The solubility product for a salt of the type AB is `4xx10^(-8)`. What is the molarity of its standard solution?

Options

A. `2xx10^(-4)mol//L`
B. `16xx10^(-16)mol//L`
C. `2xx10^(-16)mol//L`
D. `4xx10^(-4)mol//L`

Answer

To solve the problem, we need to determine the molarity of the standard solution of the salt AB given its solubility product (Ksp).

### Step-by-Step Solution:

1. **Understanding the Dissociation of the Salt**:
The salt AB dissociates in water according to the following reaction:
\[ AB \rightleftharpoons A^+ + B^- \]
This means that for every mole of AB that dissolves, it produces one mole of \( A^+ \) and one mole of \( B^- \).

2. **Defining Solubility**:
Let the solubility of the salt AB be \( S \) mol/L. This means:
- The concentration of \( A^+ \) ions in solution will also be \( S \) mol/L.
- The concentration of \( B^- \) ions in solution will also be \( S \) mol/L.

3. **Writing the Expression for Ksp**:
The solubility product (Ksp) is given by the formula:
\[ K_{sp} = [A^+][B^-] \]
Substituting the concentrations we defined:
\[ K_{sp} = S \times S = S^2 \]

4. **Substituting the Given Ksp Value**:
We know from the problem that \( K_{sp} = 4 \times 10^{-8} \). Therefore, we can write:
\[ S^2 = 4 \times 10^{-8} \]

5. **Solving for S**:
To find \( S \), we take the square root of both sides:
\[ S = \sqrt{4 \times 10^{-8}} = \sqrt{4} \times \sqrt{10^{-8}} = 2 \times 10^{-4} \]

6. **Final Answer**:
Thus, the molarity of the standard solution of the salt AB is:
\[ S = 2 \times 10^{-4} \text{ mol/L} \]

### Conclusion:
The molarity of the standard solution of the salt AB is \( 2 \times 10^{-4} \) mol/L.