Question

Solubility of AgCl is $7.2 \times 10^{-7}\ \text{mol}\ \text{dm}^{-3}$. What is its solubility product?

• A. $3.6 \times 10^{-13}$
• B. $7.2 \times 10^{-14}$
• C. $2.59 \times 10^{-14}$
• D. $5.18 \times 10^{-13}$

Hint:

For any symmetric $1:1$ ionic electrolyte (like AgCl, $\text{BaSO}_4$, or AgBr), the mathematical relationship is always simplified to $K_{sp} = S^2$. Simply squaring the value ($7^2 = 49$) allows you to instantly isolate $5.18 \times 10^{-13}$ among the choices.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The problem provides the molar solubility ($S$) of a sparingly soluble salt, silver chloride (AgCl), and requests the evaluation of its equilibrium solubility product constant ($K_{sp}$).

Step 2: Key Formula or Approach:
Silver chloride dissociates in an aqueous solution according to a $1:1$ binary electrolyte ratio:
$$ \text{AgCl}_{(s)} \rightleftharpoons \text{Ag}^+_{(aq)} + \text{Cl}^-_{(aq)} $$ If the molar solubility of the salt is denoted by $S$, the equilibrium concentrations are $[\text{Ag}^+] = S$ and $[\text{Cl}^-] = S$.
The formula for the solubility product is:
$$ K_{sp} = [\text{Ag}^+][\text{Cl}^-] = (S)(S) = S^2 $$

Step 3: Detailed Explanation:
Given the molar solubility value:
$$ S = 7.2 \times 10^{-7}\ \text{mol}\ \text{dm}^{-3} $$ Substitute this value into the derived equilibrium relationship:
$$ K_{sp} = (7.2 \times 10^{-7})^2 $$ Squaring the numerical coefficient and the power of ten independently yields:
$$ K_{sp} = (7.2)^2 \times (10^{-7})^2 $$ $$ K_{sp} = 51.84 \times 10^{-14} $$ Adjusting the decimal layout to fit standard scientific notation gives:
$$ K_{sp} = 5.184 \times 10^{-13} \approx 5.18 \times 10^{-13} $$

Step 4: Final Answer: The computed solubility product matches option (D).