Question

The $K_{sp}\left({25}^{o}C\right)$ of sparingly soluble salt $X{Y}_2$ is $3.56\times {10}^{-5}mo{l}^3{L}^{-3}$ and ${30}^{o}C$, the vapour pressure of its saturated solution in water in 31.78 mm of Hg.
$X{Y}_2+Aq.\rightleftharpoons X^{2+}(aq.)+2Y^{-}(aq.)$ (100% ionisation)
The enthalpy change of the reaction if the vapour pressure of $H_{2}O$ at ${30}^{o}C$ is $31.82$ mm in kJ (divide by 25 and write answer to the nearest integer) is_____________.

Answer 1

AT ${30}^{o}C:P^0=31.82mm$; At ${25}^{o}C:K_{sp}=3.56\times {10}^{-5}$

$(molality=molarity)$ for sparingly soluble salt

$\therefore \frac{P^0-P_S}{P_S}=\frac{n\times M\times 1000}{W\times 1000}(1-\alpha +x\alpha +y\alpha )$

$[\because \alpha =1$ and $(x+y)=3$ for $X{Y}_2]$

$\therefore \frac{31.82-31.78}{31.78}=\frac{n\times 1000\times 18}{W\times 1000}\times 3(M=18)$

$\therefore \frac{n}{W}\times 1000=\frac{0.04\times 1000}{31.78\times 3\times 18}=2.33\times {10}^{-2}$

$\therefore Solubility=2.33\times {10}^{-2}$

$\therefore K_{sp}=4s^2=4\times (2.33\times {10}^{-2}{)}^3=5.06\times {10}^{-5}$ at 303 K

Since, $2.303log\frac{K_{s{p}_2}}{K_{s{p}_1}}=\frac{\Delta H}{R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$

Since, $2.303log\frac{5.06\times {10}^{-5}}{3.56\times {10}^{-5}}=\frac{\Delta H}{8.314}\frac{[303-298]}{[303\times 298]}$

$\Delta H=52.799\times {10}^{3}J=52.799kJ$