Chemistry - Ionic Equilibrium Question with Solution | TestHub

Here's a concise solution:

Concentration of $\mathrm{Ca}(\mathrm{LaC}{)}_2$ = 0.6 mol / 2 L = 0.3 M

Since it's 90% dissociated, $[\mathrm{C}{\mathrm{a}}^{2+}]=0.3\times 0.9=0.27\mathrm{M}$ and $[\mathrm{La}{\mathrm{C}}^{-}]=2\times 0.3\times 0.9=0.54\mathrm{M}$

$\mathrm{La}{\mathrm{C}}^{-}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons \mathrm{LaC}-\mathrm{H}+\mathrm{O}{\mathrm{H}}^{-}$

$\mathrm{pOH}=5.60$, so $\mathrm{pOH}=\mathrm{p}{\mathrm{K}}_{\mathrm{b}}+\log \frac{[\mathrm{LaC}-\mathrm{H}]}{[\mathrm{La}{\mathrm{C}}^{-}]}$

Since $[\mathrm{O}{\mathrm{H}}^{-}]=10^{-5.6}$, $[\mathrm{LaC}-\mathrm{H}]\approx [\mathrm{O}{\mathrm{H}}^{-}]$

$\mathrm{p}{\mathrm{K}}_{\mathrm{b}}=\mathrm{pOH}-\log \frac{[\mathrm{LaC}-\mathrm{H}]}{[\mathrm{La}{\mathrm{C}}^{-}]}\approx 5.6-\log \frac{10^{-5.6}}{0.54}=5.6-(-5.6-\log 0.54)=\log 0.54$

$\mathrm{p}{\mathrm{K}}_{\mathrm{b}}=5.6+\log 0.54$

$\mathrm{p}{\mathrm{K}}_{\mathrm{a}}+\mathrm{p}{\mathrm{K}}_{\mathrm{b}}=14$

$\mathrm{p}{\mathrm{K}}_{\mathrm{a}}=14-\mathrm{p}{\mathrm{K}}_{\mathrm{b}}=14-(5.6+\log 0.54)=8.4-\log 0.54=2.8-\log 0.54$ (since $8.4-5.6=2.8+3.=2.8$)

Therefore, $\mathrm{p}{\mathrm{K}}_{\mathrm{a}}=2.8-\log (0.54)$