At `25^(@)C` concentration of AgCl in its saturated aqueous solution is 0.00287`g*L^(-1)`. Find its solublity product at that temperature. [Ag=108, Cl=35.5].

Explanation

AgCl, in its saturated aqueous solution gives rise to the followingg equilibrium: `AgCl(s)hArr Ag^(+)(aq)+Cl^(-)(aq)`
So, the solubility product of AgCl, `K_(sp)=[Ag^(+)]xx[Cl^(-)]`
As per given data, solubility of AgCl=`0.00287g*L^(-1)`
`therefore`Solubility of `AgCl=(0.00287)/(143.5)=2xx10^(-5)mol*L^(-1)" "M_(AgCl):143.5]`
Hence, in the saturated solution of AgCl, `[Ag^(+)]=2xx10^(-5)mol*L^(-1) and [Cl^(-)]=2xx10^(-5)mol*L^(-1)`
`therefore` Solubility product of AgCl, `K_(sp)=[Ag^(+)]xx[Cl^(-)]`
`=(2xx10^(-5))xx(2xx10^(-5))=4xx10^(-10)`.