Using Gibb's free energy change `Delta G^@=57.34kJ mol^-1` for the reaction `X_2Y_(s)=2X^(+)+Y^(2-) (aq)` Calculate the solubility product of `X_2Y` in water at 300 K

Options

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

Explanation

`57.34kJ mol^-1=-2.303 times 8.3 JK^-1mol^-1 times 300Klog K_(sp)`
`logK_(sp)=(-57.37 times 10^3 J mol^-1)/(2.303 times 8.3 JK^-1 mol^-1 times 300K)`
`log_10K_(sp)=-10`
`therefore K_(sp)=10^-10`
`DeltaG^@=-2.303 RT log K_(eq)`
`X_2Y(s) leftrightarrow 2X^(+) (aq)+Y^(2-) (aq)`
`K_(eq)=([X^+]^2[Y^(2-)])/([X_2Y])`
`K_(eq)=[X^+]^2[Y^(2-)]( thereforeX_2Y(s)=1)`
`K_(eq)=K_(sp)`