1.

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

Answer»

`10^-10`
`10^-12`
`10^-14`
can not be CALCULATED from the given data

Solution :`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)`


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