The solubility of silver chloride (AgCl) in water at `25^(@)C` ia `1.06 xx 10^(-5)` mol `L^(-1)`. Calculate the solubility product of AgCl at this temperature.

Explanation

We are given that
Solubility of AgCl in water `= 1.06 xx 10^(-5)` mol `L^(-1)`
We are to calculate the solubility product of AgCl.
Solubility equilibria for AgCl can be written as `AgCl(s) hArr Ag^(+) (aq) + Cl^(-)(aq)` ……(i)
Solubility product, `K_(sp) = [Ag^(+)(aq)][Cl^(-)(aq)]`
Since one mole of AgCl gives 1 mole of `Ag^(+)` (aq) and 1 mole of `Cl^(-)`. Therefore,
`[Ag^(+)] = 1.06 xx 10^(-5)` mol `L^(-1), [Cl] = 1.06 xx 10^(-5)` mol `L^(-1)`
So, `K_(sp) = [Ag^(+)][Cl^(-)]` Hence, `K_(sp) = [1.06 xx 10^(-5)][1.06 xx 10^(-5)] = 1.12 xx 10^(-10)`