Question

The solubility product of a sparingly soluble salt AX₂ is 3.2 ×10–11. Its solubility (in moles/liter) is

• A. 3.1×10^–4
• B. 2 × 10^–4 
• C. 4 × 10^–4
• D. 5.6 × 10^–6

Answer 1

The Correct Option is B

To find the solubility of the sparingly soluble salt AX_2 given its solubility product K_{sp} = 3.2 \times 10^{-11}, we need to understand the dissolution process and the expression for the solubility product constant.

The dissolution of AX_2 in water can be represented by the following equation:

AX_2 \rightleftharpoons A^{2+} + 2X^{-}

If the molar solubility of AX_2 is s moles/liter, at equilibrium, the concentrations of the ions will be:

• [A^{2+}] = s
• [X^{-}] = 2s

According to the definition of solubility product:

K_{sp} = [A^{2+}][X^{-}]^2

Substitute the equilibrium concentrations into the K_{sp} expression:

K_{sp} = s(2s)^2 = 4s^3

We know K_{sp} = 3.2 \times 10^{-11}, so we can set up the equation:

4s^3 = 3.2 \times 10^{-11}

Solve for s:

s^3 = \frac{3.2 \times 10^{-11}}{4} = 0.8 \times 10^{-11} = 8.0 \times 10^{-12}

Take the cube root of both sides:

s = \sqrt[3]{8.0 \times 10^{-12}}

Calculating the cube root:

s = 2 \times 10^{-4} moles/liter

Thus, the solubility of the salt AX_2 is 2 \times 10^{-4} moles/liter.

The correct answer is: 2 × 10^–4