6 × 10−6 3 × 107 pH = –log[H+] = 5.2. At pH values equal to or less than 5.2, MnS will not precipitate. 18.64 Cu(OH)2(s) Cu2+(aq) + 2 OH–(aq) K sp = Cu 2+ OH − 2 200 Chapter 18 2 4.8 × 10-20 = [ 0.10] OH − [OH–] = 6.9 × 10–10 M pOH = –log[OH–] = –log[6.9 × 10–10] = 9.2 pH = 14.00 –pOH = 4.8 4.8 is the pH below which all the Cu(OH)2 will be soluble. Mn(OH)2(s) Mn2+(aq) + 2 OH–(aq) K sp = Mn 2+ OH − 2 2 1.6 × 10-13 = [ 0.10] OH − [OH−] = 1.3 × 10–6 M pOH = –log[OH−] = –log[1.3 × 10–6] = 5.9 pH = 14.00 – pOH = 8.1 8.1 is the pH below which all the Mn(OH)2 will be soluble. Therefore, from pH = 4.8–8.1 Mn(OH)2 will be soluble, but some Cu(OH)2 will precipitate out of solution. 18.65 The following reactions are possible: CaC2O4(s) Ca2+(aq) + C2O42–(aq) MgC2O4(s) Mg2+(aq) + C2O4–2(aq) + Ksp = [Ca2+][C2O42–] = 2.3 × 10–9 Ksp = [Mg2+][C2O42–] = 4.8 × 10–6 – H + HC2 O4 – = 6.5 × 10–2 = [ H 2 C2 O4 ] H2C2O4(aq) H (aq) + HC2O4 (aq) K a1 HC2O4–(aq) H+(aq) + C2O42–(aq) H + C2 O4 2– = 6.1 × 10–5 Ka2 = HC2 O4 – 2 H + C2 O4 2– = 4.0 × 10–6 H2C2O4(aq) 2H (aq) + C2O4 (aq) KO = H C O [ 2 2 4] Assume that the concentration of H2C2O4 at the end of the process is 0.10 M First, calculate the concentration of C2O42–, at which the MgC2O4 will precipitate Ksp = [Mg2+][C2O42–] = 4.8 × 10–6 [Mg2+] = 0.10 M (0.10)(x) = 4.8 × 10–6 x = 4.8 × 10–5 M = [C2O42–] As long as the concentration of C2O42– is kept below 4.8 × 10–5 M, the Mg2+ will remain in solution In order for the concentration of oxalate to remain below = 4.8 × 10–5 M, the pH will be: + 2– 2 2H+(aq) + C2O42–(aq) H2C2O4(aq) 2 H + C2 O4 2– = 4.0 × 10–6 KO = [ H 2 C 2 O4 ] H + 4.8 × 10 –5 = 4.0 × 10–6 KO = 0.10 [ ] [H+] = 0.0913 M pH = 1.04 By keeping the pH < 1.04, the MgC2O4 will not precipitate. To precipitate the CaC2O4, a similar calculation needs to be done: 201 Chapter 18 Ksp = [Ca2+][C2O42–] = 2.3 × 10–9 [Ca2+] = 0.10 M (0.10)(x) = 2.3 × 10–9 x = 2.3 × 10–8 M = [C2O42–] As long as the concentration of C2O42– is kept above 2.3 × 10–8 M, the Ca2+ will precipitate In order for the concentration of oxalate to remain above = 2.3 × 10–8 M, the pH must be: 2 H2C2O4(aq) 2H+(aq) + C2O42–(aq) Ka2 H + C O 2– 2 4 = = 4.0 × 10–6 [ H 2 C2 O4 ] 2 H + 2.3 × 10 –8 = 4.0 × 10–6 K= [0.10] [H+] = 4.17 M pH = 0.62 18.66 Ksp = [Ag+][CO32–]= 8.5 × 10–12 Ksp = [Ni2+][CO32–] = 1.4 × 10–7 Assume the solutions are equal molar with concentration of 0.10 M. NiCO3 is more soluble and will precipitate when: K sp 1.4 × 10 7 [CO32–] = = = 1.4 × 10 6 2+ 0.10 [Ni ] Ag2CO3 will precipitate when: Ksp 8.5 × 10 12 [CO32− ] = = = 8.5 × 10 [Ag + ]2 (0.10)2 10 Ag2CO3 will precipitate and NiCO3 will not precipitate if [CO32–] > 8.5 × 10–10 and [CO32–] < 1.4 × 10–6. Now, using the equation in example 18.10 we get: 0.030 [H + ]2 = (2.4 × 10−17 ) [CO 2 − ] 3 NiCO3 will precipitate if: 0.030 −13 [H + ]2 = (2.4 × 10−17 ) = 5.5 × 10 1.4 × 10−6 [H+] = 7.4 × 10–7 pH = 6.13 Ag2CO3 will precipitate: 0.030 [H + ]2 = (2.4 × 10 −17 ) = 8.5 × 10−10 −10 8.5 × 10 [H + ] = 2.9 × 10−5 pH = 4.54 So Ag2CO3 will precipitate and NiCO3 will not if the pH is maintained between pH = 4.54 and pH = 6.13. 18.67 Ksp = [Zn2+][CO32–]= 1.5 × 10–10 Ksp = [Ni2+][CO32–] = 1.4 × 10–7 Assume the solutions are equal molar with concentration of 0.10 M. 202 Chapter 18 NiCO3 is more soluble and will precipitate when: K sp 1.4 × 10 7 [CO32–] = = = 1.4 × 10 6 0.10 [Ni 2+ ] ZnCO3 will precipitate when: Ksp 1.5 × 10 [CO32− ] = = 2+ 0.10 [Zn ] 10 = 1.5 × 10 9 ZnCO3 will precipitate and NiCO3 will not precipitate if [CO32–] > 1.5 × 10–9 and [CO32–] < 1.4 × 10–6. Now, using the equation in example 18.10 we get: 0.030 [H + ]2 = (2.4 × 10−17 ) [CO 2 − ] 3 NiCO3 will precipitate if: 0.030 [H + ]2 = (2.4 × 10−17 ) = 5.1 × 10−13 −6 1.4 × 10 [H+] = 7.2 × 10–7 pH = 6.14 ZnCO3 will precipitate: 0.030 −10 [H + ]2 = (2.4 × 10−17 ) = 4.8 × 10 1.5 × 10−9 [H + ] = 2.2 × 10−5 pH = 4.66 So ZnCO3 will precipitate and NiCO3 will not if the pH is maintained between pH = 4.66 and pH = 6.14. 18.68 18.69 (a) Cu2+(aq) + 4Cl–(aq) (b) Ag+(aq) + 2I–(aq) 3+ CuCl42–(aq) K form = AgI2–(aq) K form = 3+ Cr (aq) + 6NH3(aq) Cr(NH3)6 (aq) K form (a) Ag+(aq) + 2S2O32–(aq) Ag(S2O3)23–(aq) K form = (b) Zn2+(aq) + 4NH3(aq) 203 Cu 2+ Cl− 4 AgI2 − Ag + I− 2 Cr(NH3 )63+ = Cr3+ NH3 6 (c) Zn(NH3)42+(aq) CuCl4 2− Ag(S2 O3 )23− Ag + S2 O32− Zn(NH3 )4 2+ K form = Zn 2+ NH3 4 2 Chapter 18 (c) 18.70 (a) (b) (c) 18.71 (a) (b) (c) 18.72 18.73 (a) 4+ 2– Sn (aq) + 3S (aq) 3+ Co (aq) + 6NH3(aq) 2+ – Co(NH3)6 (aq) 2– Hg (aq) + 4I (aq) 2+ 4– Fe (aq) + 6CN (aq) Fe(CN)6 (aq) 2+ 2+ Hg (aq) + 4NH3(aq) – Sn (aq) + 6F (aq) 3+ Hg(NH3)4 (aq) 2– 3– Fe (aq) + 6CN (aq) 3+ Co(NH3)6 (aq) 2- HgI4 (aq) (c) 4– Fe(CN)6 (aq) 2+ Hg(NH3)4 (aq) Fe(CN)6 (aq) 3+ Co (aq) + 6NH3(aq) 2+ - Hg (aq) + 4I (aq) 2+ K form = K form – Fe (aq) + 6CN (aq) 2+ Hg (aq) + 4NH3(aq) 204 Sn 4+ S2− 3 HgI 42 − Hg 2+ I− 4 Fe(CN)6 4− Fe 2+ CN − 6 Hg(NH3 )4 2 + = 2+ Hg NH3 4 K form = SnF6 (aq) – K form SnS 2 − 3 Co(NH ) 3+ 3 6 = 3+ Co NH 6 3 K form = HgI4 (aq) – 4+ K form = SnS3 (aq) 3+ (b) (a) 2– K form = SnF6 2− Sn 4+ F− 6 Fe(CN)63− Fe3+ CN − 6 K inst Co3+ NH3 6 = 3 + Co(NH3 )6 K inst Hg 2+ I− = Hg(I)4 2− K inst Fe 2+ CN − = Fe(CN)6 4− K inst Hg 2+ NH3 4 = 2 + Hg(NH3 )4 4 6 Chapter 18 (b) (c) 2– SnF6 (aq) 3– Fe(CN)6 (aq) 4+ – Sn (aq) + 6F (aq) 3+ – Fe (aq) + 6CN (aq) 6 K inst Sn 4+ F− = SnF 2− 6 K inst Fe3+ CN − = Fe(CN)63− 6 18.74 Kc = Ksp × Kform = (1.7 × 10–5)(2.5 × 101) = 4.3 × 10–4 18.75 Kc = Ksp × Kform = (6.0 × 10–17)(5.3 × 1018) = 3.2 × 102 18.76 There are two events in this net process: one is the formation of a complex ion (an equilibrium which has an appropriate value for Kform), and the other is the dissolving of Fe(OH)3, which is governed by Ksp for the solid. Fe3+(aq) + 3OH–(aq) Fe(OH)3(s) 3+ – Fe (aq) + 6CN (aq) 3– Fe(CN)6 (aq) 3 K sp = Fe3+ OH − = 2.8 × 10−39 K form = Fe(CN)63− 3+ Fe CN − 6 = 1.0 × 1031 The net process is: Fe(OH)3(s) + 6CN–(aq) Fe(CN)63–(aq) + 3OH–(aq) The equilibrium constant for this process should be: Fe(CN)63− OH − Kc = 6 CN − 3 The numerical value for the above Kc is equal to the product of Ksp for Fe(OH)3(s) and Kform for Fe(CN)63–, as can be seen by multiplying the mass action expressions for these two equilibria: Kc = Kform × Ksp = 2.8 × 10–8 Because Kform is so very large, we can assume that all of the dissolved iron ion is present in solution as the complex, thus: [Fe(CN)63–] = 0.11 mol/1.2 L = 0.092 M. Also the reaction stoichiometry shows that each iron ion that dissolves gives 3 OH– ions in solution, and we have: [OH–] = 0.092 × 3 = 0.28 M. We substitute these values into the Kc expression and rearrange to get: [ CN −] = 6 Fe(CN) 3− OH − 6 Kc 3 (0.092)(0.28)3 =6 2.8 × 10−8 205 Chapter 18 Thus we arrive at the concentration of cyanide ion that is required in order to satisfy the mass action requirements of the equilibrium: [CN–] = 6.45 mol L–1. Since this concentration of CN– must be present in 1.2 L, the number of moles of cyanide that are required is: 6.45 mol L–1 × 1.2 L = 7.74 mol CN–. Additionally, a certain amount of cyanide is needed to form the complex ion. The stoichiometry requires six times as much cyanide ion as iron ion. This is 0.11 moles × 6 = 0.66 mol. This brings the total required cyanide to (7.74 + 0.66) = 8.4 mol. 8.4 mol × 49.0 g/mol = 412 g NaCN are required. 18.77 Ag+ + Br– Ksp = 5.0 × 10–13 AgBr(s) Ag+ + 2S2O32– Ag(S2O3)23– Kf = 2.0 × 1013 – 2– 3– Ag(S2O3)2 + Br Kc = Ksp*Kf = 10 AgBr(s) + 2S2O3 2– [S2O3 ] 1.20 – 2x 1.20 – 2x I C E [Br–] – +x +x 3– [Ag(S2O3)2 ] – +x +x Note: Since the AgBr(s) has a constant concentration, it may be neglected. Kc = [Ag(S2 O3 ) 23− ][Br − ] Kc = [S2 O32− ]2 = 10 x2 = 10 (1.20 − 2x)2 To solve this equation, take the square root of both sides and then solve for x. x = 0.518 M = [Ag(S2O3)23–] Since 1 mole of AgBr produces 1 mole of Ag(S2O3)23–, we can determine the number of grams of AgBr that will dissolve in 125 mL. g AgBr = (0.125 L)(0.518 moles/L)(187.77 g/mole) = 12.2 g AgBr 18.78 The applicable equilibria are as follows: AgI(s) Ag+(aq) + I–(aq) Ag+(aq) + 2I–(aq) AgI2–(aq) K sp = Ag + I − = 8.5 × 10-17 K form = AgI − 2 2 = 1 × 1011 Ag + I− When a solution of AgI2– is diluted, all of the concentrations of the species in Kform above decrease. However, the decrease of [I–] has more effect on equilibrium because its expression is squared. Hence, the denominator is decreased more than the numerator in the reaction quotient, Q. The system reacts according to Le Châtelier’s Principle, by moving to the left (toward reactants) to increase the value of [I–]. As the system moves to the left, more Ag+ is created, which has an effect on the first equilibrium above. Again, Le Châtelier’s Principle causes the reaction to move to the left to re-establish equilibrium, which produces AgI(s) precipitate. The two equations above may be combined and Kc found as follows: 206 Chapter 18 AgI − 2 AgI(s) + I (aq) AgI2 (aq) Kc = = Ksp × Kform = 8.5 × 10–6 − I To answer the second question, we make a table and fill in what we know. We begin with 1.0 M I–. This is reduced by some amount (x) as it reacts with the silver ions, and [AgI2–] is increased by the same amount: – I C E – [I–] 1.0 –x 1.0 – x [AgI2–] – +x +x Now we insert the equilibrium values into the above equation: AgI − 2 Kc = = 8.5 × 10–6 − I [ x ] = 8.5 × 10–6 Kc = [1.0 − x ] x = 8.5 × 10–6 This value represents the change in concentration of I– which, from the balanced equation, equals the change in concentration of AgI(s). The given volume is 0.100 L, which allows us to find the amount of AgI reacting: 0.100 L(8.5× 10–6 mol/L) = 8.5 × 10–7 mol AgI 8.5 × 10–7 mol AgI(234.8 g/mol) = 2.0 × 10–4 g AgI 18.79 The applicable equilibria are as follows: K sp = Ag + I− = 8.5 × 10−17 Ag ( I ) − 2 K form = = 1 × 1011 AgI2–(aq) Ag+(aq) + 2I–(aq) 2 Ag + I− The two equations above may be combined and Kc found as follows: Ag ( I ) − 2 = Ksp × Kform = 8.5 × 10–6 AgI(s) + I–(aq) Ag(I)2–(aq) K c = − I If all of the AgI dissolves, it will be in the form of Ag(I)2–, therefore the concentration of Ag(I)2– is: AgI(s) [Ag(I)2–] = Ag+(aq) + I–(aq) 0.020 mol Ag ( I )2 − = 0.200 M Ag(I)2– 0.100 L solution 0.200 [CN–] = =2.4 × 104 M −6 8.5 × 10 The amount of KI that must be added is: (2.4 × 104 M)(0.100 L) = 2400 mol KI g KI = (2400 mol KI)(166 g/mol) = 3.98 × 105 g KI 18.80 The applicable equilibria are as follows: AgI(s) Ag+(aq) + I–(aq) K sp = Ag + I− = 8.5 × 10−17 207 Chapter 18 + – Ag (aq) + 2CN (aq) – K form = Ag(CN)2 (aq) Ag ( CN ) − 2 2 = 5.3 × 1018 Ag + CN − The two equations above may be combined and Kc found as follows: Ag ( CN ) − I− 2 Kc = = Ksp × Kform = 4.5 × 102 AgI(s) + 2CN–(aq) Ag(CN)2–(aq) + I–(aq) 2 − CN We begin with 0.010 M CN–. This is reduced by some amount (2x) as it reacts with the silver ions, and [AgI2–] is increased by x: I C E [CN–] 0.010 – 2x 0.010 – 2x [Ag(CN)2–] – +x x I– – +x x Now we insert the equilibrium values into the above equation: Ag ( CN ) − I− 2 Kc = = 4.5 × 102 2 − CN x x [ ][ ] = 4.5 × 102 Kc = [ 0.010 − 2x ]2 Take the square root of both sides and solve for x: x = 4.9 × 10–3 This value represents the change in concentration of I– which, from the balanced equation, equals the change in concentration of AgI(s). The molar solubility of AgI in 0.010 M KCN is 4.9 × 10–3 M. 18.81 In case (a), the formation constant is relatively small indicating that the complex is not very stable. At the same time, the extremely small value for Ksp indicates that the ML2 solid is very stable. Consequently, the solution will contain very little M2+. In case (b), the solubility of ML2 is even smaller than in case (a). However, the large value for the formation constant indicates that any M2+ ions in solution will react with any ligand present to form the complex ion. As a result, more of the ML2 solid will dissolve increasing the amount of M2+ in solution. 18.82 Recall that Kinst = 1/ Kform. Zn(OH)2 (s) Zn2+(aq) + 2OH–(aq) 2+ Zn (aq) + 4NH3 (aq) 2 K sp = Zn 2+ OH − = 3.0 × 10-17 2+ Zn(NH3)4 (aq) K form Zn(NH ) 2+ 3 4 = =? Zn 2+ NH 4 3 Combined, this is: Zn(OH)2 (s) + 4NH3 (aq) Zn(NH3)42+(aq) + 2OH–(aq) 208 Chapter 18 Zn(NH ) 2+ OH - 3 4 Kc = 4 NH3 [NH3] 1.0 – 4x 1.0 – 4x I C E Kc = 2 [Zn(NH3)42+] – +x x [OH–] – + 2x 2x [ x ][ 2x ]2 [1.0 - 4x ]4 The problem gives the molar solubility of Zn(OH)2 as 5.7 × 10–3 M. This means in one liter of 1.0 M NH3, x = 5.7 × 10–3 moles. Substituting this value in for x, we get Kc = 8.1 × 10–7. Kc = Ksp × Kform 8.1 × 10–7 = 3.0 × 10–17 × Kform Kform = 2.7 × 1010 Kinst = 1/Kform Kinst = 1/(2.7 × 109) = 3.7 × 10–11 18.83 2 K sp = Cu 2+ OH − = 4.8 × 10-20 2+ 13 Cu(NH3)4 (aq) Kform = 1.1 x 10 Cu2+(aq) + 2OH–(aq) Cu(OH)2 (s) Cu2+(aq) + 4NH3 (aq) Combined, this is: Cu(OH)2 (s) + 4NH3 (aq) Cu(NH3)42+(aq) + 2OH–(aq) Cu(NH ) 2 + OH - 3 4 Kc = 4 NH3 [NH3] 2.0 – 4x 2.0 – 4x I C E Kc = [ x ][ 2x ]2 [ 2.0 − 4x ]4 2 = 5.3 × 10–7 [Cu(NH3)42+] – +x x [OH–] – + 2x 2x = 5.3 × 10–7 Solve for x using successive approximations. x = 1.3 × 10–2 [Cu(NH3)42+] = 1.3 × 10–2 2+ Since all of the Cu comes from the Cu(OH)2, the molar solubility of Cu(OH)2 is 1.3 × 10–2 M Additional Exercises 18.84 Assume the more soluble compound is 1.00 M. Then the less soluble compound is 0.0001 M. M1q + = 1.00 M M 2q + = 0.0001 M The counter ion has molar concentration = X q − 209 Chapter 18 K sp (1) K sp (2) 18.85 = 1.00 X q − 0.0001 X q − =1 x 104 The student would not collect PbCl2 by adding HCl to the final solution. The Pb2+ would be preciptated in step (1) when H2S was added to the acidified solution. This precipitate was separated from the remaining solution. The first precipitate would consist of PbS and CdS while the second precipitate consists of NiS. 18.86 We must first calculate the solubility in terms of # mols/L, i.e., mol = 7.05 × 10−3 g 1 mol Mg(OH)2 = 1.21 × 10−4 M L L 58.32 g Mg(OH) 2 Next, use this to establish the individual ion concentrations based on the equilibrium: Mg(OH)2 Mg2+ + 2OH– ( ) Mg 2+ = 1.21 × 10−4 M − OH = 2.42 × 10−4 M Finally, calculate Ksp using the standard expression: 2 K sp = Mg 2+ OH − = 1.21 × 10−4 2.42 × 10−4 18.87 + FeS(s) + 2H (aq) I C E K spa ( )( 2+ K spa Fe (aq) + H2S(aq) [H+] 8 8 – 2x 8 – 2x [Fe2+] – +x +x Fe 2+ [ H S] 2 = (x)(x) = 600 = 2 (8 − 2x)2 H+ ) 2 = 7.09 × 10−12 Fe2+ [ H S] 2 = 2 H + [H2S] – +x +x (from Table 18.2) x = 24.5 (8 − 2x) Solving gives x = 3.92 M. FeS is very soluble in 8 M acid. take the square root of both sides to get; 18.88 In order to answer this question, we need the [OH–] at equilibrium. Ca(OH)2 is a sparingly soluble compound. According to Table 18.1, Ksp = 5.0 × 10–6. Ca(OH)2(s) Ca2+(aq) + 2OH–(aq) K sp = Ca 2+ OH − [Mg2+] [OH–] – – I +x + 2x C +x + 2x E Ksp = (x)(2x)2 = 4x3 = 5.0 × 10–6 210 2 Chapter 18 x = 1.1 × 10–2 M, [OH–] = 2x = 2.2 × 10–2 M: The pH = 14.00 – pOH = 12.34. pOH = –log[OH–] = 1.66 18.89 The reaction that will dissolve the Mg(OH)2 is: Mg(OH)2 + 2H+ Mg2+ + 2H2O The concentration of H+ in the solution, before any reaction has occurred between the acid and the Mg(OH)2, is (0.025 L)(0.10 M HCl) = 2.5 × 10–3 mol HCl 2.5 × 10–3 mol/1.000 L = 2.5 × 10–3 M Since all of the H+ will react with the solid Mg(OH)2, the amount of Mg2+ in solution will be: 1 mol Mg 2+ 2.5 × 10–3 mol HCl = 1.3 × 10–3 mol Mg2+ 2 mol H + Find the equilibrium concentration of OH– with 1.3 × 10–3 M Mg2+ Ksp = [Mg2+][OH–]2 = 5.6 × 10–12 5.6 × 10–12 = (1.3 × 10–3)(x)2 x = 6.6 × 10–5 = [OH–] pOH = –log[OH–] = –log(6.6 × 10–5) = 4.18 pH = 14 – pOH = 14 – 4.18 = 9.82 18.90 In this problem, we have two simultaneous equilibria occurring: 2 K sp = Mn 2+ OH − = 1.6 × 10–13 1 Fe2+(aq) + 2OH–(aq) Fe(OH)2(s) = 1/(4.9 × 10–17) = 2.04 × 1016 Kc = 2 2 + Fe OH The second equilibrium represents the opposite equation from that of Ksp. Therefore, its value is 1/Ksp for Fe(OH)2. Mn(OH)2(s) Mn2+(aq) + 2OH–(aq) Combined, and omitting spectator ions, this is: Mn(OH)2(s) + Fe2+(aq) Mn2+(aq) + Fe(OH)2(s) Mn 2+ =K −13 Kc = 2.04 × 1016 = 3265 sp (Mn) ⋅ K c (Fe) = 1.6 × 10 2 + Fe ( I C E )( [Fe2+] 0.100 –x 0.100 – x [Mn2+] – +x x Mn 2+ Kc = Fe 2+ [x] 3265 = [ 0.100-x ] 326.5 – 3265x = x 326.5 = 3266x x = 0.099969 = 0.100 M Therefore, [Fe2+] = 0.100 – x = 0 M and [Mn2+] = 0.1005 M 211 ) Chapter 18 Since Ksp for Fe(OH)2 and Mn(OH)2 are so small, we assume there is almost no free hydroxide ion present and therefore the pH would remain neutral, around 7. 18.91 (a) The number of moles of the two reactants are: 0.12 M Ag+ × 0.050 L = 6.0 × 10–3 moles Ag+ 0.048 M Cl– × 0.050 L = 2.4 × 10–3 moles Cl– The precipitation of AgCl proceeds according to the following stoichiometry: Ag+ + Cl– AgCl(s). If we assume that the product is completely insoluble, then 2.4 × 10–3 moles of AgCl will be formed because Cl– is the limiting reagent (see above.) 143.3 g AgCl g AgCl = 2.4 × 10−3 mol AgCl = 0.34 g AgCl 1 mol AgCl ( (b) ) The silver ion concentration may be determined by calculating the amount of excess silver added to the solution: [Ag+] = (6.0 × 10–3 moles – 2.4 × 10–3 moles)/0.100 L = 3.6 × 10–2 M The concentrations of nitrate and sodium ions are easily calculated since they are spectators in this reaction: [NO3–] = (0.12 M)(50.0 mL)/(100.0 mL) = 6.0 × 10–2 M [Na+] = (0.048 M)(50.0 mL)/(100.0 mL) = 2.4 × 10–2 M In order to determine the chloride ion concentration, we need to solve the equilibrium expression. Specifically, we need to ask what is the chloride ion concentration in a saturated solution of AgCl that has a [Ag+] = 3.6 × 10–2 M. AgCl(s) I C E Ag+ + Cl– [Ag+] 0.036 +x 0.036 + x Ksp = 1.8 × 10–10 [Cl–] – +x +x K sp = Ag + Cl− = (0.036+x)(x) = 1.8 × 10–10 x = 5.0 × 10–9 M if we assume that x<<0.036 Therefore, [Cl–] = 5.0 × 10–9 M. (c) The percentage of the silver that has precipitated is: (2.4 × 10–3 moles)/(6.0 × 10–3 moles) × 100% = 40% 18.92 To solve this problem, recognize that for a solution having a density of 1.00 g mL–1, 1 ppm = 1 mg L–1. Therefore, the initial hard water solution has a concentration of 278 mg Ca2+ / 1 L solution. Converting to molar concentration: 278 mg Ca 2+ 1 g Ca 2+ 1 mol Ca 2+ mol Ca 2 + = 2 + 2+ L solution 1 L solution 1000 mg Ca 40.078 g Ca = 6.94 × 10−3 M Ca 2 + 212 Chapter 18 The concentration of CO32– is: mol CO32 − 1.00 g Na 2 CO3 = L solution 1 L solution 1 mol Na 2 CO3 105.99 g Na 2 CO3 1 mol CO32− 1 mol Na 2 CO3 = 9.43 × 10−3 M CO32 − Comparing the concentrations of Ca2+ and CO32–, we observe that Ca2+ is the limiting reactant. Because of the small value of Ksp, we can assume that CaCO3 will precipitate using all of the available Ca2+ and leaving 9.43 × 10–3 – 6.94 × 10–3 = 2.49 × 10–3 M CO32–. The question now becomes, how much Ca2+ will be present in a solution having a [CO32–] = 2.49 × 10–3 M? Use the solubility product constant for Ksp to answer this question. K sp = 3.4 × 10−9 = Ca 2+ CO32− −9 K 3.4 × 10 sp Ca 2+ = = = 1.4 × 10−6 M −3 CO 2− 2.49 × 10 3 Converting back to units of ppm (mg L–1) we get: 1.4 × 10−6 mol Ca 2+ ppm Ca 2+ = 1 L solution 40.08 g Ca 2+ 1 mol Ca 2+ 1000 mg 1 g = 5.5 × 10−2 ppm Ca 2+ 18.93 A saturated solution of La2(CO3)3 satisfies the following equilibrium expression: 2 K sp = 4.0 × 10−34 = La 3+ CO32− 3 If [La3+] = 0.010 M, then the carbonate concentration of a saturated solution is: CO 2− = 3 K sp 3 3+ 2 = 3 4.0 × 10−34 2 = 1.6 × 10 −10 La We do the same calculation for PbCO3: K sp = 7.4 × 10−14 = Pb2+ CO32− If [Pb2+] = 0.010 M, then the carbonate concentration of a saturated solution is: K sp 7.4 × 10−14 CO 2− = = = 7.4 × 10−12 M 3 0.010 Pb2+ Therefore, at a carbonate ion concentration between 7.4 × 10–12 M and 1.6 × 10–10 M, PbCO3 will precipitate, but La2(CO3)3 will not precipitate. The upper limit for the carbonate ion concentration is therefore 1.6 × 10–10 M. ( 0.010 ) The equilibrium we need to look at now is: H2CO3(aq) 2H+(aq) + CO32–(aq) The Ka for this reaction is the product of Ka1 and Ka2 for carbonic acid. From Table 17.4we see that Ka1 = 4.5 × 10–7 and Ka2 = 4.7 × 10–11. So the equilibrium expression and value for the reaction of interest is: 213 Chapter 18 2 H + CO 2− 3 Ka = = K a 1 × K a 2 = 2.1 × 10−17 H CO 2 3 This equation is rearranged and the values above and the values given in the problem are substituted in order to determine the pH range over which PbCO3 will selectively precipitate: H+ = K a H 2 CO3 = CO32− ( 2.1 × 10 )(3.3 × 10 ) −17 −2 CO32 − If we substitute CO32− = 7.4 × 10−12 M we determine H + = 3.1 × 10−4 M and the pH = 3.51. Substituting CO32− = 1.6 × 10−10 M , H + = 6.6 × 10−5 M and pH = 4.18. Consequently, if [H+] = 6.6 × 10–5 M (pH = 4.18), La2(CO3)3 will not precipitate but PbCO3 will precipitate. At pH = 3.51 and below, neither carbonate will precipitate. 18.94 (a) Mg(OH)2(s) Mg2+ + 2OH– + – NH4 + OH NH3 + H2O Mg(OH)2(s) + 2NH4+(aq) Mg2+(aq) + 2H2O + 2NH3(aq) (b) We want all of the Mg(OH)2 to go into solution. The NH4+ reacts with any OH– produced in the dissociation of Mg(OH)2 thereby shifting the equilibrium to the right. Using the Ksp value for Mg(OH)2, we may find the hydroxide ion concentration under these conditions: Mg2+ + 2OH– Mg(OH)2(s) 2 K sp = Mg 2+ OH − 5.6 × 10−12 = [ 0.10 ] OH − − − 6 OH = 7.5 × 10 2 Now we can use this value in the following, simultaneous equilibrium: NH4+(aq) + OH–(aq) H2O + NH3 (aq) Kc = 1/KbNH3 = 1/1.8 × 10–5 = 5.6 × 104 [ 0.20] NH3 Kc = = = 5.6 × 104 OH − NH + 7.5 × 10-6 NH + 4 4 (We know that [NH3] = 0.20 M because in the equation below 2 moles of ammonia are formed for every one mole of magnesium ion: Mg(OH)2(s) + 2NH4+(aq) Mg2+(aq) + 2H2O + 2NH3 (aq)) Solving for [NH4+], we get 0.48 M. So the total [NH3] + [NH4+] = 0.20 + 0.48 = 0.68 M One must therefore add 0.68 mol NH4Cl to a liter of solution. (c) 18.95 The resulting solution will contain 0.20 mol of NH3. Solve the weak base equilibrium problem for NH3. The pH = 11.28. There are two reactions that have to be considered here: the dissociation of CaCO3 in water, CaCO3(s) Ca2+(aq) + CO32–(aq) Ksp = [Ca2+][CO32–] = 3.4 × 10–9 214 Chapter 18 and the ionization of carbonate ion in water, CO32–(aq) + H2O HCO3–(aq) + OH–(aq) Kb = [OH − ][HCO3− ] 2− = 1.8 × 10−4 [CO3 ] Assuming that all of the CO32– reacts with the water, the net reaction is: CaCO3(s) + H2O(l) Ca2+(aq) + HCO3–(aq) + OH–(aq) Kc = Ksp × Kb = [Ca2+][HCO3–][OH–] = 6.1 × 10–13 We can obtain [OH–] from the pH: pOH = 14 – pH = 14 – 8.50 = 5.50 [OH–] = 10–pOH = 10–5.50 = 3.2 × 10–6 M Assume that [Ca2+] = [HCO3–] = x Kc = 6.1 × 10–13 = (x)(3.2 × 10–6 M)(x) = (x)2(3.2 × 10–6 M) Solve for x: x = 4.4 × 10–4 M = [Ca2+] = [HCO3–] 2+ The [Ca ] is equal to the molar solubility. Thus, the molar solubility of CaCO3 is 4.4 × 10–4 M. 18.96 Fe(OH)3(s) I C E Fe3+ + 3OH– [Fe3+] – +x +x Ksp = [Fe3+][OH–]3 = 2.8 × 10–39 [OH–] 1.0 × 10–7 + 3x 1.0 × 10–7 + 3x Assume 3x << 1.0 × 10–7, Ksp = (x)(1.0 × 10–7)3, solving for x we get, x = 2.8 × 10–18 M. Thus, 2.8 × 10–18 mol of Fe(OH)3 dissolve in 1 L of water. 18.97 The reaction for this problem is the formation of Ag(NH3)2+: Ag(NH ) + 3 2 + 2+ Ag (aq) + 2NH3(aq) Ag(NH3) K form = = 1.6 × 107 2 + Ag NH 3 We can rearrange this equation and substitute the values from the text to determine the [Ag+]: Ag(NH ) + 2.8 × 10−3 3 2 Ag + = = = 1.8 × 10−10 M 2 2 1.6 × 107 ( 1) K form NH3 ( ( 18.98 ) ) Let x = mols of PbI2 that dissolve per liter; Let y = mols of PbBr2 that dissolve per liter. Then, at equilibrium, we have [Pb2+] = x + y, [I–] = 2x and [Br–] = 2y We know: PbBr2(s) Pb2+ + 2Br– PbI2(s) Pb2+ + 2I– Ksp = 6.6 × 10–6 = [Pb2+][Br–]2 Ksp = 9.8 × 10–9 = [Pb2+][I–]2 Substituting we get: 6.6 × 10–6 = (x + y)(2y)2 and 9.8 × 10–9 = (x + y)(2x)2 Solving for x and y we find: x = 4.85 × 10–4 y = 7.91 × 10–3 Thus, [Pb2+] = 8.40 × 10–3 M, [I–] = 9.70 × 10–4 M and [Br–] = 1.58 × 10–2 M Note: [Br–] > [I–] because PbBr2 is more soluble than PbI2. 215 Chapter 18 18.99 Initially, both Ag+ and HC2H3O2 are at 1.0 M concentrations. These values will be used to determine if the AgC2H3O2 will precipitate. First, determine the concentration of the acetate ion from the equilibrium: H+ + C2H2O2– HC2H3O2 H + C H O – 2 3 2 Ka = = 1.8 × 10–5 HC2 H3O 2 [HC2H3O2] 1.0 –x 1.0 – x I C E Assume x << 1.0 [x][x] Ka = = 1.8 × 10−5 [1.0] [H3O+] – +x x [C2H3O2–] – +x x x = 4.2 × 10-3 = [C2 H3O 2 − ] Next, using the concentration of the acetate ion, determine whether or not a precipitate will form. AgC2H3O2(s) Ag+ + C2H3O2– + Ksp = [Ag ][C2H3O2–] Q = [Ag+][C2H3O2–] before equilibrium is established Q = (1.0 M)(4.2 × 103 M) = 4.2× 103 Q > Ksp therefore a precipitate will form. 18.100 [Ag+] = 0.200 M [H+] = 0.10 M First, the concentration of acetate ion needs to be determined at the point that the silver acetate precipitates: AgC2H3O2(s) Ag+ + C2H3O2– + Ksp = [Ag ][C2H3O2–] = 2.3 × 10–3 Let x = [C2H3O2–] 2.3 × 10–3 = (0.200)(x) x = 1.2 × 10–2 M = [C2H3O2–] When NaC2H3O2 is added to the solution, the C2H3O2– will react with the H+ from the nitric acid to form HC2H3O2. This will give a concentration of 0.10 M HC2H3O2. This will then come to equilibrium Now, find the concentration of HC2H3O2 using the Ka for acetic acid: HC2H3O2 H+ + C2H3O2– [HC2H3O2](0.200 L) + [C2H3O2–](0.200 L) = mole NaC2H3O2 that needs to be added. [HC2H3O2] + [H+] = 0.10 M which is from the nitric acid [HC2H3O2] = 0.10 M – [H+] H + C H O − 2 3 2 Ka = HC2 H3O 2 Let x = [H+] [ x ] 1.2 ×10 –2 1.8 × 10-5 = [ 0.10 – x ] x = 1.5 × 10–4 M = [H+] [HC2H3O2] = 0.10 M – [H+] = 0.10 – 1.5 × 10–4 M = 9.99 × 10–4 M The amount of NaC2H3O2 to be added is: [HC2H3O2](0.200 L) + [C2H3O2–](0.200 L) = mole NaC2H3O2 [9.99 × 10–4 M](0.200 L) + [1.2 × 10–2 M](0.200 L) = 2.6 × 10–3 mole NaC2H3O2 The number of grams to be added is 216 Chapter 18 82.03 g NaC2 H3O 2 g NaC2H3O2 = 2.6 × 10–3 mole NaC2H3O2 = 0.21 g NaC2H3O2 1 mol NaC2 H3O 2 18.101 First, let’s examine the question to make clear what is happening. The solution contains 0.20 M Ag+ ions and 1.34 x 10− 3 M acetate ions. The ion product of these two (2.7 x 10− 4 ) is less than Ksp for silver acetate (2.3 × 10–3), so the silver acetate remains in solution. There are also H+ ions (H3O+) in the solution as a result of the following equilibrium (OAc– will symbolize acetate): H2O + HOAc H3O+(aq) + OAc–(aq) The amount of H3O+ may be found by using the Ka for acetic acid: I C E Ka [H3O+] – +x X [HOAc] 0.10 –x 0.10 – x [OAc–] – +x x H O + OAc− 3 = [ HOAc] [ x ][ x ] 1.8 × 10-5 = [0.10 − x] –3 x ≈ 1.3 × 10 So [H3O+] = 1.3 × 10–3 M. However, when F– is added to the solution (in the form of KF), the following equilibrium takes place: F–(aq) + H3O+(aq) H2O + HF This depletes H3O+ ions from the solution, which causes the first equilibrium above to move to the right, producing more acetate ions. When the acetate ion concentration hits some minimum value (determined by Ksp) silver acetate will precipitate. That value may be found as follows: K sp = Ag + OAc− 2.3 × 10−3 = [ 0.20 ][ x ] x = 0.012 mol/L So the problem becomes…How many grams of KF must be added such that the acetate concentration increases to 0.012 M? This is now a simultaneous equilibrium problem: K a = 1.8 × 10−5 H3O+(aq) + OAc–(aq) H2O + HOAc F–(aq) + H3O+(aq) H2O + HF K c ’ = 1 1 = = 1.5 × 103 K a 6.8 × 10−4 (Note the above equation is simply the reverse of that for the Ka of HF, so Kc’ = 1/Ka.) Combined, this becomes: F–(aq) + HOAc HF + OAc–(aq) ( )( ) K c = K a ⋅ K b = 1.8 × 10−5 1.5 × 103 = 2.7 × 10−2 217 Chapter 18 Recall that we have already found the initial concentrations of OAc– and HOAc above. Using this information, and the fact that we want the final [OAc–] to be 1.2 × 10–2, we can begin to fill out the table below. I C E [F–] x – 0.011 ? [HOAc] 0.099 – 0.011 0.088 Kc = [HF] – + 0.011 0.011 [OAc–] 1.3 × 10–3 + 0.011 0.012 [ HF] OAc− F− [ HOAc ] [ 0.011][0.012] 2.7 × 10−2 = F− [ 0.088] – [F ] = 0.056 M Placing this value into the table as the equilibrium concentration of [F–], we find the initial [F–] must be 0.056 + 0.011 = 0.067 M. Therefore the amount of KF needed in the 200 mL solution is: 0.200 L(0.067 mol KF/L)(58.01 g KF/1 mol KF) = 0.78 g KF 18.102 Step 1: Determine the [OH–] from the NH3 reaction with water: NH4+ + OH– NH3 + H2O NH + OH – 4 Kb = NH 3 1.8 × 10 −5 NH + OH – 4 = NH3 I C E Assume x << 0.10 [ x ][ x ] 1.8 × 10−5 = [ 0.10] [NH3] 0.10 –x 0.10 – x [OH–] – +x x [NH4+] – +x x x = 1.3 × 10–3 = [OH–] Step 2: Find the concentration of Mg2+ at the given concentration of OH–. The concentration of OH– from the NH3 is 1.3 × 10–3, there is an additional amount of OH– from the equilibrium of the Mg(OH)2, which makes the calculation: Ksp = 5.6 × 10–12 = [Mg2+][OH–]2 5.6 × 10–12 = (x)(1.3 × 10–3 + 2x)2 The additional amount of OH– can be ignored since it will be less than 1.3× 10–3: Using the molar solubility of Mg(OH)2 in distilled water: 5.6 × 10–12 = [Mg2+][OH–]2 s = [Mg2+] and 2s = [OH–] 4s3 = 5.6 × 10–12 s = 1.1 × 10–4 218 Chapter 18 The solubility of Mg(OH)2 in distilled water is less than the amount of OH– supplied by the ammonia so we a re justified in ignoring its contribution. We may now solve for x x = 3.3 × 10–6 M = [Mg2+] The molar solubility of Mg(OH)2 = 3.3 × 10–6 M 18.103 At its simplest, this is only a Ksp problem. The concentration of Mn2+ is (0.400 L)(0.10 M Mn2+)/(0.500 L) = 0.080 M Mn2+ K sp = Mn 2+ OH − 2 1.6 × 10-13 = [ 0.080] OH − − -6 OH = 1.4 × 10 2 When [OH–] = 1.4 × 10–6 M, Mn(OH)2 precipitates. 0.100 L(2.0 mol/L) = 0.20 mol NH3 are added to 400 mL solution which would make an initial concentration of 0.20 mol/0.500 L = 0.40 M NH3. The following equilibrium is set up: K b = 1.8 × 10-5 NH4+(aq) + OH–(aq) H2O + NH3 The problem tells us that all of the Sn2+ is precipitated as Sn(OH)2: Sn2+(aq) + 2OH–(aq) Sn(OH)2(s) The concentration of the Sn2+ is (0.400 L)(0.10 M Sn2+)/(0.500 L) = 0.080 M Sn2+, just before it precipitates. The Sn2+ immediately uses the first 2(0.080 M) = 0.16 M OH– which is produced from the reaction of ammonia with water above, using 0.16 M NH3. This effectively brings our initial concentration of NH3 to 0.24 M. Now we determine how much NH3 will produce [OH–] = 1.3 × 10–6 M. K b = 1.8 × 10-5 NH4+(aq) + OH–(aq) H2O + NH3 I C E [NH3] x –1.2 × 10–6 ? [NH4+] – + 1.2 × 10–6 1.2 × 10–6 [OH–] 1.0 × 10–7 + 1.2 × 10–6 1.3 × 10–6 NH + OH − 4 Kb = NH3 1.2 × 10-6 1.3 × 10-6 1.8 × 10 = NH3 [NH3] = 8.7 × 10–8 -5 Therefore the initial [NH3] should be 8.7 × 10–8 + 1.2 × 10–6 = 1.3 × 10–6. So we want to reduce [NH3] by 0.24 – 1.3 × 10–6 = 0.23999 M, essentially by 0.24 M. This would require adding equimolar amounts of HCl, or: 0.500 L(0.24 mol NH3/L)(1 mol HCl/1 mol NH3)(36.5 g HCl/1 mol HCl) = 4.4 g HCl. 219 Chapter 18 (The difficulty here arises from the fact that such a small amount of OH– is required to precipitate the Mg2+ from solution that even a minimal amount of NH3 produces enough hydroxide ion to do so.) 18.104 Three reactions are occurring in the solution: Cu2+(aq) + 4NH3–(aq) 1) 2) Cu(NH3)42+–(aq) + NH3(aq) + H2O – NH4 (aq) + OH (aq) Cu(NH3 )4 Kform = 1.1 × 1013 = Cu 2+ NH 4 3 NH + OH – 4 Kb = 1.8 × 10 = NH3 –5 2 Ksp = 4.8 × 10–20 = Cu 2+ OH – Starting with the first reaction, using the Kform of Cu(NH3)42+, calculate the concentration of NH3. The concentration of Cu2+ before any reaction has occurred is 0.050 M, and the concentration of NH3 before any reaction has occurred is 0.50 M. Assume that all of the Cu2+ has reacted and then the reaction has come to equilibrium. Therefore, the initial concentration of Cu2+ is 0 and the initial concentration of NH3 is 0.50 M – (4 × 0.05 M) 3) I C E Cu2+ + 2OH– Cu(OH)2(s) [Cu2+] – +x + 2x [NH3] 0.30 + 4x 0.30 + 4x Cu(NH3)42+ 0.050 M –x 0.050 – x Cu(NH3 )4 Kform = 1.1 × 1013 = Cu 2+ NH 4 3 Kform = 1.1 × 1013 = [0.050 – x ] [ 2x ][0.30 + 4x ]4 assume x << 0.050 x = 2.8 × 10–13 [NH3] = 0.30 [Cu2+] = 5.6 × 10–13 Next, determine the [OH–] from the reaction of NH3 with H2O NH3(aq) + H2O I C E [NH3] 0.30 –x 0.30 – x + – NH4 (aq) + OH (aq) [NH4+] – +x x NH + OH – 4 Kb = 1.8 × 10 = NH3 –5 OH– – +x x NH + OH – 4 Kb = 1.8 × 10–5 = NH3 [ x ][ x ] Kb = 1.8 × 10–5 = [0.30 – x ] assume x << 0.30 x = 2.3 × 10–3 = [OH–] Finally, with the concentration of OH– and the concentration of Cu2+, determine if any precipitate has formed: Cu(OH)2 Ksp = 4.8 × 10–20 = Cu 2+ OH – [Cu2+] = 5.6 × 10–13 Cu2+ + 2OH– [OH–] = 2.3 × 10–3 220 2 Chapter 18 2 Q = Cu 2+ OH – = (5.6 × 10–13)(2.3 × 10–3)2 = 3.0 × 10–18 Q > Ksp therefore a precipitate forms. Assume that all of the Cu2+ has reacted with the OH– and then the solution returns to equilibrium I C E [Cu2+] – +x +x [OH–] 2.3 × 10–3 + 2x 2.3 × 10–3 + 2x Ksp = 4.8 × 10–20 = Cu 2+ OH – 2 Ksp = 4.8 × 10–20 = [ x ] 2.3 ×10 –3 + 2x –15 2+ x = 9.1 × 10 M = [Cu ] 2 x << 2.3 × 10–3 18.105 Al(OH)3 has such an exceedingly small Ksp, there is almost no dissociation in pure water. Therefore, the pH would be expected to be about 7. In doing the calculations, this is borne out: K sp = Al3+ OH − 3 3 3 × 10−34 = [ x ][ 3x ] x = 1.8 × 10–9 [OH–] = 5.5 × 10–9 [OH–] = (1 × 10–7) + (5.5 × 10–9) = 1.055 × 10–7 pOH = 6.98 pH = 7.02 18.106 The value of K is very large, therfore, all of the EDTA4− will be consumed to produce the complex, PbEDTA2−. The reaction of Pb2+ with the EDTA4− is a one-to-one stoichiometric reaction, thus, Pb2+ is in excess. After the reaction we will have the following: [Pb2+] = 0.15 M – 0.10 M = 0.05 M [EDTA4−] = 0 [PbEDTA2+] = 0.10 M In reality, the system in in equilibirum so a very small amount of EDTA4− will exist in solution but the amount is insignificant. The hydroxide concentration can be determined from the pH of the solution. [OH−] = 10(14-12.5) = 3.2 x 10−2 Q = [Pb2+][OH−]2 = 0.05 x (3.2 x 10-2)2 = 5.1 x 10−5 This value is greater than Ksp for Pb(OH)2 which is 1.4 x 10−20 (CRC Handbook of Chemistry and Physics, 91st ed.,2010-2011), therefore, Pb(OH)2 will precipitate from the reaction mixture. 18.107 To answer this question we can use cation groupings from qualitative analysis. Ag+ and Pb2+ are group I ions. These ions can be separated from the others by adding Cl− to the solution. AgCl and PbCl2 will 221 Chapter 18 precipitate and can be filtered off from the solution. AgCl can be separated from PbCl2 by adding 6 M NH3. AgCl will react forming Ag(NH3)2+ while PbCl2 will remain a solid. Cu2+ and Bi3+ can be separated by adding 0.1 M H2S and 6M HCl. This will precipitate CuS and Bi2S3 which can be filtered The solids can be converted to soluble species by adding HNO3. Then, base can be added to precipitate hydroxides; Cu(OH)2 and Bi(OH)3. The addition of 6M NH3 will convert the Cu(OH)2 to Cu(NH3)42+ leaving the solid Bi(OH)3. Then add base to the remaining solution along with H2S to precipitate CoS and MnS. Filter the solution and dissolve the solid be adding acid. Cobalt can be separated by adding KNO2 which forms a yellow precipitate while Mn2+ remains in solution. Finally, Ba2+ and Ca+ can be separated by adding acid to make a weakly acidic solution and then adding Na2CrO4. BaCrO4 will precipitate while Ca2+ will remain in solution. 18.108 (1) (2) In order, Figure C Figure B Figure E Figures A and D are excluded because PbBr2 will precipitate before PbCl2. 18.109 The volume of a cone is given by V=(1/3)π r2h 3 1 2 12 in 2.54 cm x 3.1415 x ( 5 ft ) x 16ft x x 3 ft in 7 3 V = 1.19 x 10 cm 3 V= The mass of the cone is: 1.19 x 107 cm3 x 2.71 g cm-3 = 3.22 x 107 g CaCO3 Saturated CaCO3 has a molar concentration of Ca2+ equal to: [Ca2+] = [CO32-] = Ksp1/2 = (3.4 x 10-9)1/2 = 5.8 x 10-5 M Mass of CaCO3 dissolved in a liter of saturated solution is: 5.8 x 10-5 mol L-1 x (100.09 g/ mol CaCO3) = 5.8 x 10-3 g L-1 The volume of solution required to build the cone would be equal to: 3.22 x 107 g CaCO3 x (1 L/5.8 x 10-3 g) = 5.6 x 109 L 10 drops have a volume of 1 mL. (10 drops/1 mL) x (1000 mL/L) = 10000 drops/L The number of drops required to deposit enough CaCO3 is given by: (10000 drops/L) x 5.6 x 109L = 5.6 x 1013 drops The age of the stalagmite equals: 5.6 x 1013 drops x (1s/drop) x (1 hr/3600 s) x (1 day/24 hrs) x (1yr/365 days) = 1.8 x 106 yrs 18.110 First, we must calculate the mass of CaSO4 dissolved. The volume is a cylinder, with V = hπr2: 222 Chapter 18 h = 0.50 in (2.54 cm/1 inch) = 1.27 cm r = 1/2 diameter = 0.50 cm Therefore: V = 1.0 cm3, and mass = 1.0 cm3(0.97 g/cm3) = 0.97 g However, because it is a hydrate (CaSO4·2H2O), plaster is only (136.2/172.2 = 0.79) 79% calcium sulfate, therefore the true mass of CaSO4 is: Mass = 0.97(.79) = 0.77 g CaSO4 In moles, this is 0.77 g CaSO4 x (1 mol CaSO4 /136.2g) = 0.0056 mol CaSO4 Now we must calculate the volume of water necessary to dissolve 0.77 g of CaSO4. We start by finding its molar solubility: K sp = Ca 2+ SO42− = 4.9 × 10–5 K sp = [ x ][ x ] = 4.9 × 10–5 x = 7.0 × 10–3 mol/L So the volume of water needed is: 0.0056 mol(1 L/7.0 × 10–3 mol) = 0.8 L Finally, we find the amount of time needed to produce this much water: 0.8 L (1 day/2.00 L) = 0.40 days, or about 9.6 hours. 18.111 Start by determining the [Cl–] if the [Pb2+] = 0.0050 M. K sp Cl− = = 1.7 × 10−5 = 5.8 × 10−2 0.0050 Pb 2+ Then determine the concentration of chloride in a saturated solution, Ksp = 4x3 where x = [Pb2+] and [Cl–] = Ksp = 1.6 × 10−2 M, Cl− = 3.2 × 10−2 M 4 The volume of 0.10 M HCl which needs to be added is: M 1V 1 = M 2V 2 (x mL)(0.10 M HCl) = (3.2 × 10–2 M HCl)(x + 100 mL) 0.1 x = 3.2× 10–2x + 3.2 6.8 × 10–2 x = 3.2 x = 47 mL 47 mL of 0.10 M HCl need to be added. 2x, x = 3 18.112 CaCO3(s) + H+(aq) HCO3−(aq) + H+(aq) Ca2+(aq) + HCO3− (aq) H2CO3(aq) Combining these equations and solving for [Ca2+] we obtain, 2+ [Ca ] = K sp H + 2 K a1K a 2 pH = 5.6 223 Chapter 18 [Ca2+] = 3.4 x 10−9 10−5.6 ( 4.3 x 10 −7 )( 2 5.6 x 10 −11 ) = 8.9 x 10-4 M Therefore, the solubility of CaCO3 equals 8.9 x 10-4 M pH = 4.2 3.4 x 10−9 10−4.2 2+ [Ca ] = 2 ( 4.3 x 10 )( 5.6 x 10 ) −7 −11 = 0.56 M Therefore, the solubility of CaCO3 equals 0.56 M 18.113 Mg(OH)2(s) Mg2+(aq) + 2OH−(aq) Ksp = [Mg2+][OH−]2 [OH−] = 10−(14.0−9.8) = 6.3 x 10-5 M [Mg2+] = (6.3 x 10-5 mol OH−/L) x (1 mol Mg2+/2 mol OH−) = 3.2 x 10-5 M Ksp = 3.2 x 10-5 x (6.3 x 10-5)2 = 1.3 x 10-13 18.114 Π = iMRT M = Π/(iRT) i = 2 for NiCO3 1 atm 13 torr x 760 torr M= = 3.5 x 10− L atm x 298 K 2 x 0.082 K mol 4 Ksp = (3.5 x 10-4)2 = 1.2 x 10− 7 224