Question

Calculate the solubility in mol \(\text{dm}^{-3}\) of sparingly soluble salt BA at 293 K if its solubility product is \(8.56\times 10^{-5}\) at same temperature.

• A. \(8.123\times 10^{-3}\)
• B. \(8.780\times 10^{-3}\)
• C. \(9.252\times 10^{-3}\)
• D. \(7.756\times 10^{-3}\)

Hint:

For a 1:1 salt, \(K_{sp} = S^2\), so \(S=\sqrt{K_{sp}}\).

Answer 1

The Correct Option is C

Step 1: Set up the equilibrium
Start with solid BA. At equilibrium the dissolved BA gives $S$ of each ion.

Step 2: Solve
$K_{sp} = S\times S = 8.56\times10^{-5}$.
Taking the root: $S = 9.25\times10^{-3}$ mol dm$^{-3}$, since $9.25^2 = 85.6$.
This is option (C).

Final Answer:
The solubility of BA is $9.252\times10^{-3}$ mol dm$^{-3}$, option (C). \[ \boxed{9.252\times10^{-3}} \]