The solubility of AgCl in water at 298 K is `1.06 xx 10^(-5)` mole per litre. Calculate its solubility product at this temperature.

Explanation

The solubility equilibrium in the saturated solution is :
`AgCI (s) overset(aq)(hArr) Ag^(+) (aq) + CI^(-) (aq)`
The solubility of AgCi is `1.06 xx 10^(-5)`mole per litre.
`[Ag^(+) (aq)] =1.06 xx 10^(-5) mol L^(-1)`
`[Cl^(-)(aq)] =1.06 xx 10^(-5) mol L^(-1)`
`K_(sp) =[Ag^(+) (aq)] [ Cl^(-) (aq)]`
`= (1.06 xx 10^(-5) mol L^(-1)) xx (1.06 xx 10^(-5) mol L^(-1))`
`= 1.12 xx 10^(-10) mol^(2)L^(-2)`