Which has maximum solubility `AB,AB_(2),AB_(3)` and `AB_(4)` if `K_(SP)` for all the salts are `10^(-10) :`

Options

A. `AB`
B. `AB_(2)`
C. `AB_(3)`
D. `AB_(4)`

Explanation

To determine which compound among \( AB, AB_2, AB_3, \) and \( AB_4 \) has the maximum solubility given that their solubility products (\( K_{sp} \)) are all equal to \( 10^{-10} \), we can analyze the dissociation of each compound and derive their solubility expressions.

### Step-by-Step Solution:

1. **Dissociation of \( AB \)**:
\[ AB \rightleftharpoons A^+ + B^- \]
- Let the solubility of \( AB \) be \( s \).
- The concentration of \( A^+ \) and \( B^- \) ions at equilibrium will both be \( s \).
- Therefore, the expression for \( K_{sp} \) is:
\[ K_{sp} = [A^+][B^-] = s \cdot s = s^2 \]
- Setting \( K_{sp} = 10^{-10} \):
\[ s^2 = 10^{-10} \implies s = 10^{-5} \text{ M} \]

2. **Dissociation of \( AB_2 \)**:
\[ AB_2 \rightleftharpoons A^+ + 2B^- \]
- Let the solubility of \( AB_2 \) be \( s \).
- The concentration of \( A^+ \) will be \( s \) and that of \( B^- \) will be \( 2s \).
- Therefore, the expression for \( K_{sp} \) is:
\[ K_{sp} = [A^+][B^-]^2 = s \cdot (2s)^2 = s \cdot 4s^2 = 4s^3 \]
- Setting \( K_{sp} = 10^{-10} \):
\[ 4s^3 = 10^{-10} \implies s^3 = \frac{10^{-10}}{4} = 2.5 \times 10^{-11} \implies s = (2.5 \times 10^{-11})^{1/3} \approx 2.92 \times 10^{-4} \text{ M} \]

3. **Dissociation of \( AB_3 \)**:
\[ AB_3 \rightleftharpoons A^+ + 3B^- \]
- Let the solubility of \( AB_3 \) be \( s \).
- The concentration of \( A^+ \) will be \( s \) and that of \( B^- \) will be \( 3s \).
- Therefore, the expression for \( K_{sp} \) is:
\[ K_{sp} = [A^+][B^-]^3 = s \cdot (3s)^3 = s \cdot 27s^3 = 27s^4 \]
- Setting \( K_{sp} = 10^{-10} \):
\[ 27s^4 = 10^{-10} \implies s^4 = \frac{10^{-10}}{27} \implies s = \left(\frac{10^{-10}}{27}\right)^{1/4} \approx 1.78 \times 10^{-3} \text{ M} \]

4. **Dissociation of \( AB_4 \)**:
\[ AB_4 \rightleftharpoons A^+ + 4B^- \]
- Let the solubility of \( AB_4 \) be \( s \).
- The concentration of \( A^+ \) will be \( s \) and that of \( B^- \) will be \( 4s \).
- Therefore, the expression for \( K_{sp} \) is:
\[ K_{sp} = [A^+][B^-]^4 = s \cdot (4s)^4 = s \cdot 256s^4 = 256s^5 \]
- Setting \( K_{sp} = 10^{-10} \):
\[ 256s^5 = 10^{-10} \implies s^5 = \frac{10^{-10}}{256} \implies s = \left(\frac{10^{-10}}{256}\right)^{1/5} \approx 3.98 \times 10^{-3} \text{ M} \]

### Conclusion:
After calculating the solubility for all four compounds:
- \( AB: s \approx 10^{-5} \text{ M} \)
- \( AB_2: s \approx 2.92 \times 10^{-4} \text{ M} \)
- \( AB_3: s \approx 1.78 \times 10^{-3} \text{ M} \)
- \( AB_4: s \approx 3.98 \times 10^{-3} \text{ M} \)

The compound with the maximum solubility is \( AB_4 \).