Question

The solubility of AgCl in it's solution is $1.25 \times 10^{-5}\ \text{mol dm}^{-3}$. What is solubility product of AgCl?

• A. $1.56 \times 10^{-10}$
• B. $3.50 \times 10^{-6}$
• C. $1.10 \times 10^{-5}$
• D. $2.53 \times 10^{-3}$

Hint:

For any binary electrolyte (AB type salt like $\text{AgCl, BaSO}_4$), the relationship is always simple: $K_{sp} = S^2$. Squaring $1.25$ yields $1.56$, which points immediately to option (A).

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
We are given the molar solubility ($S$) of a sparingly soluble salt, silver chloride (AgCl). We need to compute its solubility product constant ($K_{sp}$).

Step 2: Key Formula or Approach:
Silver chloride dissociates in an aqueous solution according to the equilibrium: $$ \text{AgCl}_{(s)} \rightleftharpoons \text{Ag}^{+}_{(aq)} + \text{Cl}^{-}_{(aq)} $$ If the solubility of the salt is $S$, then the equilibrium concentrations of the ions are: $$ [\text{Ag}^+] = S \quad \text{and} \quad [\text{Cl}^-] = S $$ The expression for the solubility product is: $$ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = S \times S = S^2 $$

Step 3: Detailed Explanation:
The given molar solubility is $S = 1.25 \times 10^{-5}\ \text{mol dm}^{-3}$. Substitute this value into our derived relationship: $$ K_{sp} = (1.25 \times 10^{-5})^2 $$ $$ K_{sp} = (1.25)^2 \times (10^{-5})^2 $$ $$ K_{sp} = 1.5625 \times 10^{-10} $$ Rounding to match the options gives $1.56 \times 10^{-10}$.

Step 4: Final Answer:
The solubility product of AgCl is $1.56 \times 10^{-10}$, matching option (A).