Question

Write out the stepwise \(K_{\mathrm{a}}\) reactions for the diprotic acid \(\mathrm{H}_{2} \mathrm{SO}_{3}\).

Short answer

The stepwise \(K_{a}\) reactions for the diprotic acid \(\mathrm{H}_{2}\mathrm{SO}_{3}\) are as follows: 1. \(\mathrm{H}_{2}\mathrm{SO}_{3}\, (aq) \rightleftharpoons \mathrm{H}^{+}\, (aq) + \mathrm{HSO}_{3}^{-}\, (aq)\), with \(K_{a1} = \frac{[\mathrm{H}^{+}][\mathrm{HSO}_{3}^{-}]}{[\mathrm{H}_{2}\mathrm{SO}_{3}]}\) 2. \(\mathrm{HSO}_{3}^{-}\, (aq) \rightleftharpoons \mathrm{H}^{+}\, (aq) + \mathrm{SO}_{3}^{2-}\, (aq)\), with \(K_{a2} = \frac{[\mathrm{H}^{+}][\mathrm{SO}_{3}^{2-}]}{[\mathrm{HSO}_{3}^{-}]}\)

Step-by-step solution

Write out the first reaction

For the first step, \(\mathrm{H}_{2}\mathrm{SO}_{3}\) donates its first proton. We'll write this equilibrium reaction: \[\mathrm{H}_{2}\mathrm{SO}_{3}\, (aq) \rightleftharpoons \mathrm{H}^{+}\, (aq) + \mathrm{HSO}_{3}^{-}\, (aq)\]

Write out the second reaction

Now that the first proton has been donated, the bisulfite ion (\(\mathrm{HSO}_{3}^{-}\)) can donate its second proton. We'll write this equilibrium reaction: \[\mathrm{HSO}_{3}^{-}\, (aq) \rightleftharpoons \mathrm{H}^{+}\, (aq) + \mathrm{SO}_{3}^{2-}\, (aq)\]

Write out the expressions for each \(K_{a}\)

Each reaction has an associated equilibrium constant, referred to as the acid dissociation constant, \(K_{a}\). We'll write expressions for the first \(K_{a}\) (\(K_{a1}\)) and the second \(K_{a}\) (\(K_{a2}\)) values. For the first reaction, \[K_{a1} = \frac{[\mathrm{H}^{+}][\mathrm{HSO}_{3}^{-}]}{[\mathrm{H}_{2}\mathrm{SO}_{3}]}\] For the second reaction, \[K_{a2} = \frac{[\mathrm{H}^{+}][\mathrm{SO}_{3}^{2-}]}{[\mathrm{HSO}_{3}^{-}]}\] Now, we have the stepwise \(K_{a}\) reactions and their corresponding expressions for the diprotic acid \(\mathrm{H}_{2}\mathrm{SO}_{3}\).