The solubility product of AgCl in water is `1.5xx10^(-10)` .Calculate its solubility in 0.01 M NaCl aqueous solution.

Explanation

NaCl dissociates completely as :
`NaCl to Na^(+) + Cl^-`
Concentration of `Cl^-` ion in 0.01 M NaCl solution is
`[Cl^-]` =0.01 M
Let solubility of AgCl in 0.01 M NaCl be x mol `L^(-1)` then
`AgCl(s) hArr Ag^(+)(aq)+Cl^(-)(aq)`
`[Ag^+]=" x mol " L^(-1), Cl^(-)(aq)= "x mol" L^(-1)`
Total `[Cl^-]` =0.01 + x `approx` 0.01 M (`because` x is very small )
Now, `K_(sp)=[Ag^+][Cl^-]`
=`x xx 0.01`=0.01 x
`therefore 0.01x=1.5xx10^(-10)`
or `x=(1.5xx10^(-10))/0.01 =1.5xx10^(-8) "mol L"^(-1)`