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Chapter 15: Problem 198

Sulfuric acid, \(\mathrm{H}_{2} \mathrm{SO}_{4}\), is a diprotic acid with dissociation equilibrium constants of \(K_{\mathrm{eq}}>1.0 \times 10^{3}\) and \(K_{\mathrm{eq}}=1.2 \times 10^{-2} .\) Write the two dissociation equilibrium equations, and match the proper \(K_{\text {eq }}\) to each. Which species is the weak acid?

Short Answer

Expert verified
The two dissociation equilibrium equations for sulfuric acid are: 1. \(\mathrm{H}_{2} \mathrm{SO}_{4}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HSO}_{4}^{-}(aq)\), with \(K_{eq}>1.0 \times 10^{3}\) 2. \(\mathrm{HSO}_{4}^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{SO}_{4}^{2-}(aq)\), with \(K_{eq}=1.2 \times 10^{-2}\) The weak acid species is the bisulfate ion (\(\mathrm{HSO}_{4}^{-}\)), which is involved in the second dissociation step with a low equilibrium constant.

Step by step solution

01

Write the first dissociation equation of sulfuric acid

For the first dissociation, sulfuric acid loses one hydrogen ion to become a bisulfate ion. The equilibrium equation for this process is: \[\mathrm{H}_{2} \mathrm{SO}_{4}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HSO}_{4}^{-}(aq)\]
02

Write the second dissociation equation of the bisulfate ion

For the second dissociation, the bisulfate ion loses another hydrogen ion to become the sulfate ion. The equilibrium equation for this process is: \[\mathrm{HSO}_{4}^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{SO}_{4}^{2-}(aq)\]
03

Match the given equilibrium constants to the corresponding dissociation equations

We are given two equilibrium constants, \(K_{eq}>1.0 \times 10^{3}\) (high value indicating a strong dissociation) and \(K_{eq}=1.2 \times 10^{-2}\) (low value indicating a weak dissociation). Since sulfuric acid is known to be a strong acid, its first dissociation will have a large equilibrium constant: \(\mathrm{H}_{2} \mathrm{SO}_{4}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{HSO}_{4}^{-}(aq)\), with \(K_{eq}>1.0 \times 10^{3}\) For the second dissociation, which is expected to have a smaller equilibrium constant: \(\mathrm{HSO}_{4}^{-}(aq) \rightleftharpoons \mathrm{H}^{+}(aq) + \mathrm{SO}_{4}^{2-}(aq)\), with \(K_{eq}=1.2 \times 10^{-2}\)
04

Identify the weak acid species

In the second dissociation step, the bisulfate ion, \(\mathrm{HSO}_{4}^{-}\), is acting as a weak acid because it has a low equilibrium constant. Therefore, the weak acid species is the bisulfate ion (\(\mathrm{HSO}_{4}^{-}\)).

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