Solveeit Logo
Login

Question

Chemistry Question on Solubility

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

10−4

Explanation

Solution

1. Solubility in Water:
AgCl ⇌\rightleftharpoons Ag+^+ + Cl−^- Ksp=[Ag+][Cl−]=s2K_{sp} = [Ag^+][Cl^-] = s^2 where s is the solubility in mol/L.
Given Ksp=10−10K_{sp} = 10^{-10} s=10−10=10−5s = \sqrt{10^{-10}} = 10^{-5} mol/L
2. Solubility in 0.1 M KCl Solution: In 0.1 M KCl solution, [Cl−^-] = 0.1 M (due to common ion effect).
Ksp=[Ag+][Cl−]=s′(0.1)=10−10K_{sp} = [Ag^+][Cl^-] = s'(0.1) = 10^{-10} where s' is solubility in 0.1 M KCl. s′=10−100.1=10−9s' = \frac{10^{-10}}{0.1} = 10^{-9} mol/L
3. Calculate the Ratio:
Ratio =Solubilityin0.1MKClSolubilityinwater=10−910−5=10−4= \frac{Solubility in 0.1 M KCl}{Solubility in water} = \frac{10^{-9}}{10^{-5}} = 10^{-4}