The solubility product of AgCl at a particular temperature is `1.08xx10^(-10)mol^(2)dm^(-6)`. Calculate its solubility in 0.01 M HCl.

Explanation

In the presence of a common ion with the concentration of c mol `dm^(-3)`, the solubility S of the sparingly soluble salt is generally given by `(K_(sp))/c`.
`AgClhArrunderset(S)(Ag^(+))+underset(S)(Cl^(-))`
`underset(0.01M)(HCl)tounderset(0.01M)(H^(+))+underset(0.01M)(Cl^(-))`
Solubility of AgCl, `K_(sp)=[Ag^(+)][Cl^(-)]`
`1.08xx10^(-10)=(S)(S+0.01)`
Assuming S to be small when compared to 0.01
`S+0.01~~0.01`
`therefore1.08xx10^(-10)=(S)(0.01)`
Solubility (S) = `(1.08xx10^(-10))/0.01=1.8xx10^(-8)"mol "dm^(-3)`.