Question

Titration on Diprotic Systems

$\mathbf{11}\mathbf{-}\mathbf{28}$ . A solution containing 0.100M glutamic acid (the molecule with no net charge) was titrated to its first equivalence point with0.0250MRbOH.

(a) Draw the structures of reactants and products.

(b) Calculate the at the first equivalence point.

Short answer

(a) The structures of reactants and products is

(b) The pH at the first equivalence point is 7.13

Step-by-step solution

Titration of Diprotic Systems

• 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.

Determine the structures of reactants and products

(a)

Reaction of glutamic acid and RbOH = 7.13 their structures are:

Determine the  of the first equivalence point

We have a titration of glutamic acid with a strong base ( RbOH ).

The pK_a values of glutamic acid can be found in the appendix G .

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

$$\begin{gathered} \mathrm{pH}=\frac{1}{2}·\left({\mathrm{pK}}_{\mathrm{a}2}+{\mathrm{pK}}_{\mathrm{a}3}\right) \\ \mathrm{pH}=\frac{1}{2}·\left(4.30+9.96\right) \\ \mathrm{pH}=7.13 \end{gathered}$$

The pH at the first equivalence point is 7.13