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 298 K given the solubility product (Ksp), we can follow these steps:

### Step 1: Write the dissociation equation
The dissociation of lead(II) chloride (PbCl₂) 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 PbCl₂ be \( S \) moles per liter. This means:
- 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:
\[ K_{sp} = (S)(2S)^2 \]
\[ K_{sp} = S \cdot 4S^2 = 4S^3 \]

### Step 4: Substitute the value of Ksp
We know from the problem that \( K_{sp} = 1.0 \times 10^{-6} \). Therefore, we can set up the equation:
\[ 4S^3 = 1.0 \times 10^{-6} \]

### Step 5: Solve for S
To find the solubility \( S \), we rearrange the equation:
\[ S^3 = \frac{1.0 \times 10^{-6}}{4} \]
\[ S^3 = 2.5 \times 10^{-7} \]

Now, take the cube root of both sides:
\[ S = (2.5 \times 10^{-7})^{1/3} \]

### Step 6: Calculate the value of S
Calculating the cube root:
\[ S \approx 0.25 \times 10^{-2} \]
\[ S \approx 2.5 \times 10^{-2} \text{ moles per liter} \]

### Final Answer
The solubility of PbCl₂ in moles per liter at 298 K is approximately:
\[ S \approx 0.025 \text{ moles per liter} \]

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