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

Step 1: Write the equilibrium expressions.
For reaction (i), \[K_{S1} = \frac{[H^+][HSO_3^-]}{[SO_{2(aq)}]}\] and for reaction (ii), \[K_{S2} = \frac{[H^+][SO_3^{2-}]}{[HSO_3^-]}\]

Step 2: Find the hydrogen ion concentration from the pH.
Since \(pH = 4.0\), \[[H^+] = 10^{-4}\ M\]

Step 3: Solve for the intermediate species \([HSO_3^-]\) using reaction (i).
Rearranging the expression for \(K_{S1}\): \[[HSO_3^-] = \frac{K_{S1}[SO_{2(aq)}]}{[H^+]} = \frac{1.3\times10^{-2}\times1.0}{10^{-4}} = 130\ M\]

Step 4: Solve for \([SO_3^{2-}]\) using reaction (ii).
Rearranging the expression for \(K_{S2}\): \[[SO_3^{2-}] = \frac{K_{S2}[HSO_3^-]}{[H^+]} = \frac{6.6\times10^{-8}\times130}{10^{-4}} = 8.58\times10^{-2}\ M\]

Step 5: Convert to mM.
\[8.58\times10^{-2}\ M = 85.8\ mM\] which sits neatly in the expected 80-90 mM range.