The solubility of AgCl is `1xx10^(-5)mol//L`. Its solubility in 0.1 molar sodium chloride solution is

Options

A. `1xx10^(-10)`
B. `1xx10^(-5)`
C. `1xx10^(-9)`
D. `1xx10^(-4)`

Answer

C

Explanation

`K_(sp)` of AgCl=(solublility of `AgCl)^(2)`
`=(1xx10^(-5))^(2)=1xx10^(-10)`
Suppose its solubility in 0.1 M NaCl is mol/L.
`AgCl Leftrightarrow underset(x)(Ag^(+))+underset(x)(Cl^(-))`
`NaCl Leftrightarrow underset(0.1M)(Na^(+))+underset(0.1M)(Cl^(-))`
`[Cl^(-)]=(x+0.1)M`
`k_(sp)"of AgCl"=[Ag^(+)][Cl^(-)]`
`=x xx(x+0,1)`
`1xx10^(-10)=x^(2)+0,1x`
Higher power of x are neglected `1xx10^(-10)=0.1x`
`x=1xx10^(-9)M`