Solubility product `(k_(sp))` of saturated `PbCl_(2)` in water is `1.8xx10^(-4)"mol"^(3)"dm"^(-9)`. What is the concentration of `Pb^(2+)` in the solution?

Options

A. `(0.45xx10^(-4))^((1)/(3))" mol dm"^(-3)`
B. `(1.8xx10^(-4))^((1)/(3))" mol dm"^(-3)`
C. `(0.4xx10^(-4))^((1)/(3))" mol dm"^(-3)`
D. `(2.0xx10^(-4))^((1)/(3))" mol dm"^(-3)`

Answer

To find the concentration of \( \text{Pb}^{2+} \) in a saturated solution of \( \text{PbCl}_2 \), we can follow these steps:

### Step 1: Write the Dissociation Equation
The dissociation of lead(II) chloride in water can be represented as:
\[ \text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{Cl}^- (aq) \]

### Step 2: Define Solubility
Let the solubility of \( \text{PbCl}_2 \) in water be \( S \) (in mol/dm³). From the dissociation equation, we can see that:
- The concentration of \( \text{Pb}^{2+} \) will be \( S \).
- The concentration of \( \text{Cl}^- \) will be \( 2S \).

### Step 3: Write the Expression for \( K_{sp} \)
The solubility product \( K_{sp} \) is given by the expression:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \]
Substituting the concentrations in terms of \( S \):
\[ K_{sp} = (S)(2S)^2 = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the Given \( K_{sp} \)
We know that \( K_{sp} \) for saturated \( \text{PbCl}_2 \) is \( 1.8 \times 10^{-4} \, \text{mol}^3/\text{dm}^9 \). Therefore, we can set up the equation:
\[ 4S^3 = 1.8 \times 10^{-4} \]

### Step 5: Solve for \( S \)
To find \( S \), rearrange the equation:
\[ S^3 = \frac{1.8 \times 10^{-4}}{4} \]
Calculating the right-hand side:
\[ S^3 = 0.45 \times 10^{-4} \]
Now, take the cube root:
\[ S = \left(0.45 \times 10^{-4}\right)^{1/3} \]

### Step 6: Calculate \( S \)
Using a calculator:
\[ S \approx 0.77 \times 10^{-2} \, \text{mol/dm}^3 \]

### Step 7: Find the Concentration of \( \text{Pb}^{2+} \)
Since the concentration of \( \text{Pb}^{2+} \) is equal to \( S \):
\[ [\text{Pb}^{2+}] = S \approx 0.77 \times 10^{-2} \, \text{mol/dm}^3 \]

### Final Answer
The concentration of \( \text{Pb}^{2+} \) in the solution is approximately \( 0.77 \times 10^{-2} \, \text{mol/dm}^3 \).

---