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.

The solubility product of AgCl at a particular temperature is 1.08xx10

1.	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.
Answer» Solution :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)`.