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 = ) [2024]

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
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 = $10^{- 10}$ ) [2024]

1 $10^{- 4}$

2 $10^{- 6}$

3 $10^{- 9}$

4 $10^{- 5}$

Ans.
(1)

Let the solubility of AgCl in water be $s$ and in KCl be $S$ .

$\begin{matrix} AgCl & \to & {Ag}^{+} & + & {Cl}^{-} \\ Solubility in water : & s & s \\ Solubility in KCl : & S & S + 0.1 \end{matrix}$

$K_{s p} = [Ag] [Cl] = 10^{- 10}$

For water, $K_{s p} = s^{2} \Rightarrow s = 10^{- 5}$

For KCl, $K_{s p} = S (S + 0.1)$

$S (0.1) = 10^{- 10}$

$\Rightarrow S = 10^{- 9} (as, S < < < < 0.1)$

$\frac{Solubility of AgCl in KCl}{Solubility of AgCl in water} = \frac{10^{- 9}}{10^{- 5}} = 10^{- 4}$

Want Us to Email you About Special Offers & Updates?