Which of the following compounds will be more soluble in acidic solution than in water: (a) CuS, (b) AgCl, (c) $PbSO_{4}$ ?

Strategy In each case, write the dissociation reaction of the salt into its cation and anion. The cation will not interact with the $H^{+}$ ion because they both bear positive charges. The anion will act as a proton acceptor only if it is the conjugate base of a weak acid. How would the removal of the anion affect the solubility of the salt?

Question

Which of the following compounds will be more soluble in acidic solution than in water: (a) CuS, (b) AgCl, (c) [latex]PbSO_{4}[/latex]?

Strategy In each case, write the dissociation reaction of the salt into its cation and anion. The cation will not interact with the [latex]H^{+}[/latex] ion because they both bear positive charges. The anion will act as a proton acceptor only if it is the conjugate base of a weak acid. How would the removal of the anion affect the solubility of the salt?

Accepted Answer

(a) The solubility equilibrium for CuS is

[latex]CuS(s) \xrightleftharpoons[]{} Cu^{2+}(aq) + S^{2-}(aq)[/latex]

The sulfide ion is the conjugate base of the weak acid [latex]HS^{-}[/latex]. Therefore, the [latex]S^{2-}[/latex] ion reacts with the [latex]H^{+}[/latex] ion as follows:

[latex]S^{2-}(aq)+H^+(aq)→HS^-(aq)[/latex]

This reaction removes the [latex]S^{2-}[/latex] ions from solution. According to Le Châtelier’s principle, the equilibrium will shift to the right to replace some of the [latex]S^{2-}[/latex] ions that were removed, thereby increasing the solubility of CuS.

(b) The solubility equilibrium is

[latex]AgCl(s) \xrightleftharpoons[]{} Ag^{+}(aq) + Cl^{-}(aq)[/latex]

Because [latex]Cl^{-}[/latex] is the conjugate base of a strong acid (HCl), the solubility of AgCl is not affected by an acid solution.

(c) The solubility equilibrium for [latex]PbSO_{4}[/latex] is

[latex]PbSO_{4}(s)\xrightleftharpoons[]{} Pb^{2+}(aq)+SO^{2-}_{4}(aq) [/latex]

The sulfate ion is a weak base because it is the conjugate base of the weak acid

[latex]HSO^{-}_{4}[/latex] Therefore, the [latex]SO^{2-}_{4}[/latex] ion reacts with the [latex]H^{+}[/latex] ion as follows:

[latex]SO^{2-}_{4}(aq)+H^{+}(aq)\rightarrow HSO^{-}_{4}(aq)[/latex]

This reaction removes the [latex]SO^{2-}_{4}[/latex] ions from solution. According to Le Châtelier’s principle, the equilibrium will shift to the right to replace some of the [latex]SO^{2-}_{4}[/latex] ions that were removed, thereby increasing the solubility of [latex]PbSO_{4}[/latex].

Question 16.13: Which of the following compounds will be more soluble in aci...

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