The solubility of a sparingly soluble salt `A_xB_y` in water at `25^@C=1.5xx10^(-4)M`.The solubility product is `1.1xx10^(-11)` The possibilities are

Options

A. x=1 ,y=2
B. x=2,y=1
C. x=1,y=3
D. x=3,y=1

Explanation

To solve the problem, we need to determine the correct values of \(X\) and \(Y\) for the sparingly soluble salt \(A_xB_y\) given its solubility and solubility product (Ksp).

### Step-by-Step Solution:

1. **Understanding the Dissociation of the Salt**:
The salt \(A_xB_y\) dissociates in water as follows:
\[ A_xB_y \rightleftharpoons xA^{y+} + yB^{x-} \]
Here, \(x\) and \(y\) are the stoichiometric coefficients for ions \(A\) and \(B\) respectively.

2. **Defining Solubility**:
The solubility \(S\) of the salt is given as \(1.5 \times 10^{-4} \, M\). This means that at equilibrium, the concentration of \(A^{y+}\) will be \(xS\) and the concentration of \(B^{x-}\) will be \(yS\).

3. **Setting Up the Ksp Expression**:
The solubility product \(K_{sp}\) can be expressed as:
\[ K_{sp} = [A^{y+}]^x \cdot [B^{x-}]^y = (xS)^x \cdot (yS)^y \]
Substituting \(S = 1.5 \times 10^{-4}\):
\[ K_{sp} = (x(1.5 \times 10^{-4}))^x \cdot (y(1.5 \times 10^{-4}))^y \]

4. **Calculating Ksp**:
Given \(K_{sp} = 1.1 \times 10^{-11}\), we can write:
\[ K_{sp} = (1.5 \times 10^{-4})^{x+y} \cdot (x^x \cdot y^y) \]

5. **Testing Each Option**:
We will test each option to see which one satisfies the Ksp value.

- **Option A**: \(X = 1, Y = 2\)
\[ K_{sp} = (1.5 \times 10^{-4})^{1+2} \cdot (1^1 \cdot 2^2 = 4) \]
\[ K_{sp} = (1.5 \times 10^{-4})^3 \cdot 4 = 3.375 \times 10^{-12} \cdot 4 = 1.35 \times 10^{-11} \quad \text{(approximately equal)} \]

- **Option B**: \(X = 2, Y = 1\)
\[ K_{sp} = (1.5 \times 10^{-4})^{2+1} \cdot (2^2 \cdot 1^1 = 4) \]
\[ K_{sp} = (1.5 \times 10^{-4})^3 \cdot 2 = 3.375 \times 10^{-12} \cdot 2 = 6.75 \times 10^{-12} \quad \text{(not equal)} \]

- **Option C**: \(X = 1, Y = 3\)
\[ K_{sp} = (1.5 \times 10^{-4})^{1+3} \cdot (1^1 \cdot 3^3 = 27) \]
\[ K_{sp} = (1.5 \times 10^{-4})^4 \cdot 27 \quad \text{(not equal)} \]

- **Option D**: \(X = 3, Y = 1\)
\[ K_{sp} = (1.5 \times 10^{-4})^{3+1} \cdot (3^3 \cdot 1^1 = 27) \]
\[ K_{sp} = (1.5 \times 10^{-4})^4 \cdot 27 \quad \text{(not equal)} \]

6. **Conclusion**:
The only option that gives a value close to \(K_{sp} = 1.1 \times 10^{-11}\) is **Option A**: \(X = 1, Y = 2\).