Which of the following salts is the most basic in aqueous solution?

We begin by recalling the rule that decides whether the aqueous solution of a salt is acidic, basic or neutral.

Statement of the rule: A salt obtained from

$$\text{(i) strong\;acid\;+\;strong\;base}\;\;\; \Rightarrow \;\;\; \text{neutral solution},$$

$$\text{(ii) weak\;acid\;+\;strong\;base}\;\;\; \Rightarrow \;\;\; \text{basic solution},$$

$$\text{(iii) strong\;acid\;+\;weak\;base}\;\;\; \Rightarrow \;\;\; \text{acidic solution},$$

$$\text{(iv) weak\;acid\;+\;weak\;base}\;\;\; \Rightarrow \;\;\; \text{pH depends on }K_a\text{ and }K_b.$$

The reason is simple: in water the ions coming from the weak component undergo hydrolysis, while the ions coming from the strong component do not.

Now we examine every given salt one by one.

Option A: $$\text{Pb(CH}_3\text{COO)}_2$$

This salt is produced by the combination

$$\text{Pb(OH)}_2\;(\text{weak base}) + 2\,\text{CH}_3\text{COOH}\;(\text{weak acid}) \rightarrow \text{Pb(CH}_3\text{COO)}_2 + 2\,\text{H}_2\text{O}$$

Both parent acid and base are weak, so the solution may be slightly acidic, slightly basic or nearly neutral depending on the relative magnitudes of $$K_a$$ and $$K_b$$. Therefore it cannot be confidently labelled “most basic”.

Option B: $$\text{Al(CN)}_3$$

This salt comes from

$$\text{Al(OH)}_3\;(\text{weak base}) + 3\,\text{HCN}\;(\text{weak acid}) \rightarrow \text{Al(CN)}_3 + 3\,\text{H}_2\text{O}$$

The cation $$\text{Al}^{3+}$$ is a highly charged, small ion that undergoes pronounced hydrolysis producing $$\text{H}^+$$:

$$\text{Al}^{3+} + 3\,\text{H}_2\text{O} \rightleftharpoons \text{Al(OH)}_3 + 3\,\text{H}^+$$

This hydrolysis drives the pH to the acidic side. At the same time the anion $$\text{CN}^-$$ is basic, but the strong acidity introduced by $$\text{Al}^{3+}$$ dominates. Thus the overall solution is acidic rather than basic.

Option C: $$\text{CH}_3\text{COOK}$$

This salt is obtained from

$$\text{CH}_3\text{COOH}\;(\text{weak acid}) + \text{KOH}\;(\text{strong base}) \rightarrow \text{CH}_3\text{COOK} + \text{H}_2\text{O}$$

Here the cation $$\text{K}^+$$ comes from a strong base and therefore does not hydrolyse. The anion $$\text{CH}_3\text{COO}^-$$ is the conjugate base of the weak acid acetic acid, so it undergoes hydrolysis according to

$$\text{CH}_3\text{COO}^- + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COOH} + \text{OH}^-$$

The production of $$\text{OH}^-$$ raises the pH; mathematically,

$$[\text{OH}^-] \gt [\text{H}^+] \;\; \Longrightarrow \;\; \text{pH} \gt 7.$$

Therefore the solution of $$\text{CH}_3\text{COOK}$$ is basic.

Option D: $$\text{FeCl}_3$$

This salt originates from

$$\text{Fe(OH)}_3\;(\text{weak base}) + 3\,\text{HCl}\;(\text{strong acid}) \rightarrow \text{FeCl}_3 + 3\,\text{H}_2\text{O}$$

Because the parent acid is strong and the parent base is weak, the salt is acidic. Indeed, $$\text{Fe}^{3+}$$ hydrolyses vigorously:

$$\text{Fe}^{3+} + 3\,\text{H}_2\text{O} \rightleftharpoons \text{Fe(OH)}_3 + 3\,\text{H}^+,$$

creating an acidic medium.

Now we compare all four salts. Only Option C, $$\text{CH}_3\text{COOK}$$, fulfils the criterion of being formed from a weak acid and a strong base, leading unambiguously to a basic solution. The other salts are either acidic or, at best, less basic because of competing acidic hydrolysis by the metal cation.

Hence, the correct answer is Option C.