Titration on Diprotic Systems

${}^{\mathbf{11}\mathbf{-}\mathbf{26}}$. Calculate the ${}^{\mathbf{pH}}$whenrole="math" localid="1655019122863" ${}^{\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{mL}}$of ${}^{\mathbf{0}\mathbf{.}\mathbf{0200}\mathbf{M}}$$\mathbf{-}\mathbf{2}$aminophenol have been titrated with role="math" localid="1655019100358" ${}^{\mathbf{10}\mathbf{.}\mathbf{9}\mathbf{mL}}$of${}^{\mathbf{0}\mathbf{.}\mathbf{0150}{\mathbf{MHCIO}}_{\mathbf{4}}}$ .

• 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 the task we have a titration of a weak diprotic base ( ${}^{\mathbf{2}}$-aminophenol) with a strong acid${}^{\left({\mathrm{HCIO}}_4\right)}$ .

To make the calculations easier we can find the equivalent point.

That is the volume of acid needed to neutralize the base:

$\mathbf{V}\left({\mathrm{HCIO}}_4\right)\mathbf{=}\frac{\mathbf{c}\left(2‐\mathrm{aminophenol}\right)\mathbf{.}\mathbf{}\mathbf{v}\left(2‐\mathrm{aminophenol}\right)}{\mathbf{c}\left({\mathrm{HCIO}}_4\right)}$

$\mathbf{V}\mathbf{\left(}{\mathbf{HCIO}}_{\mathbf{4}}\mathbf{\right)}\mathbf{=}\frac{\mathbf{25}\mathbf{.}\mathbf{0}\mathbf{mL}\mathbf{}\mathbf{·}\mathbf{0}\mathbf{.}\mathbf{0200}\mathbf{M}}{\mathbf{0}\mathbf{.}\mathbf{0150}\mathbf{M}}$

$\mathbf{V}\mathbf{\left(}{\mathbf{HCIO}}_{\mathbf{4}}\mathbf{\right)}\mathbf{=}\mathbf{33}\mathbf{.}\mathbf{33}\mathit{m}\mathit{L}$

Since the volume of ${}^{{\mathbf{HCIO}}_{\mathbf{4}}}$needed to reach the equivalence point is${}^{\mathbf{33}\mathbf{.}\mathbf{33}\mathbf{mL}}$, when we add${}^{\mathbf{10}\mathbf{.}\mathbf{9}\mathbf{mL}}$of it, the remaining volume of ${}^{\mathbf{2}}$aminophenol is ${}^{\mathbf{22}\mathbf{.}\mathbf{43}\mathbf{mL}}$and the volume of ${}^{{\mathbf{BH}}^{\mathbf{+}}}$is ${}^{\mathbf{10}\mathbf{.}\mathbf{9}\mathbf{mL}}$ since its equivalent to the volume of the added acid.

To calculate ${}^{{\mathbf{pK}}_{\mathbf{b}\mathbf{1}}}$value we can use the appendix ${}^{\mathbf{G}}$:

${\mathbf{pK}}_{\mathbf{b}\mathbf{1}}\mathbf{=}\mathbf{14}\mathbf{-}{\mathbf{pK}}_{\mathbf{a}\mathbf{1}}$

$\mathbf{=}\mathbf{14}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{70}$

$\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{30}$

Now we calculate the pH before the first equivalence point by using the Henderson-Hasselbalch equation:

$\mathbf{pOH}\mathbf{=}{\mathbf{pK}}_{\mathbf{b}\mathbf{1}}\mathbf{+}\mathbf{log}\frac{\left[{\mathrm{BH}}^{+}\right]}{\left[B\right]}$

At ${}^{\mathbf{10}\mathbf{.}\mathbf{9}\mathbf{mL}}$

$\mathbf{=}\mathbf{9}\mathbf{.}\mathbf{30}\mathbf{+}\mathbf{log}\frac{\mathbf{10}\mathbf{.}\mathbf{9}\mathbf{/}\mathbf{33}\mathbf{.}\mathbf{33}}{\mathbf{22}\mathbf{.}\mathbf{43}\mathbf{/}\mathbf{33}\mathbf{.}\mathbf{33}}$

$\mathbf{=}\mathbf{8}\mathbf{.}\mathbf{99}$

$\mathbf{pH}\mathbf{=}\mathbf{14}\mathbf{-}\mathbf{8}\mathbf{.}\mathbf{99}$

$\mathbf{=}\mathbf{5}\mathbf{.}\mathbf{01}$

Therefore, the at the first equivalence point is ${}^{\mathbf{5}\mathbf{.}\mathbf{01}}$.