In 1 L saturated solution of AgCl (`K_(sp)` of `AgCl = 1.6 xx 10^(-10)`), 0.1 mol of CuCl (`K_(sp)` of `CuCl = 1.0 xx 10^(-6)`) is added. The resultant concentration of `Ag^(o+)` in the solution is `1.6 xx 10^(-x)`. Calculate the value of x.

Answer

7

Explanation

`AgCl hArr underset(S)(Ag^(o+)) + underset(S)(Cl^(ɵ))(K_(sp) = 1.6 xx 10^(-10)), CuCl hArr underset(x)(Cu^(o+)) + underset((x+S))(Cl^(ɵ))(K_(sp) =1 xx 10^(-6))`
`x(x+S.) = 10^(-6), (s.)/(x) = 1.6 xx 10^(-4) therefore S. = (1.6 xx 10^(-4)) x rArr (x+ 1.6 xx 10^(-4)x) = 10^(-6) rArr x = 10^(-3)`
`therefore S. = 1.6 xx 10^(-7) = 1.6 xx 10^(-x)` (The value of x = 7)
Alternative method :
`Cl^(ɵ)` will be from CuCl (Since CuCl is more soluble than AgCl as it is evident from their `K_(sp)` value )
`K_(sp) = [Cu^(o+)][Cl^(ɵ)] = 10^(-6) = x xx x rArr x = 10^(-3)`
Since AgCl is sparingly soluble `K_(sp)` of `AgCl = [Ag^(o+)](10^(-3))`
`1.6 xx 10^(-10) = [Ag^(o+)](10^(-3)) therefore [Ag^(o+)] = 1.6 xx 10^(-7) = 1.6 xx 10^(-x)` (the value of x = 7)