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 euquilibrium is important in understanding effects such as the influence of pH, complex fomation and common ion cffect. Equilibrium constant in solution should be written correctly using activities and not concentrations. The difference between thesc 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^(-10)` at `18^(@)`C, and M.W of AB is 143.5 g/mol.
The solubility in g/lit of AB is

Options

A. 14.35 gm/lit
B. `2.87xx10^(-3)` gm/lit
C. 1.43 gm / lit
D. 28.7 gm/lit

Answer

To solve the problem of finding the solubility of the AB type salt given its solubility product (Ksp) and molecular weight, we can follow these steps:

### Step-by-Step Solution:

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

2. **Write the Expression for Ksp**:
The solubility product constant \( K_{sp} \) for the reaction is given by:
\[ K_{sp} = [A^+][B^-] \]
If we let the solubility of AB be \( s \) (in moles per liter), then at equilibrium:
\[ [A^+] = s \quad \text{and} \quad [B^-] = s \]
Therefore, the expression for \( K_{sp} \) becomes:
\[ K_{sp} = s \cdot s = s^2 \]

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

4. **Solve for Solubility (s)**:
To find \( s \), take the square root of both sides:
\[ s = \sqrt{4 \times 10^{-10}} = 2 \times 10^{-5} \text{ moles per liter} \]

5. **Convert Solubility to grams per liter**:
Now, to convert the solubility from moles per liter to grams per liter, we use the molecular weight of AB:
\[ \text{Molecular Weight} = 143.5 \text{ g/mol} \]
Therefore, the solubility in grams per liter is:
\[ \text{Solubility (g/L)} = s \times \text{Molecular Weight} = (2 \times 10^{-5} \text{ mol/L}) \times (143.5 \text{ g/mol}) \]
\[ = 2 \times 143.5 \times 10^{-5} \text{ g/L} = 287 \times 10^{-5} \text{ g/L} = 0.0287 \text{ g/L} \]

6. **Final Result**:
The solubility of AB in grams per liter is:
\[ 28.7 \text{ g/L} \]

### Summary of the Solution:
The solubility of the AB type salt, given the solubility product \( K_{sp} = 4 \times 10^{-10} \) and the molecular weight of 143.5 g/mol, is \( 28.7 \text{ g/L} \).