91Ó°ÊÓ

• Titration of a diprotic acid with a specified quantity of ${}^{\mathbf{NaOH}}$solution

• The molecular weight (or molar mass) of diprotic acid is measured in grams per mole.

• Weighing the original acid sample will give you its mass in grams.

• The volume of ${}^{\mathbf{NaOH}}$titrant required to achieve the first equivalence point can be used to calculate moles.

In this task, we have a titration of a diprotic salt (dipotassium oxalate) with a strong acid (${}^{{\mathbf{HCIO}}_{\mathbf{4}}}$ ).

First we need to calculate how many molls of ${}^{{\mathbf{HCIO}}_{\mathbf{4}}}$would be added to the solution

$\mathbf{n}\left({\mathrm{HCIO}}_4\right)\mathbf{}\mathbf{=}\mathbf{c}\left({\mathrm{HCIO}}_4\right)\mathbf{}\mathbf{·}\mathbf{}\mathbf{V}\left({\mathrm{HCIO}}_4\right)$

$\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{800}\mathbf{M}\mathbf{·}\mathbf{20}\mathbf{.}\mathbf{0}\mathbf{mL}$

$\mathbf{=}\mathbf{16}\mathbf{mmol}$

By looking at the${}^{{\mathbf{pK}}_{\mathbf{a}}}$values of oxalic acid in the appendix${}^{\mathbf{G}}$, we can see that the${}^{\mathbf{pH}}$of${}^{\mathbf{4}\mathbf{.}\mathbf{40}}$is higher than both of the${}^{{\mathbf{pK}}_{\mathbf{a}}}$values so we can conclude that the oxalate anion has not been neutralized completely and that there is still some dipotassium oxalate left in the solution and we can write the equilibrium equation ${\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{}\mathbf{+}\mathbf{}{\mathbf{H}}^{\mathbf{+}}\mathbf{⇌}{\mathbf{HC}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}^{\mathbf{-}}$

The initial numbers of moles are:

$\left[{\mathrm{HC}}_2{O}_4^{-}\right]\mathbf{=}\mathbf{0}$

$\mathbf{\left[}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}^{\mathbf{2}\mathbf{-}}\mathbf{\right]}\mathbf{=}\mathbf{x}\left[H^{+}\right]$

$\mathbf{=}\mathbf{16}$

The final numbers of moles are:

$\left[H_2{O}_4^{-}\right]\mathbf{}\mathbf{=}\mathbf{16}\mathbf{}\left[C_2{O}_4^{2-}\right]$

$\mathbf{=}\mathbf{x}\mathbf{-}\mathbf{16}\left[H^{+}\right]$

$\mathbf{=}\mathbf{0}$

By putting the final numbers of moles in the Henderson-Hasselbalch equation we get:

$\mathbf{pH}\mathbf{=}{\mathbf{pK}}_{\mathbf{a}\mathbf{2}}\mathbf{+}\mathbf{log}\frac{\left[C_2{O}_4^{2-}\right]}{\left[{\mathrm{HC}}_2{O}_4^{-}\right]}$

${\mathbf{pK}}_{\mathbf{a}\mathbf{2}}\mathbf{+}\mathbf{log}\mathbf{}\left(\frac{x-16}{16}\right)$

By rearranging the equation in the step ${}^{\mathbf{4}}$we get:

$\mathbf{log}\left(\frac{x-16}{16}\right)\mathbf{=}\mathbf{pH}\mathbf{-}{\mathbf{pK}}_{\mathbf{a}\mathbf{2}}\frac{\mathbf{x}\mathbf{-}\mathbf{16}}{\mathbf{16}}$

${\mathbf{10}}^{\mathbf{pH}\mathbf{-}{\mathbf{pK}}_{\mathbf{a}\mathbf{2}}}\mathbf{}\mathbf{x}\mathbf{-}\mathbf{16}$

${\mathbf{10}}^{\mathbf{pH}\mathbf{-}{\mathbf{pK}}_{\mathbf{a}\mathbf{2}}}\mathbf{}\mathbf{}\mathbf{·}\mathbf{16}\mathit{x}$

$\mathbf{=}\mathbf{16}\mathbf{+}\left({10}^{\mathrm{pH}-{\mathrm{pK}}_{a2}}·16\right)\mathbf{x}$

$\mathbf{=}\mathbf{16}\mathbf{+}\mathbf{(}{\mathbf{10}}^{\mathbf{4}\mathbf{.}\mathbf{40}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{266}}\mathbf{}\mathbf{·}\mathbf{16}\mathbf{)}\mathbf{x}$

$\mathbf{n}\left(K_2{C}_2{O}_4\right)$

$\mathbf{=}\mathbf{37}\mathbf{.}\mathbf{78}\mathbf{mmol}$

Lastly, we can calculate the mass of ${}^{{\mathbf{K}}_{\mathbf{2}}{\mathbf{C}}_{\mathbf{2}}{\mathbf{O}}_{\mathbf{4}}}$:

$\mathbf{m}\left(K_2{C}_2{O}_4\right)\mathbf{=}\mathbf{n}\left(K_2{C}_2{O}_4\right)\mathbf{}\mathbf{·}\mathbf{m}\left(K_2{C}_2{O}_4\right)\mathbf{}$

$\mathbf{=}\mathbf{37}\mathbf{.}\mathbf{78}\mathbf{×}{\mathbf{10}}^{\mathbf{-}\mathbf{3}}\mathbf{}\mathbf{mol}\mathbf{·}\mathbf{166}\mathbf{.}\mathbf{22}\mathbf{mol}\mathbf{/}\mathbf{g}$

$\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{28}\mathbf{}\mathbf{g}$

Therefore, the mass of${}^{\mathbf{m}\left(K_2{C}_2{O}_4\right)}$is${}^{\mathbf{6}\mathbf{.}\mathbf{28}\mathbf{}\mathbf{g}}$ .