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: Understanding the Question:
We are given the molar solubility ($S$) of a sparingly soluble salt, silver oxalate ($\text{Ag}_2\text{C}_2\text{O}_4$). We need to determine its corresponding solubility product constant ($K_{sp}$).

Step 2: Key Formula or Approach:
1. Write out the balanced equilibrium dissolution equation for the salt: $$\text{Ag}_2\text{C}_2\text{O}_4(s) \rightleftharpoons 2\text{Ag}^+(aq) + \text{C}_2\text{O}_4^{2-}(aq)$$ 2. If the solubility of the salt is $S$, then the equilibrium concentrations of the constituent ions are: $$[\text{Ag}^+] = 2S \quad \text{and} \quad [\text{C}_2\text{O}_4^{2-}] = S$$ 3. The solubility product expression is: $$K_{sp} = [\text{Ag}^+]^2 [\text{C}_2\text{O}_4^{2-}] = (2S)^2 \cdot (S) = 4S^3$$

Step 3: Detailed Explanation:
Given that $S = 2 \times 10^{-4}\text{ mol L}^{-1}$, substitute this value into our algebraic expression: $$K_{sp} = 4 \times (2 \times 10^{-4})^3$$ First, cube the term inside the parenthesis: $$(2 \times 10^{-4})^3 = 2^3 \times (10^{-4})^3 = 8 \times 10^{-12}$$ Now, multiply this by the coefficient of 4: $$K_{sp} = 4 \times (8 \times 10^{-12}) = 32 \times 10^{-12}$$ Convert this into proper standard scientific notation format: $$K_{sp} = 3.2 \times 10^{-11}$$

Step 4: Final Answer:
The solubility product constant of the salt is $3.2 \times 10^{-11}$, which matches option (B).