Question

Solubility of a salt $A_{2}B_{3}$ is $1\times10^{-3}\text{ mol dm}^{-3}$. What is the value of its solubility product?

• A. $1.08\times10^{-13}$
• B. $8.1\times10^{-15}$
• C. $2.7\times10^{-15}$
• D. $2.0\times10^{-13}$

Hint:

Logic Tip: The relationships between $K_{sp}$ and $S$ for common salt types are highly standardized. For $AB$ type ($x=1,y=1$), $K_{sp} = S^2$. For $AB_2$ type ($x=1,y=2$), $K_{sp} = 4S^3$. For $A_2B_3$ type ($x=2,y=3$), $K_{sp} = 108S^5$. Memorizing these forms saves valuable time during calculation.

Answer 1

The Correct Option is A

Concept:
For a sparingly soluble salt of the general formula $A_x B_y$, the dissociation equilibrium in water is given by: $$A_x B_{y(s)} \rightleftharpoons xA^{y+}_{(aq)} + yB^{x-}_{(aq)}$$ If the molar solubility of the salt is $S$, the equilibrium concentrations of the ions will be $[A^{y+}] = xS$ and $[B^{x-}] = yS$. The solubility product constant ($K_{sp}$) formula is $K_{sp} = [A^{y+}]^x [B^{x-}]^y$, which simplifies to: $$K_{sp} = x^x y^y S^{(x+y)}$$
Step 1: Identify the coefficients x and y for the given salt.
The salt is $A_2 B_3$. Dissociation: $A_2 B_3 \rightleftharpoons 2A^{3+} + 3B^{2-}$ Here, $x = 2$ and $y = 3$.
Step 2: Set up the $K_{sp}$ expression in terms of S.
Substitute $x$ and $y$ into the general formula: $$K_{sp} = (2)^2 (3)^3 S^{(2+3)}$$ $$K_{sp} = (4)(27) S^5$$ $$K_{sp} = 108 S^5$$
Step 3: Calculate the numerical value of $K_{sp}$.
We are given the solubility $S = 1 \times 10^{-3} \text{ mol dm}^{-3}$. Substitute $S$ into the expression: $$K_{sp} = 108 \times (1 \times 10^{-3})^5$$ $$K_{sp} = 108 \times 10^{-15}$$ Convert to standard scientific notation: $$K_{sp} = 1.08 \times 10^{-13}$$