Titration on Diprotic Systems

$\mathbf{11}\mathbf{-}\mathbf{29}$ Find the pH of the solution when 0.0100M tyrosine is titrated to the equivalence point with0.00400MHCIO₄.

• Titration of a diprotic acid with a specified quantity of NaOH solution
• Diprotic acid's molecular weight (or molar mass) is expressed in grams per mole.
• You can determine the original acid sample's mass in grams by weighing it.
• The volume of NaOH titrant required to achieve the first equivalence point can be used to calculate moles.

As we can see in the task, the molarity of tyrosine is two and a half times higher than the molarity HCIO₄ of so we can put their volumes in the following relation:

$\mathrm{V}\left(\mathrm{tyrosine}\right)=\frac{5}{2}\mathrm{V}\left({\mathrm{HCIO}}_4\right)$

To calculate at the equivalence point we need to write the reaction of weak acid with water:

${\mathrm{H}}_3{\mathrm{T}}^{+}\rightleftharpoons {\mathrm{H}}_2\mathrm{T}+{\mathrm{H}}^{+}$

The concentrations are:

$$\begin{gathered} \left[{\mathrm{H}}_3{\mathrm{T}}^{+}\right]=\mathrm{F}\text{'}-\mathrm{x} \\ \left[{\mathrm{H}}_2\mathrm{T}\right]=\mathrm{x} \\ \left[{\mathrm{H}}^{+}\right]=\mathrm{x} \end{gathered}$$

The only thing we need to do before putting the concentrations in the weak acid equation is to calculate the formal concentration of tyrosine, F' :

$$\begin{gathered} \mathrm{F}\text{'}=\mathrm{concentration}\mathrm{of}\mathrm{initial}\mathrm{tyrosine}\frac{\mathrm{initial}\mathrm{volume}\mathrm{of}\mathrm{tyrosine}}{\mathrm{total}\mathrm{volume}\mathrm{of}\mathrm{the}\mathrm{solution}} \\ \mathrm{F}\text{'}=0.0100\mathrm{M}·\frac{\mathrm{V}\left(\mathrm{tyrosine}\right)}{\mathrm{V}\left(\mathrm{tyrosine}\right)+\mathrm{V}\left({\mathrm{HCIO}}_4\right)} \\ \mathrm{F}\text{'}=0.0100\mathrm{M}·\frac{1}{7/2} \\ \mathrm{F}\text{'}=0.00286\mathrm{M} \end{gathered}$$

We can calculate the ${\mathrm{K}}_{\mathrm{a}1}$value of tyrosine by using the appendix G :

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}1}={10}^{-2.41} \\ =3.9\times {10}^{-3} \end{gathered}$$

Now we can insert it in the equation for the equilibrium of the weak acid:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}1}=\frac{\left[{\mathrm{H}}_2\mathrm{T}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{H}}_3{\mathrm{T}}^{+}\right]} \\ =\frac{{\mathrm{x}}^2}{0.00286-\mathrm{x}} \end{gathered}$$

By rearranging the equation we get a quadratic equation:

${\mathrm{x}}^2+3.9\times {10}^{-3}\mathrm{x}-1.12\times {10}^{-5}=0$

By solving the quadratic equation we get the pH at 31.9mL :

$$\begin{gathered} \mathrm{x}=\left[{\mathrm{H}}^{+}\right] \\ =0.00192\mathrm{M} \\ \mathrm{pH}=2.72 \end{gathered}$$

Therefore, the pH of the solution is 2.72.