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)M` to `10^(-7)M`
B. between `10^(-7) M` to `10^(-6)M`
C. `gt 10^(-5) M`
D. `lt 10^(-8) M`

Explanation

To determine the concentration of \( B^- \) ions at which the salt \( AB \) will precipitate, we can use the concept of the solubility product constant (\( K_{sp} \)).

### Step-by-Step Solution:

1. **Understanding the Solubility Product**:
The solubility product \( K_{sp} \) for the salt \( AB \) can be expressed as:
\[ K_{sp} = [A^+] [B^-] \]
where \( [A^+] \) is the concentration of \( A^+ \) ions and \( [B^-] \) is the concentration of \( B^- \) ions.

2. **Given Values**:
From the problem, we know:
- \( K_{sp} = 1 \times 10^{-8} \, M^2 \)
- \( [A^+] = 10^{-3} \, M \)

3. **Substituting Known Values**:
We can substitute the known values into the \( K_{sp} \) expression:
\[ 1 \times 10^{-8} = (10^{-3}) [B^-] \]

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

5. **Conclusion**:
The salt \( AB \) will begin to precipitate when the concentration of \( B^- \) ions is equal to or exceeds \( 1 \times 10^{-5} \, M \).

### Final Answer:
The salt will precipitate when the concentration of \( B^- \) ions is kept at \( 1 \times 10^{-5} \, M \) or higher.