Consider an ionic solid that dissolves in water according to the equation:
`M_(n)X_(m(s)) nM_(aq)^(m+)n + mX_(aq)^(n-)` . The equilibrium constant for this reaction, `K _(sp)=[M^(m+)]^(n) [X^(n-)]^(m) ` is known as the solubility product of `M_(n), X_(m)`. The form of this equilibrium is important in understanding effects such as the influence of pH, complex formation and common ion effect. Equilibrium constant in solution should be written correctly using activities and not concentrations. The difference between these quantities is large in concentrated ionic solutions and `K_(sp)` is quantitatively reliable as a guide of solubilities only for very dilute solutions, If solubility product of AB type salt is `4xx 10^(-8)` at `18^(@)`C, and M.W of AB is 143.5 g/mol.
What is the molarity of its standard solution?

Options

A. `2X10^(-4)` mol/lit
B. `9X10^(-4)`mol/lit
C. `14 X 10^(-4)` mol/lit
D. `8 X 10^(-4)` mol/lit

Answer

To find the molarity of the standard solution of the AB type salt with a given solubility product (Ksp), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Dissolution Reaction:**
The dissolution of the salt AB can be represented as:
\[ AB_{(s)} \rightleftharpoons A^{+}_{(aq)} + B^{-}_{(aq)} \]
When one mole of AB dissolves, it produces one mole of \(A^{+}\) and one mole of \(B^{-}\).

2. **Defining Solubility (S):**
Let the solubility of AB be \(S\) mol/L. Therefore, at equilibrium:
- The concentration of \(A^{+}\) will be \(S\) mol/L.
- The concentration of \(B^{-}\) will also be \(S\) mol/L.

3. **Writing the Expression for Ksp:**
The solubility product \(K_{sp}\) for the dissolution of AB is given by:
\[ K_{sp} = [A^{+}][B^{-}] \]
Substituting the concentrations in terms of solubility \(S\):
\[ K_{sp} = S \cdot S = S^2 \]

4. **Substituting the Given Ksp Value:**
We are given that \(K_{sp} = 4 \times 10^{-8}\). Therefore, we can set up the equation:
\[ S^2 = 4 \times 10^{-8} \]

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

6. **Calculating Molarity of the Standard Solution:**
The molarity of the standard solution of the salt AB is equal to the solubility \(S\):
\[ \text{Molarity} = 2 \times 10^{-4} \text{ mol/L} \]

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

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