2024 Re-NEET EquilibriumSolubility product and common ion effect Medium Question 15 of 104 in Equilibrium

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 $10^{-5}$

The answer is A.

Solution

In 0.1 M KCl: $K_{sp}=s(s+0.1)\approx s\times0.1$ (since $s\ll0.1$), so $s=\dfrac{10^{-10}}{0.1}=10^{-9}$ M. In water: $K_{sp}=s^2\Rightarrow s=\sqrt{10^{-10}}=10^{-5}$ M. Ratio $=\dfrac{10^{-9}}{10^{-5}}=10^{-4}$.

Concept Solubility product and common ion effect

Equilibrium in past papers

104 questions from this chapter have appeared across 27 exam years.