Solubility product of a salt AB is `1 xx 10^(-8)` in a solution in which concentration of A is `10^(-3) M`. The salt will precipitate when the concentration of B becomes more than

Options

A. `10^(-4) M`
B. `10^(-7) M`
C. `10^(-6) M`
D. `10^(-5) M`

Explanation

To solve the problem, we need to determine the concentration of ion B (B⁻) at which the salt AB will start to precipitate in a solution where the concentration of ion A (A⁺) is already given as \(10^{-3} \, M\). The solubility product (Ksp) of the salt AB is given as \(1 \times 10^{-8}\).

### Step-by-Step Solution:

1. **Understand the Dissociation of the Salt:**
The salt AB dissociates in water as follows:
\[ AB \rightleftharpoons A^+ + B^- \]
The solubility product (Ksp) expression for this dissociation is:
\[ K_{sp} = [A^+][B^-] \]

2. **Substitute the Known Values:**
We know that:
\[ K_{sp} = 1 \times 10^{-8} \]
and the concentration of A⁺ is given as:
\[ [A^+] = 10^{-3} \, M \]

3. **Set Up the Ksp Equation:**
Substitute the known concentration of A⁺ into the Ksp expression:
\[ 1 \times 10^{-8} = (10^{-3})[B^-] \]

4. **Solve for [B⁻]:**
Rearranging the equation to find the concentration of B⁻:
\[ [B^-] = \frac{1 \times 10^{-8}}{10^{-3}} = 1 \times 10^{-5} \, M \]

5. **Determine the Precipitation Condition:**
The salt will start to precipitate when the ionic product exceeds the Ksp. Therefore, precipitation will occur when:
\[ [B^-] > 1 \times 10^{-5} \, M \]

### Conclusion:
The salt AB will start to precipitate when the concentration of B⁻ exceeds \(1 \times 10^{-5} \, M\).