Question

The ionization constant of chloroacetic acid $\left({\mathrm{ClCH}}_2\mathrm{COOH}\right)$in water is role="math" localid="1663723728735" $\mathbf{1}\mathbf{.}\mathbf{528}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$at $\mathbf{0}\mathbf{°}\mathit{C}$and $\mathbf{1}\mathbf{.}\mathbf{230}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$at $\mathbf{40}\mathbf{°}\mathit{C}$. Calculate the enthalpy of ionization of the acid in the water, assuming that $\mathbf{∆}\mathbf{H}$and $\mathbf{∆}\mathbf{S}$ are constant over this temperature range.

Short answer

The enthalpy of ionization of the acid in water over this temperature range is $∆\mathrm{H}°=-3853.22\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{mol}$}\right.$.

Step-by-step solution

Van’t Hoff equation

Using the van't Hoff equation, calculate the enthalpy of ionization is

$\ln \left(\frac{{\mathrm{K}}_2}{{\mathrm{K}}_1}\right)=\frac{-∆\mathrm{H}°}{R}\times \left(\frac{1}{{\mathrm{T}}_2}-\frac{1}{{\mathrm{T}}_1}\right)$

Step 2: Calculation

When combining the values,

$$\begin{gathered} \ln \left(\frac{1.230\times {10}^{-3}}{1.528\times {10}^{-3}}\right)=\frac{-∆\mathrm{H}°}{8.314{\displaystyle \raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.}\times \mathrm{mol}}\times \left(\frac{1}{313}\mathrm{K}-\frac{1}{273}\mathrm{K}\right) \\ -0.2169=\frac{{\displaystyle -∆\mathrm{H}°}}{{\displaystyle 8.314\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.\times \mathrm{mol}}}\times \left(-4.68\times {10}^{-4}\mathrm{K}\right) \\ ∆\mathrm{H}°=\frac{0.2169\times 8.314\raisebox{1ex}{$\mathrm{J}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{K}$}\right.\times \mathrm{mol}}{4.68\times {10}^{-4}\mathrm{K}} \\ ∆\mathrm{H}°=-3853.22\raisebox{1ex}{${\displaystyle \mathrm{J}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \mathrm{mol}}$}\right. \end{gathered}$$

Therefore, as a result, the enthalpy of ionization of the acid in water is $∆\mathrm{H}°=-3853.22\raisebox{1ex}{${\displaystyle \mathrm{J}}$}\!\left/ \!\raisebox{-1ex}{${\displaystyle \mathrm{mol}}$}\right.$.