Question

Consider the titration of 50.0 mL of 1.0 M glycine hydrochloride [(H₃NCH₂COOH)Cl], with 1.0 M NaOH.For ⁺H₃NCH₂COOH, Ka for the carboxylic acid groupis $\mathbf{4}\mathbf{.3}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{3}}$and K_b for the amino group is $\mathbf{6}\mathbf{.0}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{5}}$.

a. Calculate the pH after 25.0 mL, 50.0 mL, and75.0 mL of NaOH has been added. Hint: Refer toSection 8.7 of the text, and treat the titration of⁺H₃NCH₂COOH as a diprotic acid titration.

b. Sketch the titration curve and indicate the majoramino acid species present after 0.0 mL, 25.0 mL,50.0 mL, 75.0 mL, and 100.0 mL of 1.0 M NaOHhave reacted completely. Assume the initial pH is1.2 and the pH at the second equivalence pointis 11.7.

c. At what pH do the majority of amino acid molecules have a net charge of zero? This pH is calledthe isoelectric point.

d. At what pH is the net charge of the major aminoacid species present equal to$\mathbf{+}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{?}\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{?}$

Short answer

a.pH after adding 25.0 mL, 50.0 mL and 75.0 mL NaOH solution will be 2.37,6.07 and 9.77 respectively.

(b)

The major amino acid species present are:

point A (0.0 mL OH⁻): ⁺NH₃CH₂COOH

point B (25.0 mL OH⁻): ⁺NH₃CH₂COOH and ⁺NH₃CH₂COO⁻

point C (50.0 mL OH⁻): ⁺NH₃CH₂COO⁻

point D (75.0 mL OH⁻): ⁺NH₃CH₂COO⁻ and NH₂CH₂COO⁻

point E (100.0 mL OH⁻): NH₂CH₂COO

(c) Isoelectric point is at pH = 6.07

(d) The net charge is$\frac{(+1+0)}{2}=+\frac{1}{2}$. This occurs at the first halfway point to equivalence, where $pH=p_{K_{a_1}}=2.37$. The net charge is = $\frac{(0-1)}{2}=-\frac{1}{2}$. This occurs at the second halfway point to equivalence, where$pH=p_{K_{a_2}}=9.77$

Step-by-step solution

Step 1:Determine the pH

Titration of a diprotic acid $\left(H_{2}A\right)$

Given,

${}^{+}N{H}_{3}C{H}_{2}COOH\rightleftharpoons {}^{+}N{H}_{3}C{H}_{2}CO{O}^{-}+H^{+}$

$\begin{array}{c}{K}_{a_1}=\frac{\left[N{H}_{3}C{H}_{2}CO{O}^{-}\right]\left[H^{+}\right]}{\left[{}^{+}N{H}_{3}C{H}_{2}COOH\right]}\\ =4.3\times {10}^{-3}\end{array}$

${}^{+}N{H}_{3}C{H}_{2}CO{O}^{-}\rightleftharpoons N{H}_{2}C{H}_{2}CO{O}^{-}+H^{+}$

$\begin{array}{c}{K}_{a_2}=\frac{\left[N{H}_{2}C{H}_{2}CO{O}^{-}\right]\left[H^{+}\right]}{\left[{}^{+}N{H}_{3}C{H}_{2}CO{O}^{-}\right]}\\ =\frac{K_w}{K_b}\\ =\frac{1.0\times {10}^{-14}}{6.0\times {10}^{-5}}\\ =1.7\times {10}^{-10}\end{array}$

As OH⁻ is added, it reacts completely with the best acid present. From 0−50.0 mL of OH⁻added, the reaction is:

${}^{+}N{H}_{3}C{H}_{2}COOH+O{H}^{-}\rightleftharpoons {}^{+}N{H}_{3}C{H}_{2}CO{O}^{-}+H_{2}O$

At 50.0 mL OH⁻ added (the first equivalence point), all of the ⁺NH₃CH₂COOH has beenconverted into ⁺NH₃CH₂COO⁻. This is an amphoteric species. To determine the pH whenan amphoteric species is the major species present, we use the formula

$pH=\frac{p_{K_{a_1}}+p_{K_{a_2}}}{2}$

From 50.1−100.0 mL of OH⁻ added, the reaction that occurs is

${}^{+}N{H}_{3}C{H}_{2}CO{O}^{-}+O{H}^{-}\rightleftharpoons N{H}_{2}C{H}_{2}CO{O}^{-}+H_{2}O$

100.0 mL of OH− added represents the second equivalence point whereH₂NCH₂COO⁻ is the major amino acid species present.

25.0 mL of OH⁻ added represents the first halfway point to equivalence. Here,

[⁺NH₃CH₂COOH] = [⁺NH₃CH₂COO⁻]. This is a buffer solution where

At 25.0 mL OH⁻ added:

$\begin{array}{c}pH=p{K}_{a_1}\\ =-\log (4.3\times {10}^{-3})\\ =2.37\end{array}$

50.0 mL of OH⁻ added represents the first equivalence point. Here, ⁺NH₃CH₂COO⁻ isthe major amino acid species present. This is amphoteric species. At 50.0 mL OH⁻added:

$\begin{array}{c}pH=\frac{p_{K_{a_1}}+p_{K_{a_2}}}{2}\\ =\frac{2.37-\log (1.7\times {10}^{-10})}{2}\\ =\frac{2.37+9.77}{2}\\ =6.07\end{array}$

75.0 mL of OH⁻ added represents the second halfway point to equivalence. Here,

[⁺NH₃CH₂COO⁻] = [NH₂CH₂COO⁻] and . At 75.0 mL OH⁻ added.

$\begin{array}{c}pH=p_{K_{a_2}}\\ =-\log (1.7\times {10}^{-10})\\ =9.77\end{array}$

Titration curve and major amino acid species

The major amino acid species present are:

point A (0.0 mL OH⁻): ⁺NH₃CH₂COOH

point B (25.0 mL OH⁻): ⁺NH₃CH₂COOH and ⁺NH₃CH₂COO⁻

point C (50.0 mL OH⁻): ⁺NH₃CH₂COO⁻

point D (75.0 mL OH⁻): ⁺NH₃CH₂COO⁻ and NH₂CH₂COO⁻

point E (100.0 mL OH⁻): NH₂CH₂COO

Determine the isoelectric point

The charged amino acid species are shown below:

⁺NH₃CH₂COOH net charge = +1

⁺NH₃CH₂COO⁻ net charge = 0

NH₂CH₂COO⁻ net charge = −1

The net charge is zero at the pH when the major amino acid species present in ⁺NH₃CH₂COO⁻ form; this occurs at the first equivalence point in the titrationproblem.

From part a, this occurs at pH = 6.07 (the isoelectric point).

Determine pH for part (d)

The net charge is +1/2 when = [⁺NH₃CH₂COOH] = [⁺NH₃CH₂COO⁻]; net charge =$\frac{(+1+0)}{2}=+\frac{1}{2}$. This occurs at the first halfway point to equivalence, where$pH=p_{K_{a_1}}=2.37$ . The net charge is $-\frac{1}{2}$ when [⁺NH₃CH₂COO⁻] = [NH₂CH₂COO⁻]; netcharge = $\frac{(0-1)}{2}=-\frac{1}{2}$ . This occurs at the second halfway point to equivalence, where$pH=p_{K_{a_2}}=9.77$