The solubility of AgCl in 0.2 M NaCl is `[K_(sp) AgCl =1.8 xx 10^(-10)]`

Options

A. `9 xx10^(-10)M`
B. `3.6 xx 10^(-10)M`
C. `1.8 xx 10^(-11)M`
D. `7.2 xx 10^(-10)M`

Answer

A

Explanation

Since NaCl is a strong electrolyte , therefore the entire conc. of `Cl^(-)` ions comes from NaCl
`K_(sp)AgCl=[Ag^(+)][Cl^(-)]`
If solubility of `Ag^(+)=s` in NaCl then
`K_(sp)=[s][Cl^(-)]`
`1.8 xx 10^(-10)=s xx 0.2`
or `s=(1.8 xx 10^(-10))/(0.2 )=9 xx 10^(-10)M`