At 298 K, the solubility of `PbCl_(2)` is `2 xx 10^(-2)` mol/lit, then `K_(sp)=`

Options

A. `1 xx 10^(-7)`
B. `3.2 xx 10^(-7)`
C. `1 xx 10^(-5)`
D. `3.2 xx 10^(-5)`

Explanation

To find the solubility product constant (Ksp) of lead(II) chloride (PbCl2) at 298 K, 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 PbCl2 be \( S \) mol/L. According to the dissociation equation:
- For every 1 mole of PbCl2 that dissolves, 1 mole of Pb²⁺ and 2 moles of Cl⁻ are produced.
- Therefore, at equilibrium:
- The concentration of Pb²⁺ will be \( S \) mol/L.
- The concentration of Cl⁻ will be \( 2S \) mol/L.

### Step 3: Substitute the given solubility
From the problem, we know that the solubility \( S = 2 \times 10^{-2} \) mol/L.

### Step 4: Write the expression for Ksp
The solubility product constant (Ksp) is given by the expression:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \]
Substituting the concentrations we found:
\[ K_{sp} = (S)(2S)^2 \]

### Step 5: Simplify the expression
Now we can simplify the expression:
\[ K_{sp} = S \cdot (4S^2) = 4S^3 \]

### Step 6: Substitute the value of S
Now substitute \( S = 2 \times 10^{-2} \) mol/L into the Ksp expression:
\[ K_{sp} = 4(2 \times 10^{-2})^3 \]

### Step 7: Calculate the value
Calculating \( (2 \times 10^{-2})^3 \):
\[ (2 \times 10^{-2})^3 = 8 \times 10^{-6} \]

Now substituting this back into the Ksp expression:
\[ K_{sp} = 4 \times 8 \times 10^{-6} = 32 \times 10^{-6} \]

### Step 8: Final answer
Thus, we can express Ksp in scientific notation:
\[ K_{sp} = 3.2 \times 10^{-5} \]

### Conclusion
The value of \( K_{sp} \) for PbCl2 at 298 K is \( 3.2 \times 10^{-5} \).

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