Chapter 18: Solubility and Complex-Ion Equilibria 943 93.(E)The answer is (d). See the reasoning below. (a)Wrong, because the stoichiometry is wrong (b)Wrong, because Ksp= [Pb2+]·[I-]2(c)Wrong, because [Pb2+] = Ksp/[ I-]2(d)Correct because of the [Pb2+]:2[I-] stoichiometry 94.(E)The answer is (a). 2244BaSOs BaSO. If Na2SO4is added, the common–ion effect forces the equilibrium to the left and reduces [Ba2+]. 95.(E)The answer is (c). Choices (b) and (d) reduce solubility because of the common–ion effect. In the case of choice (c), the diverse non-common–ion effect or the “salt effect” causes more Ag2CrO4to dissolve. 96.(E)The answer is (b). The sulfate salt of Cu is soluble, whereas the Pb salt is insoluble. 97.(M)The answers are (c) and (d). Adding NH3causes the solution to become basic (adding OH–). Mg, Fe, Cu, and Al all have insoluble hydroxides. However, only Cu can form a complex ion with NH3, which is soluble. In the case of (NH4)2SO4, it is slightly acidic and dissolved readily in a base. 98.(M)The answer is (a). CaCO3is slightly basic, so it is more soluble in an acid. The only option for an acid given is NH4Cl. 99.(M)The answer is (c). Referring to Figure 18-7, it is seen that ammonia is added to an aqueous H2S solution to precipitate more metal ions. Since ammonia is a base, increasing the pH should cause more precipitation. 100.(M)(a)H2C2O4is a moderately strong acid, so it is more soluble in a basic solution. (b)MgCO3is slightly basic, so it is more soluble in an acidic solution. (c)CdS is more soluble in acidic solutions, but the solubility is still so small that it is essentially insoluble even in acidic solutions. (d)KCl is a neutral salt, and therefore its solubility is independent of pH. (e)NaNO3is a neutral salt, and therefore its solubility is independent of pH. (f)Ca(OH)2, a strong base, is more soluble in an acidic solution. 101.(E)The answer is NH3. NaOH(aq) precipitates both, and HCl(aq) precipitates neither. Mg(OH)2precipitates from an NH3(aq) solution but forms the soluble complex Cu(NH3)4(OH)2.