The sp cond of a saturated solution of `AgCI` at `25^(@)C` after substractin the sp conductances of conductivity of water is `2.28 xx 10^(-6) mho cm^(-1)`. Find the solubility product of `AgCI` at `25^(@)C(lambda_(AgCI)6(oo) = 138.3 mho cm^(2))`

Answer

`2.70 xx 10^(-10)("mole"//"litre")^(2)`

Explanation

For equilibrium,
`AgCl = Ag^(+) + Cl^(-) K_(sp)=[Ag^(+)][CL^(-)]`
If solubility of `AgCl` in water is, say, x moles/L or x.eq/L
`K_(sp) = x.x =x^(2)`
`:.` volume containing 1 eq. of `AgCl = (1000)/(x)`
`Lamda_(AgCl) = "sp. Cond." x V = 2.28 xx 10^(-6) xx(1000)/(x)`
Since `AgCl` is sparingly soluble in water,
`Lamda_(AgCl)=Lamda_(AgCl)^(@)=138.3`
`:. 2.28 xx 10^(-6) xx (1000)/(x)=138.3`
or `x =1.644 xx 10^(-5)` eq./L or mole/L
`K_(sp)=x^(2)=(1.644xx 10^(-5))^(-2)=2.70 xx 10^(-10)("mole/L")^(2)`.