If the maximum concentration of `PbCl_2` in water is 0.01 M at `25^@C`, its maximum concentration in 0.1 M NaCl will be

Options

A. `2 xx 10^(-3) M`
B. `1.6 xx 10^(-2) M`
C. `1 xx 10^(-4) M`
D. `4 xx 10^(-4)M`

Answer

To solve the problem, we need to determine the maximum concentration of lead chloride (PbCl₂) in a 0.1 M NaCl solution, given that its maximum concentration in pure water is 0.01 M at 25°C.

### Step-by-Step Solution:

1. **Understanding the Dissociation of PbCl₂**:
- When PbCl₂ dissolves in water, it dissociates into its ions:
\[ \text{PbCl}_2 \rightleftharpoons \text{Pb}^{2+} + 2\text{Cl}^- \]
- For every mole of PbCl₂ that dissolves, it produces 1 mole of Pb²⁺ ions and 2 moles of Cl⁻ ions.

2. **Calculating Ion Concentrations in Pure Water**:
- Given that the maximum concentration of PbCl₂ in pure water is 0.01 M, the concentration of Pb²⁺ will also be 0.01 M.
- The concentration of Cl⁻ ions produced will be:
\[ 2 \times 0.01 \, \text{M} = 0.02 \, \text{M} \]

3. **Calculating the Solubility Product (Ksp)**:
- The solubility product (Ksp) expression for PbCl₂ is:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \]
- Substituting the concentrations:
\[ K_{sp} = (0.01)(0.02)^2 = (0.01)(0.0004) = 4 \times 10^{-6} \]

4. **Considering the Effect of Common Ion (Cl⁻) from NaCl**:
- When PbCl₂ is added to a solution containing 0.1 M NaCl, the concentration of Cl⁻ ions increases due to the NaCl dissociation:
\[ \text{NaCl} \rightleftharpoons \text{Na}^+ + \text{Cl}^- \]
- The concentration of Cl⁻ ions in the solution is now 0.1 M.

5. **Setting Up the Ksp Equation in NaCl Solution**:
- The Ksp expression remains the same:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \]
- Substituting the known values:
\[ 4 \times 10^{-6} = [\text{Pb}^{2+}](0.1)^2 \]
- This simplifies to:
\[ 4 \times 10^{-6} = [\text{Pb}^{2+}](0.01) \]

6. **Solving for [Pb²⁺]**:
- Rearranging the equation to find the concentration of Pb²⁺:
\[ [\text{Pb}^{2+}] = \frac{4 \times 10^{-6}}{0.01} = 4 \times 10^{-4} \, \text{M} \]

### Final Answer:
The maximum concentration of PbCl₂ in 0.1 M NaCl is **4 x 10⁻⁴ M**.