Question

The $K_{sp}\left({25}^{o}C\right)$ of sparingly salt $X{Y}_2\left(s\right)$ is $3.56\times {10}^{-5}{\left(mol{L}^{-1}\right)}^3$ and at ${30}^{o}C$,the vapour pressure of its saturated solution in water is 31.78 mm of Hg.Calculate the enthalpy change of the reaction.( write your answer to nearest integer)
$X{Y}_2\left(s\right)\rightleftharpoons X^{2+}\left(aq\right)+2Y^{\ominus }\left(aq\right)$
Given : Vapour pressure of pure water =31.82 mm of Hg

Answer 1

Let the solubility $X{Y}_2\left(s\right)$ at ${30}^{o}C$ be $smol{L}^{-1}$.
$X{Y}_2\left(s\right)\rightleftharpoons X^{2+}\left(aq\right)+2Y^{\ominus }\left(aq\right)$
S 2S
$\frac{P^o-P_S}{P_S}=\frac{n}{N}$ (i=3,f 100% ionization)
$\frac{31.82-31.78}{31.78}=3{\chi }_2,{\chi }_2=0.0004,\frac{n_2}{n_1}=0.004$
$n_2=55.6\times 0.004=0.0233$
$S=0.2333mol{L}^{-1}$
$K_{sp}\left(at{30}^{o}C\right)=\left(0.0233\right){(2\times 0.0233)}^2$
$=5.05\times {10}^{-5}{\left(mol{L}^{-1}\right)}^3$
$K_{sp}\left(at{25}^{o}C\right)=3.56\times {10}^{-5}{\left(mol{L}^{-1}\right)}^3$
Now,$log\frac{K_{sp}\left({35}^{o}C\right)}{K_{sp}\left({25}^{o}C\right)}=\frac{\Delta H}{2.303R}\left(\frac{T_2-T_1}{T_1{T}_2}\right)$
$log(\frac{5.05\times {10}^{-5}}{3.56\times {10}^{-5}}=\frac{\Delta H}{2.303\times 8.314}\left(\frac{5}{303\times 298}\right)$
$\Delta H=52.5kJ{mol}^{-1}$