What is the solubility of `PbSO_(4)` in `0.01M Na_(2)SO_(4)` solution if `K_(sp)` for `PbSO_(4) = 1.25 xx 10^(-9)`?

Options

A. `1.25 xx 10^(-7) mol L^(-1)`
B. `1.25 xx 10^(-9)mol L^(-1)`
C. `1.25 xx 10^(-10) mol L^(-1)`
D. `0.10 mol L^(-1)`

Explanation

To find the solubility of `PbSO4` in a `0.01 M Na2SO4` solution, we can follow these steps:

### Step 1: Write the dissociation equation for `PbSO4`
`PbSO4` dissociates in water as follows:
\[ PbSO_4 (s) \rightleftharpoons Pb^{2+} (aq) + SO_4^{2-} (aq) \]

### Step 2: Define the solubility
Let the solubility of `PbSO4` in the solution be \( S \). This means:
- The concentration of \( Pb^{2+} \) ions will be \( S \).
- The concentration of \( SO_4^{2-} \) ions from `PbSO4` will also be \( S \).

### Step 3: Consider the contribution of `Na2SO4`
Since we are adding `0.01 M Na2SO4` to the solution, it will dissociate completely:
\[ Na_2SO_4 (s) \rightleftharpoons 2Na^+ (aq) + SO_4^{2-} (aq) \]
From this dissociation:
- The concentration of \( SO_4^{2-} \) from `Na2SO4` will be \( 0.01 M \).

### Step 4: Calculate the total concentration of \( SO_4^{2-} \)
The total concentration of \( SO_4^{2-} \) in the solution will be the sum of the contributions from both `PbSO4` and `Na2SO4`:
\[ [SO_4^{2-}] = S + 0.01 \]
However, since \( S \) is expected to be very small compared to \( 0.01 \), we can approximate:
\[ [SO_4^{2-}] \approx 0.01 \]

### Step 5: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) for `PbSO4` is given by:
\[ K_{sp} = [Pb^{2+}][SO_4^{2-}] \]
Substituting the concentrations we have:
\[ K_{sp} = S \times 0.01 \]

### Step 6: Substitute the value of \( K_{sp} \)
Given that \( K_{sp} = 1.25 \times 10^{-9} \):
\[ 1.25 \times 10^{-9} = S \times 0.01 \]

### Step 7: Solve for \( S \)
Rearranging the equation to solve for \( S \):
\[ S = \frac{1.25 \times 10^{-9}}{0.01} \]
\[ S = 1.25 \times 10^{-7} \]

### Conclusion
The solubility of `PbSO4` in `0.01 M Na2SO4` solution is:
\[ S = 1.25 \times 10^{-7} \, M \]