The solubility of `PbCl_(2)` is

Options

A. `sqrt(K_(sp))`
B. `3sqrt(K_(sp))`
C. `3sqrt((K_(sp))/(4))`
D. `sqrt(8K_(sp))`

Explanation

To find the solubility of \( \text{PbCl}_2 \), we can follow these steps:

### Step 1: Write the Dissociation Equation
When lead(II) chloride (\( \text{PbCl}_2 \)) dissolves in water, it dissociates into lead ions and chloride ions:
\[ \text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{Cl}^- (aq) \]

### Step 2: Define Solubility
Let the solubility of \( \text{PbCl}_2 \) be \( S \) moles per liter. This means that when \( \text{PbCl}_2 \) dissolves:
- The concentration of \( \text{Pb}^{2+} \) ions will be \( S \) mol/L.
- The concentration of \( \text{Cl}^- \) ions will be \( 2S \) mol/L (since 2 moles of chloride ions are produced for every mole of \( \text{PbCl}_2 \)).

### Step 3: Write the Expression for Ksp
The solubility product constant (\( K_{sp} \)) for the dissociation of \( \text{PbCl}_2 \) can be expressed as:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (S)(2S)^2 \]
\[ K_{sp} = S \cdot 4S^2 = 4S^3 \]

### Step 4: Solve for Solubility (S)
To find the solubility \( S \), we rearrange the equation:
\[ S^3 = \frac{K_{sp}}{4} \]
Taking the cube root gives us:
\[ S = \sqrt[3]{\frac{K_{sp}}{4}} \]

### Conclusion
Thus, the solubility of \( \text{PbCl}_2 \) in terms of \( K_{sp} \) is:
\[ S = \sqrt[3]{\frac{K_{sp}}{4}} \]