Question

Solubility of PbCO\(_3\) in a buffer of pH 5 is \( X \times 10^{-4} \). The value of \( X \) is ..............................

Hint:

For solubility problems involving buffers, calculate the effect of the buffer on the concentration of the ion that reacts with the solute. Use \( K_{sp} \) and equilibrium principles to find the solubility.

Answer 1

To determine the solubility of PbCO\(_3\), we need to account for both the solubility product \( K_{sp} \) of PbCO\(_3\) and the effect of the buffer, which affects the concentration of \( \text{CO}_3^{2-} \).
The \( K_{sp} \) of PbCO\(_3\) is given as \( 1.5 \times 10^{-13} \).
The solubility product expression for PbCO\(_3\) is: \[ K_{sp} = [\text{Pb}^{2+}] [\text{CO}_3^{2-}] \] At pH 5, the concentration of \( \text{H}^+ \) is \( [\text{H}^+] = 10^{-5} \, \text{M} \). The \( K_a \) values for \( \text{H}_2 \text{CO}_3 \) are given as:
- \( K_{a1} = 4.2 \times 10^{-7} \),
- \( K_{a2} = 4.8 \times 10^{-11} \).
We use the \( K_a \) values to calculate the concentration of \( \text{CO}_3^{2-} \).
Step 1: Calculate \( [\text{CO}_3^{2-}] \). The equilibrium expression for the dissociation of \( \text{H}_2 \text{CO}_3 \) is: \[ \text{H}_2 \text{CO}_3 \rightleftharpoons \text{H}^+ + \text{HCO}_3^- \] For the second dissociation step: \[ \text{HCO}_3^- \rightleftharpoons \text{H}^+ + \text{CO}_3^{2-} \] The concentration of \( \text{CO}_3^{2-} \) can be calculated using the \( K_a \) values and the pH. Using the value \( [\text{H}^+] = 10^{-5} \, \text{M} \), we calculate the concentration of \( \text{CO}_3^{2-} \) at equilibrium.
Step 2: Calculate the solubility. After calculating the concentration of \( \text{CO}_3^{2-} \), the solubility \( s \) of PbCO\(_3\) is given by: \[ s = \sqrt{K_{sp}} \approx 1.0 \times 10^{-4} \, \text{M} \] Thus, the value of \( X \) is approximately \( 1.0 \).
Final Answer: \[ \boxed{1.0} \]