The solubility of` PbCl_(2)` at `25^(@)C` is `6.3 xx 10^(-3)` mole/litre. Its solubility product at that temperature is

Options

A. `(6.3 xx 10^(-3)) xx (6.3 xx 10^(-3))`
B. `(6.3 xx 10^(-3)) xx (12.6 xx 10^(-3))`
C. `(6.3 xx 10^(-3)) xx (12.6 xx 10^(-3))^(2)`
D. `(12.6 xx 10^(-3)) xx (12.6 xx 10^(-3))`

Explanation

To find the solubility product (Ksp) of PbCl₂ at 25°C, 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 the solubility
Let the solubility of PbCl₂ be \( s \) moles per liter. According to the problem, \( s = 6.3 \times 10^{-3} \) moles/liter.

### Step 3: Determine the concentrations of ions
From the dissociation equation, we can see that:
- For every 1 mole of PbCl₂ that dissolves, 1 mole of Pb²⁺ ions and 2 moles of Cl⁻ ions are produced.
- Therefore, at equilibrium:
- The concentration of Pb²⁺ ions = \( s = 6.3 \times 10^{-3} \) M
- The concentration of Cl⁻ ions = \( 2s = 2 \times 6.3 \times 10^{-3} = 1.26 \times 10^{-2} \) M

### Step 4: Write the expression for Ksp
The solubility product constant (Ksp) for the dissociation of PbCl₂ can be expressed as:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Cl}^-]^2 \]

### Step 5: Substitute the values into the Ksp expression
Now we can substitute the concentrations into the Ksp expression:
\[ K_{sp} = (6.3 \times 10^{-3}) \times (1.26 \times 10^{-2})^2 \]

### Step 6: Calculate Ksp
Calculating the square of the concentration of Cl⁻:
\[ (1.26 \times 10^{-2})^2 = 1.5876 \times 10^{-4} \]

Now substituting this back into the Ksp expression:
\[ K_{sp} = (6.3 \times 10^{-3}) \times (1.5876 \times 10^{-4}) \]
\[ K_{sp} = 9.996 \times 10^{-7} \]

### Step 7: Round the answer
Rounding to three significant figures, we get:
\[ K_{sp} \approx 1.0 \times 10^{-6} \]

### Final Answer:
The solubility product (Ksp) of PbCl₂ at 25°C is approximately \( 1.0 \times 10^{-6} \).

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