If `K_(sp) (PbSO_(4)) =1.8 xx 10^(-8)` and `K_(a) (HSO_(4)^(Theta)) = 1.0 xx 10^(-2)` the equilibrium constant for the reaction.
`PbSO_(4)(s) +H^(o+)(aq) hArr HSO_(4)^(Theta)(aq) +Pb^(2+)(aq)` is

Options

A. `1.8 xx 10^(-6)`
B. `1.8 xx 10^(-10)`
C. `2.8 xx 10^(-10)`
D. `1.0 xx 10^(-2)`

Explanation

To find the equilibrium constant for the reaction

\[ \text{PbSO}_4(s) + \text{H}^+(aq) \rightleftharpoons \text{HSO}_4^-(aq) + \text{Pb}^{2+}(aq) \]

we will use the given solubility product constant \( K_{sp} \) for lead(II) sulfate (\( \text{PbSO}_4 \)) and the acid dissociation constant \( K_a \) for bisulfate (\( \text{HSO}_4^- \)).

### Step-by-Step Solution:

1. **Write the dissociation reaction for \( \text{PbSO}_4 \)**:
\[ \text{PbSO}_4(s) \rightleftharpoons \text{Pb}^{2+}(aq) + \text{SO}_4^{2-}(aq) \]
The solubility product constant \( K_{sp} \) for this reaction is given by:
\[ K_{sp} = [\text{Pb}^{2+}][\text{SO}_4^{2-}] \]
Given \( K_{sp} = 1.8 \times 10^{-8} \).

2. **Write the dissociation reaction for \( \text{HSO}_4^- \)**:
\[ \text{HSO}_4^-(aq) \rightleftharpoons \text{H}^+(aq) + \text{SO}_4^{2-}(aq) \]
The acid dissociation constant \( K_a \) for this reaction is given by:
\[ K_a = \frac{[\text{H}^+][\text{SO}_4^{2-}]}{[\text{HSO}_4^-]} \]
Given \( K_a = 1.0 \times 10^{-2} \).

3. **Determine the equilibrium constant \( K \) for the target reaction**:
The equilibrium constant for the reaction can be expressed in terms of \( K_{sp} \) and \( K_a \):
\[ K = \frac{K_{sp}}{K_a} \]
Substituting the known values:
\[ K = \frac{1.8 \times 10^{-8}}{1.0 \times 10^{-2}} \]

4. **Perform the calculation**:
\[ K = 1.8 \times 10^{-8} \times 10^{2} = 1.8 \times 10^{-6} \]

5. **Final answer**:
The equilibrium constant for the reaction is:
\[ K = 1.8 \times 10^{-6} \]