Chapter 4, Problem 6P

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.

1. Calculating Concentrations of Species in Amino Acid Solutions

(Integrates with Chapter 2.) Calculate the concentrations of all ionic species in a 0.25 M solution of histidine at pH 2, pH 6.4, and pH 9.3.

Interpretation Introduction

To propose:

The concentrations of all the species in 0.25 M solution of histidine at pH 2, pH 6.4, and pH 9.3.

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

To find out the concentrations of the all ionic species at various pH values. First write the Henderson-Hasselbalch equation for each dissociation:

$\begin{array}{l}{\text{pH =pK}}_1+\log \frac{\left[{\text{His}}^{+}\right]}{\left[{\text{His}}^{2+}\right]}\\ {\text{pH =pK}}_2+\log \frac{\left[{\text{His}}^0\right]}{\left[{\text{His}}^{+}\right]}\\ {\text{pH =pK}}_3+\log \frac{\left[{\text{His}}^{-}\right]}{\left[{\text{His}}^0\right]}\end{array}$

The total concentration of all species in solution will always be 0.25 M.

  $\left[{\text{His}}^{2+}\right]+\left[{\text{His}}^{+}\right]+\left[{\text{His}}^0\right]+\left[{\text{His}}^{-}\right]=0.25\text{M}$

A pH of 2 is closet to pK₁and the Henderson-Hasselbalch equation for the first disassociation will be used:

  $\begin{array}{l}\text{pH}-{\text{pK}}_1=\text{log}\frac{\left[{\text{His}}^{+}\right]}{\left[{\text{His}}^{2+}\right]}\\ {10}^{\text{pH}-{\text{pK}}_{\text{1}}}=\frac{\left[{\text{His}}^{+}\right]}{\left[{\text{His}}^{2+}\right]}\\ \left[{\text{His}}^{2+}\right]{10}^{\text{pH}-{\text{pK}}_{\text{1}}}=\left[{\text{His}}^{+}\right]\end{array}$

The value of histidine is 1.8

  $\begin{array}{l}\left[{\text{His}}^{2+}\right]{10}^{\text{2}-\text{1}\text{.8}}=\left[{\text{His}}^{+}\right]\\ \left[{\text{His}}^{2+}\right]{10}^{0.2}=\left[{\text{His}}^{+}\right]\\ 1.58\left[{\text{His}}^{2+}\right]=\left[{\text{His}}^{+}\right]\end{array}$

At pH 2, [His²⁺] and [His⁺] will be the main species in solution:

  $\begin{array}{l}\left[{\text{His}}^{2+}\right]+\left[{\text{His}}^{+}\right]=0.25\text{M}\\ \left[{\text{His}}^{2+}\right]+1.58\left[{\text{His}}^{2+}\right]=0.25\text{M}\\ 2.58\left[{\text{His}}^{2+}\right]=0.25\text{M}\\ 2.58\left[{\text{His}}^{2+}\right]=0.25\text{M}\\ \left[{\text{His}}^{2+}\right]=\boxed{0.097\text{M}}\end{array}$

The concentration of His²+ can be used to determine the concentration of His+

$\begin{array}{l}\text{1}\text{.58(0}\text{.097M)=}\left[{\text{His}}^{+}\right]\\ 0.153\text{M=}\left[{\text{His}}^{+}\right]\end{array}$

The value of [His⁰] could be determined from [His⁺] using the Henderson-Hasselbalch equation for second dissociation. The value of pK₂for histidine is 6.0.

  $\begin{array}{l}\text{ pH}-{\text{pK}}_2=\text{log}\frac{\left[{\text{His}}^0\right]}{\left[{\text{His}}^{+}\right]}\\ {10}^{\text{2}-\text{6}\text{.0}}=\frac{\left[{\text{His}}^{\text{0}}\right]}{\text{0}\text{.153M}}\\ 0.153\text{M}\left({10}^{-4}\right)=\left[{\text{His}}^0\right]\\ \boxed{1.53\times {10}^{-5}}=\left[{\text{His}}^0\right]\end{array}$

The value of [His^-] using the Henderson-Hasselbalch equation for the third dissociation. The value of pK₃for histidine is 9.2

  $\begin{array}{l}\text{ pH}-{\text{pK}}_3=\text{log}\frac{\left[{\text{His}}^{-}\right]}{\left[{\text{His}}^0\right]}\\ {10}^{\text{2}-\text{9}\text{.2}}=\frac{\left[{\text{His}}^{\text{-}}\right]}{\text{1}\text{.53}\times {\text{10}}^{-5}\text{M}}\\ \left(1.53\times {10}^{-5}\right)\left({10}^{-7.2}\right)=\left[{\text{His}}^{-}\right]\\ \boxed{9.6\times {10}^{-13}}=\left[{\text{His}}^{-}\right]\end{array}$

These calculations can be repeated to solve for the concentration of all of the histidine species at pH 6.4. A pH of 6.4 is closet to pK₂and the Henderson-Hasselbalch equation for the second dissociation will be used.

  $\begin{array}{l}\text{pH}-{\text{pK}}_2=\text{log}\frac{\left[{\text{His}}^0\right]}{\left[{\text{His}}^{+}\right]}\\ {10}^{\text{pH}-{\text{pK}}_{\text{2}}}=\frac{\left[{\text{His}}^0\right]}{\left[{\text{His}}^{+}\right]}\\ \left[{\text{His}}^{+}\right]{10}^{\text{pH}-{\text{pK}}_{\text{2}}}=\left[{\text{His}}^0\right]\end{array}$

The value of pK₂for histidine is 6.0.

  $\begin{array}{l}\left[{\text{His}}^{+}\right]{10}^{\text{6}\text{.4}-\text{6}\text{.0}}=\left[{\text{His}}^0\right]\\ \left[{\text{His}}^{+}\right]{10}^{0.4}=\left[{\text{His}}^0\right]\\ 2.51\left[{\text{His}}^{+}\right]=\left[{\text{His}}^0\right]\end{array}$

At pH 6.4, [His⁺] and [His⁰] will be the main species in solution:

  $\begin{array}{l}\left[{\text{His}}^{+}\right]+\left[{\text{His}}^0\right]=0.25\text{M}\\ \left[{\text{His}}^{+}\right]+2.51\left[{\text{His}}^{+}\right]=0.25\text{M}\\ 3.51\left[{\text{His}}^{+}\right]=0.25\text{M}\\ \left[{\text{His}}^{+}\right]=\boxed{0.071\text{M}}\end{array}$

The concentration of His+ can be used to determine the concentration of His⁰$\begin{array}{l}\text{2}\text{.51(0}\text{.071M)=}\left[{\text{His}}^0\right]\\ \boxed{0.179\text{M}}\text{=}\left[{\text{His}}^{+}\right]\end{array}$

The value of [His²⁺] could be determined from [His⁺] using the Henderson-Hasselbalch equation for first dissociation. The value of pK₁for histidine is 1.8.

  $\begin{array}{l}\text{pH}-{\text{pK}}_1=\text{log}\frac{\left[{\text{His}}^{+}\right]}{\left[{\text{His}}^{2+}\right]}\\ {10}^{\text{6}\text{.4}-\text{1}\text{.8}}=\frac{0.071\text{M}}{\left[{\text{His}}^{2+}\right]}\\ \left[{\text{His}}^{2+}\right]=\frac{0.071\text{M}}{\left[{\text{10}}^{4.6}\right]}\\ \left[{\text{His}}^{2+}\right]=\boxed{\text{1}\text{.78}\times {\text{10}}^{-6}\text{M }}\end{array}$

The value of [His^-] could be determined from [His⁰] using the Henderson-Hasselbalch equation for third dissociation. The value of pK₃for histidine is 9.2.

  $\begin{array}{l}\text{ pH}-{\text{pK}}_3=\text{log}\frac{\left[{\text{His}}^{-}\right]}{\left[{\text{His}}^0\right]}\\ {10}^{\text{6}\text{.4}-\text{9}\text{.2}}=\frac{\left[{\text{His}}^{-}\right]}{0.0179\text{M}}\\ \left(0.0179\text{M}\right)\left({10}^{-2.8}\right)=\left[{\text{His}}^{-}\right]\\ \boxed{\text{2}\text{.84}\times {\text{10}}^{-4}}=\left[{\text{His}}^{-}\right]\end{array}$

These calculations can be repeated to solve for the concentration of all of the histidine species at pH 9.3. A pH of 9.3 is closet to pK₃and the Henderson-Hasselbalch equation for the third dissociation will be used.

  $\begin{array}{l}\text{ pH}-{\text{pK}}_3=\text{log}\frac{\left[{\text{His}}^{-}\right]}{\left[{\text{His}}^0\right]}\\ {10}^{\text{pH}-{\text{pK}}_{\text{3}}}=\frac{\left[{\text{His}}^{-}\right]}{\left[{\text{His}}^0\right]}\\ \left[{\text{His}}^0\right]{10}^{\text{pH}-{\text{pK}}_{\text{3}}}=\left[{\text{His}}^{-}\right]\end{array}$

The value of pK₃for histidine is 9.2.

  $\begin{array}{l}\left[{\text{His}}^0\right]{10}^{\text{9}\text{.3}-\text{9}\text{.2}}=\left[{\text{His}}^{-}\right]\\ \left[{\text{His}}^0\right]{10}^{0.1}=\left[{\text{His}}^{-}\right]\\ 1.26\left[{\text{His}}^0\right]=\left[{\text{His}}^{-}\right]\end{array}$

At pH 9.3, [His⁰] and [His^-] will be the main species in solution:

  $\begin{array}{l}\left[{\text{His}}^0\right]+\left[{\text{His}}^{-}\right]=0.25\text{M}\\ \left[{\text{His}}^0\right]+1.26\left[{\text{His}}^0\right]=0.25\text{M}\\ 2.26\left[{\text{His}}^0\right]=0.25\text{M}\\ \left[{\text{His}}^0\right]=\boxed{0.111\text{M}}\end{array}$

The concentration of His⁰can be used to determine the concentration of His-

$\begin{array}{l}\text{1}\text{.26(0}\text{.111M)=}\left[{\text{His}}^{-}\right]\\ \boxed{0.139\text{M}}\text{=}\left[{\text{His}}^{-}\right]\end{array}$

The value of [His⁺] could be determined from [His⁰] using the Henderson-Hasselbalch equation for second dissociation. The value of pK₂for histidine is 6.0.

  $\begin{array}{l}\text{pH}-{\text{pK}}_2=\text{log}\frac{\left[{\text{His}}^0\right]}{\left[{\text{His}}^{+}\right]}\\ {10}^{\text{9}\text{.3}-\text{6}\text{.0}}=\frac{0.111}{\left[{\text{His}}^{+}\right]}\\ \left[{\text{His}}^{+}\right]=\frac{0.111}{{10}^{\text{3}\text{.3}}}\\ \left[{\text{His}}^{+}\right]=\boxed{\text{5}\text{.56}\times {10}^{-5}\text{M}}\end{array}$

The value of [His²⁺] could be determined from [His⁺] using the Henderson-Hasselbalch equation for first dissociation. The value of pK₁for histidine is 1.8.

  $\begin{array}{l}\text{pH}-{\text{pK}}_1=\text{log}\frac{\left[{\text{His}}^{+}\right]}{\left[{\text{His}}^{2+}\right]}\\ {10}^{\text{9}\text{.3}-\text{1}\text{.8}}=\frac{5.56\times {10}^{-5}\text{M}}{\left[{\text{His}}^{2+}\right]}\\ \left[{\text{His}}^{2+}\right]=\frac{5.56\times {10}^{-5}\text{M}}{{10}^{\text{7}\text{.5}}}\\ \left[{\text{His}}^{2+}\right]=\boxed{1.75\times {10}^{-12}\text{M}}\end{array}$