Chapter 4, Problem 9P

Answers to all problems are at the end of this book. Detailed solutions are available in the Student Solutions Manual, Study Guide, and Problems Book.

Calculating pH in Amino Acid Solutions III (Integrates with Chapter 2.) Calculate the pH of a 0.3 M solution of (a) leucine hydrochloride, (b) sodium leucinate, and (c) isoelectric leucine.

Interpretation Introduction

(a)

To calculate:

The pH of 0.3 M solution of leucine hydrochloride.

Introduction:

The Henderson-Hasselbalch equation for the initial disassociation is:

${\text{pH=pK}}_1+\log \frac{\left[{\text{Leu}}^0\right]}{\left[{\text{Leu}}^{+}\right]}$

Here, [Leu⁰] is the concentration of the Leucine’s neutral form and [Leu⁺] is the concentration of leucine’s fully protonated form.

Explanation of Solution

Leucine hydrochloride is so called because it has the hydrochloric acid and amino acid leucine. Hydrochloric acid will lower the solution’s pH and form it acidic. This means that within the solution most of the leucine will be in the protonated form or in the neutral form.

The Henderson-Hasselbalch equation for the initial disassociation is:

${\text{pH=pK}}_1+\log \frac{\left[{\text{Leu}}^0\right]}{\left[{\text{Leu}}^{+}\right]}$

Here, [Leu⁰] is the concentration of the Leucine’s neutral form and [Leu⁺] is the concentration of leucine’s fully protonated form.

The concentration of Leucine of the solution is known. The mass balance of leucine in the solution is:

  $\left[{\text{Leu}}^{+}\right]+\left[{\text{Leu}}^0\right]=0.3\text{M}$

The charges of all the species in the solution must also be equivalent. The solution contains four charges species: COO- and NH₃+ of leucine and H+ and Cl- from hydrochloric acid.

$\left[{\text{NH}}_3^{+}\right]+\left[{\text{H}}^{+}\right]=\left[{\text{COO}}^{-}\right]\left[{\text{Cl}}^{-}\right]$

Since the solution is acidic, all the amino groups of leucine are likely to be protonated. This means that NH₃+ is equal to the concentration of leucine. The concentration of leucine and the concentration of chloride ions are both 0.3 M, so $\left[{\text{NH}}_3^{+}\right]$= $\left[{\text{Cl}}^{-}\right]$.

The charge balance equation can now be simplified.

$\left[{\text{H}}^{+}\right]$= $\left[{\text{COO}}^{-}\right]$

The deprotonated carboxyl group will only be present in Leu⁰, so [COO^-] = [Leu⁰]. The mass balance equation can now be rearranged

  $\begin{array}{l}\left[{\text{Leu}}^{+}\right]=0.3\text{M}-\left[{\text{Leu}}^0\right]\\ =0.3\text{M}-\left[{\text{COO}}^{-}\right]\end{array}$

The equation can be substituted into the Henderson-Hasselbalch equation and [H⁺]can be substituted for [COO^-].

  $\begin{array}{l}{\text{pH=pK}}_1+\log \frac{\left[{\text{COO}}^{-}\right]}{0.3\text{M}-\left[{\text{COO}}^{-}\right]}\\ {\text{ = pK}}_1\text{+}\log \frac{\left[{\text{H}}^{+}\right]}{0.3\text{M}-\left[{\text{H}}^{+}\right]}\end{array}$

The Henderson-Hasselbalch equation can now be simplified if the assumption is made that the value of [H⁺] is significantly less than 0.3 M. This would allow the [H⁺] term in the denominator to be dropped.

  $\begin{array}{l}{\text{pH = pK}}_1\text{+}\log \frac{\left[{\text{H}}^{+}\right]}{0.3\text{M}}\\ {\text{ = pK}}_{\text{1}}+\log \left[{\text{H}}^{+}\right]-\log \left(0.3\text{M}\right)\\ =2.4+\log \left[{\text{H}}^{+}\right]+0.52\\ \text{ = 2}\text{.92+}\log \left[{\text{H}}^{+}\right]\end{array}$

Note: - pH = -log[H⁺], thus, log[H⁺] = - pH.

  $\begin{array}{l}\text{pH = 2}\text{.92}-\text{pH}\\ \text{2pH = 2}\text{.92}\\ \text{ pH = }\boxed{\text{1}\text{.46}}\end{array}$

Interpretation Introduction

(b)

To determine:

Determine the pH in amino acid solutions.

Introduction:

Amino acids are so named because they all are weak acids. All amino acids have at least two dissociable hydrogens, (diprotic), and some have three dissociable hydrogens and are triprotic.

Explanation of Solution

Sodium leucinate is so called since it has sodium ions in solution having the negative form of leucine. This denotes that the pH of the solution is basic. Most of the leucine in solution will be in the fully deprotonated form of the neutral form.

The Henderson-Hasselbalch equation for the second disassociation is:

${\text{pH=pK}}_2+\log \frac{\left[{\text{Leu}}^{-}\right]}{\left[{\text{Leu}}^0\right]}$

Here, [Leu^-] is the concentration of the Leucine’s fully deprotonated form and [Leu⁰] is the concentration of leucine’s neutral form.

The concentration of Leucine of the solution is known. The mass balance of leucine in the solution is:

  $\left[{\text{Leu}}^{-}\right]+\left[{\text{Leu}}^0\right]=0.3\text{M}$

The charges of all the species in the solution must also be equivalent. The solution contains four charges species: COO- and NH₃+ of leucine and Na+ and OH- in solution.

$\left[{\text{NH}}_3^{+}\right]+\left[{\text{Na}}^{+}\right]=\left[{\text{COO}}^{-}\right]\left[{\text{OH}}^{-}\right]$

Since the solution is acidic, all the carboxyl group of leucine are expected to be deprotonated. This means that COO- is equal to the concentration of leucine. The concentration of leucine and the concentration of sodium ions are both 0.3 M, so $\left[{\text{Na}}^{+}\right]$= $\left[{\text{COO}}^{-}\right]$.

The charge balance equation can now be simplified.

$\left[{\text{NH}}_3^{+}\right]$= $\left[{\text{OH}}^{-}\right]$

The protonated amino group will only be present in Leu⁰, so [NH₃⁺] = [Leu⁰]. The mass balance equation can now be rearranged

  $\begin{array}{l}\left[{\text{Leu}}^{-}\right]=0.3\text{M}-\left[{\text{Leu}}^0\right]\\ =0.3\text{M}-\left[{\text{NH}}_3^{+}\right]\end{array}$

The equation can be substituted into the Henderson-Hasselbalch equation and [OH^-]can be substituted for [NH₃⁺].

  $\begin{array}{l}{\text{pH=pK}}_2+\log \frac{0.3\text{M}-\left[{\text{NH}}_3^{+}\right]}{\left[{\text{NH}}_3^{+}\right]}\\ {\text{ = pK}}_2\text{+}\log \frac{0.3\text{M}-\left[{\text{OH}}^{-}\right]}{\left[{\text{OH}}^{-}\right]}\end{array}$

The Henderson-Hasselbalch equation can now can be simplified if the assumption is made that the value of [OH^-] is significantly less than 0.3 M. This would allow the [OH^-] term in the numerator to be dropped.

  $\begin{array}{l}{\text{pH = pK}}_2\text{+}\log \frac{0.3\text{M}}{\left[{\text{OH}}^{-}\right]}\\ {\text{ = pK}}_{\text{2}}+\log \left(0.3\text{M}\right)-\log \left[{\text{OH}}^{-}\right]\\ =9.6-0.52-\log \left[{\text{OH}}^{-}\right]\\ \text{ = 9}\text{.08}-\log \left[{\text{OH}}^{-}\right]\end{array}$

Note: - pOH = -log [OH^-], and pOH =14-pH, thus, -log [OH^-] = 14 - pH.

  $\begin{array}{l}\text{pH = 9}\text{.08+14}-\text{pH}\\ \text{2pH = 23}\text{.08}\\ \text{ pH = }\boxed{\text{11}\text{.54}}\end{array}$

Interpretation Introduction

(c)

To determine:

Determine the pH in amino acid solutions.

Introduction:

Amino acids are so named because they all are weak acids. All amino acids have at least two dissociable hydrogens, (diprotic), and some have three dissociable hydrogens and are triprotic.

Explanation of Solution

The isoelectric point of leucine is a set pH value which is independent of concentration. The isoelectric point is the pH value at which the solution of leucine is neutral. The value is determined by averaging the pKa values.

$\begin{array}{l}\text{pH = }\frac{{\text{pK}}_{\text{1}}+{\text{pK}}_{\text{2}}}{2}\\ \text{ = }\frac{2.4+9.6}{2}\\ \text{ = }\frac{\text{12}\text{.0}}{2}\\ \text{ = 6}\text{.0}\end{array}$