The solubility of `AgCl_(s)` with solubility product `1.6 xx 10^(-10)` in 0.1 M NaCl Solution is `x xx 10^(-9)` M. The value of x is_____

Answer

1.6

Explanation

`underset("M")("Concentration ") underset()(AgCl_(s))^(-) underset(S)(Ag^(+)(aq))S + underset(S + 0.1)(Cl^(-)(aq))`
`K_(sp) = [Ag^(+)][Cl^(-)]`
`therefore 1.6 xx 10^(-10) = (S) (S + 0.1)`
`therefore 1.6 xx 10^(-10) = S xx 0.1 (therefore S lt lt 0.1))`
`therefore S = 1.6 xx 10^(-9) M`