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

Explanation

`AgCl(s)hArrAg^(+)+Cl^(-)`
`{:(1,0,0),(1-s,s,s):}`
`K_(sp)=[Ag^(+)][Cl^(-)]`
`=sxxs=s^(2)`
`=(1.06xx10^(-5))^(2)`
`=1.12xx10^(-10)mol^(2)L^(-2)`