Solubility product of `PbC_(2)` at 298K is `1.0 xx 10^(-6)` At this temperature solubility of `PbCI_(2)` in moles per litre is

Options

A. `(1.0 xx 10^(6))^(1//2)`
B. `(1.0 xx 10^(-6))^(1//3)`
C. `(0.25 xx 10^(-6))^(1//3)`
D. `(0.25 xx 10^(-6))^(1//2)`

Answer

To find the solubility of PbCl₂ at 298K given its solubility product (Ksp) of 1.0 x 10^(-6), we can follow these steps:

### Step 1: Write the dissociation equation
PbCl₂ dissociates in water as follows:
\[ \text{PbCl}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{Cl}^- (aq) \]

### Step 2: Define solubility
Let the solubility of PbCl₂ be \( S \) moles per liter. Thus, at equilibrium:
- The concentration of Pb²⁺ ions will be \( S \) moles per liter.
- The concentration of Cl⁻ ions will be \( 2S \) moles per liter (since there are two chloride ions for each formula unit of PbCl₂).

### Step 3: Write the expression for Ksp
The solubility product (Ksp) expression for PbCl₂ is given by:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \]
Substituting the concentrations from step 2, we get:
\[ K_{sp} = S \cdot (2S)^2 \]

### Step 4: Substitute Ksp value
Now substituting the value of Ksp:
\[ 1.0 \times 10^{-6} = S \cdot (2S)^2 \]
\[ 1.0 \times 10^{-6} = S \cdot 4S^2 \]
\[ 1.0 \times 10^{-6} = 4S^3 \]

### Step 5: Solve for S
Now, we can solve for \( S \):
\[ S^3 = \frac{1.0 \times 10^{-6}}{4} \]
\[ S^3 = 2.5 \times 10^{-7} \]
To find \( S \), take the cube root:
\[ S = \sqrt[3]{2.5 \times 10^{-7}} \]

### Step 6: Calculate the value
Calculating the cube root:
\[ S \approx 0.29 \times 10^{-2} \]
\[ S \approx 2.9 \times 10^{-3} \text{ moles per liter} \]

Thus, the solubility of PbCl₂ at 298K is approximately \( 2.9 \times 10^{-3} \) moles per liter.

### Final Answer
The solubility of PbCl₂ at 298K is approximately **2.9 x 10^(-3) moles per liter**.

---