Using Gibb's free energy change, `DeltaG^@=57.34 "kJ mol"^(-1)`, for the reaction,`X_2Y_((s))hArr2X^(+)+Y^2(aq)` calculate the solubility product of `X_2 Y_2` in water at 300 K `(R = 8.3 "J K"^(-1)"Mol"^(-1))`

Options

A. `10^(-10)`
B. `10^(-12)`
C. `10^(-14)`
D. can not be calculated from the given data

Explanation

`57.32 "kJ mol"^(-1) =-2.303 xx8.3 "JK"^(-1) "mol"^(-1) xx300K log K_(sp)`
`logK_(sp) = (-57.32 xx10^3 " Jmol"^(-1))/(2.303 xx8.3"J K"^(-1)" mol"^(-1) xx300K)`
`log_(10)K_(sp) =-10`
`:. K_(sp) =10^(-10)`