Question

The following two reactions describe the chemistry of S(IV) in an aquatic system.

Under ideal conditions, the equilibrium constants for reactions (i) and (ii) are \(1.3\times10^{-2}\ M\) (\(K_{S1}\)) and \(6.6\times10^{-8}\ M\) (\(K_{S2}\)), respectively.

If the pH of the system is 4.0 and the equilibrium concentration of \(SO_{2(aq)}\) is 1.0 M, the equilibrium concentration of \(SO_3^{2-}\) is ______ mM (rounded off to one decimal place).

\[SO_{2(aq)} \rightleftharpoons H^+ + HSO_3^- \quad (i)\]

\[HSO_3^- \rightleftharpoons H^+ + SO_3^{2-} \quad (ii)\]

Hint:

Use the first equilibrium to find [HSO3-] from [SO2(aq)] and [H+], then feed that into the second equilibrium to get [SO3(2-)]; remember [H+] comes straight from the pH.

Answer 1

Correct Answer: 85.8

Instead of solving the two equilibria one after another, it is quicker to combine them into a single overall reaction and use a combined equilibrium constant.

Adding reaction (i) and reaction (ii) together gives the overall dissociation $$SO_{2(aq)} \rightleftharpoons 2H^+ + SO_3^{2-}$$

and the equilibrium constant for this combined reaction is simply the product of the two individual constants: $$K_{overall} = K_{S1} \times K_{S2} = (1.3\times10^{-2})(6.6\times10^{-8}) = 8.58\times10^{-10}$$

The equilibrium expression for the combined reaction is $$K_{overall} = \frac{[H^+]^2[SO_3^{2-}]}{[SO_{2(aq)}]}$$

Rearranging for $[SO_3^{2-}]$ and substituting $[H^+] = 10^{-4}$ M and $[SO_{2(aq)}] = 1.0$ M: $$[SO_3^{2-}] = \frac{K_{overall}\,[SO_{2(aq)}]}{[H^+]^2} = \frac{8.58\times10^{-10}\times1.0}{(10^{-4})^2} = \frac{8.58\times10^{-10}}{10^{-8}} = 8.58\times10^{-2}\ M$$

Converting to millimolar units gives the same result as the stepwise method, which is reassuring since both are algebraically equivalent. \[\boxed{[SO_3^{2-}] \approx 85.8\ mM}\]