The `K_(sp)` of `PbCrO_4` is `1.0xx10^(-16)` . Then the molar solubility of `PbCrO_4` is

Options

A. `1.0xx10^(-6)`
B. `1.0xx10^(-4)`
C. `1.0xx10^(-16)`
D. `1.0xx10^(-8)`

Answer

To find the molar solubility of \( \text{PbCrO}_4 \) given its solubility product constant \( K_{sp} = 1.0 \times 10^{-16} \), we can follow these steps:

### Step 1: Write the dissociation equation for \( \text{PbCrO}_4 \)
The dissociation of lead(II) chromate in water can be represented as:
\[ \text{PbCrO}_4 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + \text{CrO}_4^{2-} (aq) \]

### Step 2: Define the molar solubility
Let the molar solubility of \( \text{PbCrO}_4 \) be \( S \). When \( \text{PbCrO}_4 \) dissolves, it produces:
- \( S \) moles of \( \text{Pb}^{2+} \)
- \( S \) moles of \( \text{CrO}_4^{2-} \)

### Step 3: Write the expression for \( K_{sp} \)
The solubility product constant \( K_{sp} \) is given by the product of the concentrations of the ions at equilibrium:
\[ K_{sp} = [\text{Pb}^{2+}][\text{CrO}_4^{2-}] \]
Substituting in the values from our dissociation:
\[ K_{sp} = (S)(S) = S^2 \]

### Step 4: Substitute the known value of \( K_{sp} \)
We know that \( K_{sp} = 1.0 \times 10^{-16} \):
\[ S^2 = 1.0 \times 10^{-16} \]

### Step 5: Solve for \( S \)
To find \( S \), take the square root of both sides:
\[ S = \sqrt{1.0 \times 10^{-16}} = 1.0 \times 10^{-8} \]

### Conclusion
The molar solubility of \( \text{PbCrO}_4 \) is:
\[ S = 1.0 \times 10^{-8} \, \text{mol/L} \]

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