91Ó°ÊÓ

• Titration of a diprotic acid with a specified quantity of $\mathrm{NaOH}$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 $\mathrm{NaOH}$ titrant required to achieve the first equivalence point can be used to calculate moles.

(a)

We have a titration of cysteine $\left({\mathrm{H}}_2\mathrm{C}\right)$with a strong acid$\left({\mathrm{HCIO}}_4\right)$.

The values of cysteine can be found in the appendix.

We can calculate at the first equivalence point by using the following equation

$$\begin{gathered} \mathrm{pH}=\frac{1}{2}·\left({\mathrm{pK}}_{\mathrm{a}2}+{\mathrm{pK}}_{\mathrm{a}3}\right) \\ \mathrm{pH}=\frac{1}{2}·\left(8.36+10.74\right) \\ \mathrm{pH}=9.55 \end{gathered}$$

Therefore, the pH at the first equivalence point is $\mathrm{pH}=9.55$

(b)

To calculate the quotient presented in the task, we need to calculate at the first equivalent point and use that to calculate the value of the quotient $\left[{\mathrm{C}}^{2-}\right]/\left[{\mathrm{HC}}^{-}\right]$.

First we can write the reaction of weak acid with water

${\mathrm{H}}_3{\mathrm{C}}^{+}⇌{\mathrm{H}}_2\mathrm{C}+{\mathrm{H}}^{+}$

The concentrations are:

$$\begin{gathered} \left[{\mathrm{H}}_3{\mathrm{C}}^{+}\right]=\mathrm{F}\text{'}-\mathrm{x} \\ \left[{\mathrm{H}}_2\mathrm{C}\right]=\mathrm{x} \\ \left[{\mathrm{H}}^{+}\right]=\mathrm{x} \end{gathered}$$

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

$$\begin{gathered} {\mathrm{K}}_{\mathrm{a}1}=\frac{\left[{\mathrm{H}}_2\mathrm{C}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{H}}_3{\mathrm{C}}^{+}\right]} \\ =\frac{{\mathrm{x}}^2}{0.0500-\mathrm{x}} \end{gathered}$$

The ${\mathrm{K}}_{\mathrm{a}1}$value of cysteine can be found in the appendix G :

${\mathrm{K}}_{\mathrm{a}1}\left(\mathrm{cysteine}\right)=0.02$

By rearranging the equation we get a quadratic equation:

$x^2+0.02x-1×{10}^{-3}=0$

By solving the quadratic equation we get the at first equivalence point:

$$\begin{gathered} \mathrm{x}=\left[{\mathrm{H}}^{+}\right] \\ =0.023\mathrm{M} \\ \mathrm{pH}=1.64 \end{gathered}$$

To calculate the value of the quotient $\left[{\mathrm{C}}^{2-}\right]/\left[{\mathrm{HC}}^{-}\right]$we can use the following Henderson-Hasselbalch equation:

$$\begin{gathered} \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}3}+\log \frac{\left[{\mathrm{C}}^{2-}\right]}{\left[{\mathrm{HC}}^{-}\right]} \\ =10.74+\log \frac{\left[{\mathrm{C}}^{2-}\right]}{\left[{\mathrm{HC}}^{-}\right]} \end{gathered}$$

By rearranging the equation we get:

$$\begin{gathered} \frac{\left[{\mathrm{C}}^{2-}\right]}{\left[{\mathrm{HC}}^{-}\right]}={10}^{1.64-10.74} \\ =7.9×{10}^{-10} \end{gathered}$$

Therefore, the value of the quotient is $\left[{\mathrm{C}}^{2-}\right]/\left[{\mathrm{HC}}^{-}\right]=7.9×{10}^{-10}$.