The solubility of `AgCl` in water at `298` K is `1.06xx10^(-5)` mol per litre. Calculate its solubility product at this temperature.

Explanation

The solubility equilibrium of `AgCl` is :
`AgCl(s)hArr Ag^(+)(aq)+Cl^(-)(aq)`
One mole of `AgCl` in solution gives `1` mole of `Ag^+` ions and `1` mole of `Cl^-` ions. Since the solubility of `AgCl` is `1.06 xx 10^(-5) mol L^(-1)`, it will give `1.06 xx 10^(-5) mol L^(-1)` of `Ag^+` ions and `1.06 xx 10 mol L^(-1)`of `Cl^-` ions. Therefore,
`[Ag^+]=1.06xx10^(-5) "mol L"^(-1)`
`[Cl^-]=1.06xx10^(-5) "mol L"^(-1)`
Now , `K_(sp)=[Ag^+][Cl^-]`
`=(1.06xx10^(-5))xx(1.06xx10^(-5))`
`=1.12xx10^(-10)`