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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 \(\mathrm{pH}\) of the following solutions. a. \(0.10 \mathrm{M} \mathrm{NaOH}\) b. \(1.0 \times 10^{-10} \mathrm{M} \mathrm{NaOH}\) c. \(2.0 \mathrm{M} \mathrm{NaOH}\)

At \(25^{\circ} \mathrm{C}\), a saturated solution of benzoic acid \(\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) has a pH of \(2.80\). Calculate the water solubility of benzoic acid in moles per liter.

Consider \(1000 . \mathrm{mL}\) of a \(1.00 \times 10^{-4} M\) solution of a certain acid HA that has a \(K_{\mathrm{a}}\) value equal to \(1.00 \times 10^{-4} .\) How much water was added or removed (by evaporation) so that a solution remains in which \(25.0 \%\) of HA is dissociated at equilibrium? Assume that HA is nonvolatile.

A solution is prepared by dissolving \(0.56 \mathrm{~g}\) benzoic acid \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}, K_{\mathrm{a}}=6.4 \times 10^{-5}\right)\) in enough water to make \(1.0 \mathrm{~L}\) of solution. Calculate \(\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2} \mathrm{H}\right],\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{CO}_{2}^{-}\right],\left[\mathrm{H}^{+}\right],\left[\mathrm{OH}^{-}\right]\), and the \(\mathrm{pH}\) of this solution.

Papaverine hydrochloride (abbreviated papH \(^{+} \mathrm{Cl}^{-} ;\) molar mass \(=\) \(378.85 \mathrm{~g} / \mathrm{mol}\) ) is a drug that belongs to a group of medicines called vasodilators, which cause blood vessels to expand, thereby increasing blood flow. This drug is the conjugate acid of the weak base papaverine (abbreviated pap; \(K_{\mathrm{b}}=8.33 \times 10^{-9}\) at \(35.0^{\circ} \mathrm{C}\) ). Calculate the \(\mathrm{pH}\) of a \(30.0 \mathrm{mg} / \mathrm{mL}\) aqueous dose of papH \(^{+} \mathrm{Cl}^{-}\) prepared at \(35.0^{\circ} \mathrm{C} . K_{\mathrm{w}}\) at \(35.0^{\circ} \mathrm{C}\) is \(2.1 \times 10^{-14}\).
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