Question

The solubility of $\text{Ag}_2\text{C}_2\text{O}_4$ is $2 \times 10^{-4}$ mol L$^{-1}$ at 298 K. What is it's solubility product?

• A. $1.6 \times 10^{-6}$
• B. $3.2 \times 10^{-11}$
• C. $1.6 \times 10^{-11}$
• D. $3.2 \times 10^{-6}$

Hint:

For any electrolyte salt of the formula structural type $\text{A}_2\text{B}$ or $\text{AB}_2$, the mathematical relationship between the solubility product and molar solubility is always fixed as $K_{sp} = 4S^3$. Memorizing these standard algebraic forms ($4S^3$ for ternary salts) eliminates errors and saves precious setup time.

Answer 1

The Correct Option is B

Step 1: Write the dissolution equilibrium.
$\text{Ag}_2\text{C}_2\text{O}_4(s) \rightleftharpoons 2\text{Ag}^+ + \text{C}_2\text{O}_4^{2-}$.
Step 2: Express ion concentrations from solubility S.
Each formula unit gives 2 silver ions and 1 oxalate ion, so $[\text{Ag}^+] = 2S$ and $[\text{C}_2\text{O}_4^{2-}] = S$.
Step 3: Build the K$_{sp}$ expression.
$K_{sp} = [\text{Ag}^+]^2[\text{C}_2\text{O}_4^{2-}] = (2S)^2(S) = 4S^3$.
Step 4: Insert the value of S.
$S = 2 \times 10^{-4}$ mol L$^{-1}$, so $K_{sp} = 4(2 \times 10^{-4})^3$.
Step 5: Cube and multiply.
$(2 \times 10^{-4})^3 = 8 \times 10^{-12}$, then $4 \times 8 \times 10^{-12} = 32 \times 10^{-12}$.
Step 6: Express in standard form.
$K_{sp} = 3.2 \times 10^{-11}$, which is option (2).
\[ \boxed{K_{sp} = 3.2 \times 10^{-11}} \]