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Chapter 14: Problem 103

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

Short Answer

Expert verified
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

01

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: \[\mathrm{H}_{2}\mathrm{SO}_{3}\, (aq) \rightleftharpoons \mathrm{H}^{+}\, (aq) + \mathrm{HSO}_{3}^{-}\, (aq)\]
02

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: \[\mathrm{HSO}_{3}^{-}\, (aq) \rightleftharpoons \mathrm{H}^{+}\, (aq) + \mathrm{SO}_{3}^{2-}\, (aq)\]
03

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 \(\mathrm{H}_{2}\mathrm{SO}_{3}\).

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Most popular questions from this chapter

Calculate the \(\mathrm{pH}\) and \(\left[\mathrm{S}^{2-}\right]\) in a \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{~S}\) solution. Assume \(K_{\mathrm{a}_{1}}=1.0 \times 10^{-7} ; K_{\mathrm{a}_{2}}=1.0 \times 10^{-19}\)

Trichloroacetic acid \(\left(\mathrm{CCl}_{3} \mathrm{CO}_{2} \mathrm{H}\right)\) is a corrosive acid that is used to precipitate proteins. The \(\mathrm{pH}\) of a \(0.050 \mathrm{M}\) solution of trichloroacetic acid is the same as the \(\mathrm{pH}\) of a \(0.040 \mathrm{M} \mathrm{HClO}_{4}\) solution. Calculate \(K_{\mathrm{a}}\) for trichloroacetic acid.

Monochloroacetic acid, \(\mathrm{HC}_{2} \mathrm{H}_{2} \mathrm{ClO}_{2}\), is a skin irritant that is used in "chemical peels" intended to remove the top layer of dead skin from the face and ultimately improve the complexion. The value of \(K_{\mathrm{a}}\) for monochloroacetic acid is \(1.35 \times 10^{-3}\). Calculate the \(\mathrm{pH}\) of a \(0.10 \mathrm{M}\) solution of monochloroacetic acid.

Arsenic acid \(\left(\mathrm{H}_{3} \mathrm{AsO}_{4}\right)\) is a triprotic acid with \(K_{\mathrm{a}_{1}}=5 \times 10^{-3}\) \(K_{\mathrm{a}_{2}}=8 \times 10^{-8}\), and \(K_{\mathrm{a}_{3}}=6 \times 10^{-10} \cdot\) Calculate \(\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\) \(\left[\mathrm{H}_{3} \mathrm{AsO}_{4}\right],\left[\mathrm{H}_{2} \mathrm{AsO}_{4}^{-}\right],\left[\mathrm{HAsO}_{4}^{2-}\right]\), and \(\left[\mathrm{AsO}_{4}^{3-}\right]\) in a \(0.20 \mathrm{M}\) arsenic acid solution.

A \(0.20 \mathrm{M}\) sodium chlorobenzoate \(\left(\mathrm{NaC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution has a \(\mathrm{pH}\) of \(8.65 .\) Calculate the \(\mathrm{pH}\) of a \(0.20 \mathrm{M}\) chlorobenzoic acid \(\left(\mathrm{HC}_{7} \mathrm{H}_{4} \mathrm{ClO}_{2}\right)\) solution.
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