Solubility product of a salt of AB is `1 xx 10^(-8) M ^(2)` in a solution in which the concentration of `A^(+)` ions is `10^(-3)M`. The salt will precipitate when the concentration of `B^(-)` ions is kept

Options

A. Between `10^(-8)` to `10^(-7)` M
B. Between `10^(-7)` to `10^(-8)` M
C. `gt 10^(-5) M`
D. `lt 10^(-8) M`

Explanation

To solve the problem, we need to determine the concentration of \( B^{-} \) ions at which the salt \( AB \) will start to precipitate. We are given the solubility product \( K_{sp} \) and the concentration of \( A^{+} \) ions.

### Step-by-Step Solution:

1. **Identify the given values:**
- The solubility product \( K_{sp} \) of salt \( AB \) is \( 1 \times 10^{-8} \, M^{2} \).
- The concentration of \( A^{+} \) ions is \( 10^{-3} \, M \).

2. **Write the expression for \( K_{sp} \):**
The dissociation of the salt \( AB \) can be represented as:
\[ AB \rightleftharpoons A^{+} + B^{-} \]
The solubility product \( K_{sp} \) is given by:
\[ K_{sp} = [A^{+}][B^{-}] \]

3. **Substitute the known values into the \( K_{sp} \) expression:**
We know \( K_{sp} = 1 \times 10^{-8} \, M^{2} \) and \( [A^{+}] = 10^{-3} \, M \). Thus, we can write:
\[ 1 \times 10^{-8} = (10^{-3})[B^{-}] \]

4. **Solve for \( [B^{-}] \):**
Rearranging the equation gives:
\[ [B^{-}] = \frac{1 \times 10^{-8}}{10^{-3}} = 1 \times 10^{-5} \, M \]

5. **Determine the condition for precipitation:**
The salt \( AB \) will start to precipitate when the product of the concentrations of \( A^{+} \) and \( B^{-} \) ions exceeds \( K_{sp} \). Therefore, precipitation occurs when:
\[ [A^{+}][B^{-}] > K_{sp} \]
Since we have calculated that \( [B^{-}] \) must be greater than \( 1 \times 10^{-5} \, M \) for precipitation to occur.

### Conclusion:
The salt \( AB \) will precipitate when the concentration of \( B^{-} \) ions is greater than \( 1 \times 10^{-5} \, M \).