Question

The solubility product of a sparingly soluble salt $\mathrm{AX}_2$ is $3.2 \times 10^{-8}$. What is its solubility in $\mathrm{mol\ dm}^{-3}$?

• A. $2.8 \times 10^{-4}$
• B. $1.6 \times 10^{-5}$
• C. $2.0 \times 10^{-3}$
• D. $4.0 \times 10^{-4}$

Hint:

For any salt of type $\mathrm{AX}_2$ or $\mathrm{A}_2\mathrm{X}$, the $K_{sp}$ relation is always $4S^3$. When dealing with powers that aren't multiples of 3 (like $10^{-8}$), adjust the decimal system to make the power perfectly divisible by 3 ($10^{-9}$) for quick mental math!

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The problem provides the solubility product constant ($K_{sp} = 3.2 \times 10^{-8}$) for a binary-ternary electrolyte salt with formula $\mathrm{AX}_2$. We need to compute its molar solubility ($S$) in units of $\mathrm{mol\ dm}^{-3}$.

Step 2: Key Formula or Approach:
Write out the balanced dissociation equilibrium for the sparingly soluble salt $\mathrm{AX}_2$: $$\mathrm{AX}_2(s) \rightleftharpoons \mathrm{A}^{2+}(aq) + 2\mathrm{X}^{-aq}$$ If the molar solubility of the salt is $S$, the equilibrium ion concentrations are: $$[\mathrm{A}^{2+}] = S \quad \text{and} \quad [\mathrm{X}^{-}] = 2S$$ The solubility product expression is defined as: $$K_{sp} = [\mathrm{A}^{2+}][\mathrm{X}^{-}]^2 = S \cdot (2S)^2 = 4S^3$$ Rearranging to calculate $S$: $$S = \sqrt[3]{\frac{K_{sp}}{4}}$$

Step 3: Detailed Explanation:
Substitute the given value of $K_{sp}$ into our derived equilibrium expression: $$4S^3 = 3.2 \times 10^{-8}$$ Isolate $S^3$ by dividing both sides by 4: $$S^3 = \frac{3.2 \times 10^{-8}}{4} = 0.8 \times 10^{-8}$$ To make taking the cube root straightforward, convert the decimal value into standard scientific power notation: $$S^3 = 8.0 \times 10^{-9}$$ Take the cube root of both sides to get the final solubility $S$: $$S = \sqrt[3]{8.0 \times 10^{-9}} = 2.0 \times 10^{-3}\ \mathrm{mol\ dm}^{-3}$$

Step 4: Final Answer:
The molar solubility is $2.0 \times 10^{-3}\ \mathrm{mol\ dm}^{-3}$, which matches option (C).