Question

The solubility product of salt $B_{2}A$ is $3.2\times10^{-11}$ at 298 K. What is solubility of the salt at same temperature?

• A. $5.52\times10^{-5}moldm^{-3}$
• B. $4.92\times10^{-4}moldm^{-3}$
• C. $2.00\times10^{-4}moldm^{-3}$
• D. $3.52\times10^{-5}moldm^{-3}$

Hint:

For $X_2Y$ or $XY_2$ type salts, $K_{sp} = 4s^3$.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
We need to calculate the molar solubility (\( s \)) from the solubility product constant (\( K_{sp} \)) for a salt of the type \( \text{B}_2\text{A} \).

Step 2: Key Formula or Approach:
For \( \text{B}_2\text{A} \rightarrow 2\text{B}^+ + \text{A}^{2-} \):
If solubility is \( s \), then \( [\text{B}^+] = 2s \) and \( [\text{A}^{2-}] = s \).
\[ K_{sp} = [B^+]^2 [A^{2-}] = (2s)^2(s) = 4s^3 \]

Step 3: Detailed Explanation:
Given: \( K_{sp} = 3.2 \times 10^{-11} \).
\[ 4s^3 = 3.2 \times 10^{-11} \]
\[ s^3 = \frac{3.2 \times 10^{-11}}{4} = 0.8 \times 10^{-11} \]
To take the cube root easily, write as:
\[ s^3 = 8.0 \times 10^{-12} \]
\[ s = \sqrt[3]{8.0 \times 10^{-12}} = 2.0 \times 10^{-4} \text{ mol dm}^{-3}. \]

Step 4: Final Answer:
The solubility is \( 2.00 \times 10^{-4} \text{ moldm}^{-3} \).