Question

The ratio of solubility of AgCl in 0.1 M KCl solution to the solubility of AgCl in water is:
(Given: Solubility product of AgCl = 10⁻¹⁰)

• A. 10⁻⁴
• B. 10⁻⁶
• C. 10⁻⁹
• D. 10⁻⁵

Answer 1

The Correct Option is A

1. Water Solubility:
AgCl $\rightleftharpoons$ Ag$^+$ + Cl$^-$ $K_{sp} = [Ag^+][Cl^-] = s^2$, with s representing solubility in mol/L.
With $K_{sp} = 10^{-10}$, $s = \sqrt{10^{-10}} = 10^{-5}$ mol/L.
2. Solubility in 0.1 M KCl Solution: In a 0.1 M KCl solution, [Cl$^-$] is 0.1 M (due to the common ion effect).
$K_{sp} = [Ag^+][Cl^-] = s'(0.1) = 10^{-10}$, where s' is the solubility in 0.1 M KCl. Thus, $s' = \frac{10^{-10}}{0.1} = 10^{-9}$ mol/L.
3. Calculate the Ratio:
Ratio $= \frac{Solubility in 0.1 M KCl}{Solubility in water} = \frac{10^{-9}}{10^{-5}} = 10^{-4}$