If the solubility of `Pbbr_(2)` is S g-mole per litre, its solubility product, considering it to be 80% ionized, is

Options

A. `2.048 S^(2)`
B. `20.48 S^(3)`
C. `2.048 S^(3)`
D. `2.048 S^(4)`

Explanation

To find the solubility product (Ksp) of PbBr2, given that it is 80% ionized and its solubility is S g-mole per liter, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Dissociation Equation:**
The dissociation of PbBr2 in water can be represented as:
\[ \text{PbBr}_2 (s) \rightleftharpoons \text{Pb}^{2+} (aq) + 2 \text{Br}^- (aq) \]

2. **Define the Solubility:**
Let the solubility of PbBr2 be \( S \) g-moles per liter. This means that initially, we have \( S \) moles of PbBr2 that can dissociate.

3. **Calculate the Ionization:**
Since it is given that PbBr2 is 80% ionized, we can calculate the concentration of ions produced:
- The concentration of Pb²⁺ ions will be \( 0.8S \) (80% of S).
- The concentration of Br⁻ ions will be \( 2 \times 0.8S = 1.6S \) (since two Br⁻ ions are produced for every PbBr2 that dissolves).

4. **Write the Expression for Ksp:**
The solubility product (Ksp) expression for PbBr2 is:
\[ K_{sp} = [\text{Pb}^{2+}][\text{Br}^-]^2 \]
Substituting the concentrations we found:
\[ K_{sp} = (0.8S)(1.6S)^2 \]

5. **Calculate Ksp:**
Now, simplify the expression:
\[ K_{sp} = (0.8S)(1.6^2S^2) \]
\[ K_{sp} = (0.8S)(2.56S^2) = 2.048S^3 \]

6. **Final Result:**
Therefore, the solubility product of PbBr2, considering it to be 80% ionized, is:
\[ K_{sp} = 2.048S^3 \]