Question

The solubility product of $Pb{I}_2$ is $7.47\times {10}^{-9}$ at ${15}^{o}C$ and $1.39\times {10}^{-8}$ at ${25}^{o}C$. Calculate the molar heat of solution of $Pb{I}_2$.

Answer 1

$K_{sp1}=7.47\times {10}^{-9},K_{sp2}=1.39\times {10}^{-8}$

$T_1=15+273=288K,T_2=25+273=298K$

Formula used:

$\log \frac{K_{sp2}}{K_{sp1}}=\frac{\Delta H}{2303R}(\frac{1}{T_1}-\frac{1}{T_2})\log \frac{1.39\times {10}^{-8}}{7.47\times {10}^{-9}}=\frac{\Delta H}{2.303\times 8.314}(\frac{1}{288}-\frac{1}{298})0.2695=\frac{\Delta H}{2.303\times 8.314}\left(0.0001165\right)\Delta H=44295.301=44.3KJ/mol$