A 0.50 mol kg^-1 aqueous solution of a weak monobasic acid HA (Ka=4.0×10^-4) is prepared. Assuming that molality can be used in place of molarity in…

First find the degree of dissociation. For $\mathrm{HA\rightleftharpoons H^{+}+A^{-}}$ with total concentration $m=0.50$, $K_a=\dfrac{m\alpha^2}{1-\alpha}$, i.e. $0.50\alpha^2+4.0\times10^{-4}\alpha-4.0\times10^{-4}=0$. The positive root is $\alpha=\dfrac{-4.0\times10^{-4}+\sqrt{(4.0\times10^{-4})^2+4(0.50)(4.0\times10^{-4})}}{2(0.50)}=0.02789$ (the approximation $\alpha\approx\sqrt{K_a/m}=0.0283$ is close but slightly high). The van't Hoff factor for one particle dissociating into two is $i=1+\alpha=1.0279$. The freezing point depression is $\Delta T_f=iK_fm=1.0279\times1.86\times0.50=0.956\ \mathrm{K}$, which rounds to $0.96\ \mathrm{K}$. Ignoring dissociation would give $0.93\ \mathrm{K}$, so the ionisation contributes about 3%. Note that the extra particles from ionisation are what raise the colligative effect above the value for a non-electrolyte of the same molality.