Question

The solubility product of $PbI _{2}$ is $7.47 \times 10^{-9}$ at $15^{\circ} C$ and $1.39 \times 10^{8}$ at $25^{\circ} C$. The molar heat of solution of $PbI _{2}$ is (use log $1.86=0.2695$ )

• A. 44.29 kJ/mol
• B. 46.25 kJ/mol
• C. 29.37 kJ/mol
• D. 21.15 kJ/mol

Answer 1

Answer: (A)

$\log \frac{\left( K _{ sp }\right)_{2}}{\left( K _{ sp }\right)_{2}}=\frac{\Delta H }{2.303 R }\left(\frac{ T _{2}- T _{1}}{ T _{1} T _{2}}\right)$
$\log \frac{1.39 \times 10^{-8}}{7.47 \times 10^{-9}}=\frac{\Delta H }{2.303 \times 8.314}\left(\frac{298-288}{288 \times 298}\right)$
$\Delta H =44.29\, kJ $ / mole