Chapter 5, Problem 5.1PR

Interpretation Introduction

Interpretation:

The appropriate speciation diagram of lysine has to be drawn.

Concept Introduction:

Speciation:

Speciation is the specification of the fractional composition of ions in a solution.  The speciation of a diprotic acid depends on $pH$ of the solution.

Consider each successive conjugate acid–base pair, with acidity constant $K_{an}$; with $n=1,2,\dots$ then:

• The acid form is dominant for $pH<p{K}_{an}$
• The conjugate pair have equal concentrations at $pH=p{K}_{an}$
• The base form is dominant for $pH>p{K}_{an}$

Explanation of Solution

Lysine is a triprotic amino acid which implies that it has three dissociable protons.

$K_{a_1}=\frac{\left[H_2{A}^{-}\right]\left[H_3{O}^{+}\right]}{\left[H_{3}A\right]}$

$K_{a_2}=\frac{\left[H{A}^{-}\right]\left[H_3{O}^{+}\right]}{\left[H_2{A}^{-}\right]}$

$K_{a_3}=\frac{\left[A^{-}\right]\left[H_3{O}^{+}\right]}{\left[H{A}^{-}\right]}$

The total concentration of acid and its successive deprotonated forms,

$\left[H_{3}A\right]+\left[H_2{A}^{-}\right]+\left[H{A}^{-}\right]+\left[A^{-}\right]=A$

And H can be written as

$H={\left[H_3{O}^{+}\right]}^3+K_{a_1}{\left[H_3{O}^{+}\right]}^2+K_{a_2}\left[H_3{O}^{+}\right]+K_{a_1}{K}_{a_2}{K}_{a_3}$

Where, in each case $\left[H_3{O}^{+}\right]={10}^{-pH}$

Then, fractions of each species present in the solution are,

$\begin{array}{l}f\left(H_{3}A\right)=\frac{{\left[H_3{O}^{+}\right]}^3}{H}\\ \\ f\left(H_2{A}^{-}\right)=\frac{K_{a_1}{\left[H_3{O}^{+}\right]}^2}{H}\\ \\ f\left(H{A}^{-}\right)=\frac{K_{a_1}{K}_{a_2}\left[H_3{O}^{+}\right]}{H}\\ \\ f\left(A^{-}\right)=\frac{\left(K_{a_1}\right)\left(K_{a_2}\right)\left(K_{a_3}\right)}{H}\end{array}$

According to the question,

$\begin{array}{l}p{K}_{a_1}=-\log K_{a_1}\\ \\ K_{a_1}={10}^{-p{K}_{a_1}}\\ \\ ={10}^{-\left(2.18\right)}=\mathrm{0.0066.}\\ \\ Similarly,\\ \\ K_{a_2}=1.1220\times {10}^{-9}\\ \\ K_{a_3}=2.9512\times {10}^{-11}.\end{array}$

Substituting the values of $K_{a_1}$, $K_{a_2}$ and $K_{a_3}$ in above expressions for $pH=0-14$,we will get a set of results which can be plotted as below

Calculation of $f\left(H_{3}A\right)$:

At $pH=0$:

$\begin{array}{l}f\left(H_{3}A\right)=\frac{{\left[H_3{O}^{+}\right]}^3}{H}\\ \\ =\frac{1}{1+K_{a_1}+K_{a_2}+K_{a_1}{K}_{a_2}{K}_{a_3}}\\ \\ =\frac{1}{1+0.0066+1.1220\times {10}^{-9}+2.1854\times {10}^{-22}}\\ \\ =\mathrm{0.9934.}\end{array}$

At $pH=2$:

$\begin{array}{l}f\left(H_{3}A\right)=\frac{{\left[H_3{O}^{+}\right]}^3}{H}\\ \\ =\frac{{10}^{-6}}{{10}^{-6}+\left(0.0066\times {10}^{-4}\right)+\left(1.1220\times {10}^{-11}\right)+\left(2.1854\times {10}^{-22}\right)}\\ \\ =\mathrm{0.6024.}\end{array}$

At $pH=4$:

$\begin{array}{l}f\left(H_{3}A\right)=\frac{{\left[H_3{O}^{+}\right]}^3}{H}\\ \\ =\frac{{10}^{-12}}{{10}^{-12}+\left(0.0066\times {10}^{-8}\right)+\left(1.1220\times {10}^{-13}\right)+\left(2.1854\times {10}^{-22}\right)}\\ \\ =\mathrm{0.0149.}\end{array}$

At $pH=6$:

$\begin{array}{l}f\left(H_{3}A\right)=\frac{{\left[H_3{O}^{+}\right]}^3}{H}\\ \\ =\frac{{10}^{-18}}{{10}^{-18}+\left(0.0066\times {10}^{-12}\right)+\left(1.1220\times {10}^{-15}\right)+\left(2.1854\times {10}^{-22}\right)}\\ \\ =\mathrm{0.0001.}\end{array}$

Similarly, the fraction of $H_{3}A$ can be calculated for rest of the $pH$.

Calculation of $f\left(H_2{A}^{-}\right)$:

At $pH=0$:

$\begin{array}{l}f\left(H_2{A}^{-}\right)=\frac{K_{a_1}{\left[H_3{O}^{+}\right]}^2}{H}\\ \\ =\frac{K_{a_1}}{1+K_{a_1}+K_{a_2}+K_{a_1}{K}_{a_2}{K}_{a_3}}\\ \\ =\frac{1}{1+0.0066+1.1220\times {10}^{-9}+2.1854\times {10}^{-22}}\\ \\ =\mathrm{0.0059.}\end{array}$

At $pH=2$:

$\begin{array}{l}f\left(H_2{A}^{-}\right)=\frac{K_{a_1}{\left[H_3{O}^{+}\right]}^2}{H}\\ \\ =\frac{0.0066\times {10}^{-4}}{{10}^{-4}+\left(0.0066\times {10}^{-4}\right)+\left(1.1220\times {10}^{-11}\right)+\left(2.1854\times {10}^{-22}\right)}\\ \\ =\mathrm{0.3975.}\end{array}$

At $pH=4$:

$\begin{array}{l}f\left(H_2{A}^{-}\right)=\frac{K_{a_1}{\left[H_3{O}^{+}\right]}^2}{H}\\ \\ =\frac{0.0066\times {10}^{-8}}{{10}^{-12}+\left(0.0066\times {10}^{-8}\right)+\left(1.1220\times {10}^{-13}\right)+\left(2.1854\times {10}^{-22}\right)}\\ \\ =\mathrm{0.9834.}\end{array}$

At $pH=6$:

$\begin{array}{l}f\left(H_2{A}^{-}\right)=\frac{K_{a_1}{\left[H_3{O}^{+}\right]}^2}{H}\\ \\ =\frac{0.0066\times {10}^{-8}}{{10}^{-12}+\left(0.0066\times {10}^{-8}\right)+\left(1.1220\times {10}^{-13}\right)+\left(2.1854\times {10}^{-22}\right)}\\ \\ =\mathrm{0.8545.}\end{array}$

Similarly, the fraction of $H_2{A}^{-}$ can be calculated for rest of the $pH$.

Calculation of $f\left(H{A}^{-}\right)$:

At $pH=0$:

$\begin{array}{l}f\left(H{A}^{-}\right)=\frac{K_{a_1}{K}_{a_2}\left[H_3{O}^{+}\right]}{H}\\ \\ =\frac{K_{a_1}{K}_{a_2}}{1+K_{a_1}+K_{a_2}+K_{a_1}{K}_{a_2}{K}_{a_3}}\\ \\ =\frac{0.0066\times 1.1220\times {10}^{-9}}{1+0.0066+1.1220\times {10}^{-9}+2.1854\times {10}^{-22}}\\ \\ =7.3566\times {10}^{-12}.\end{array}$

At $pH=2$:

$\begin{array}{l}f\left(H{A}^{-}\right)=\frac{K_{a_1}{K}_{a_2}\left[H_3{O}^{+}\right]}{H}\\ \\ =\frac{0.0066\times 1.1220\times {10}^{-11}}{{10}^{-6}+0.0066\times {10}^{-4}+1.1220\times {10}^{-11}+2.1854\times {10}^{-22}}\\ \\ =4.4609\times {10}^{-8}.\end{array}$

At $pH=4$:

$\begin{array}{l}f\left(H{A}^{-}\right)=\frac{K_{a_1}{K}_{a_2}\left[H_3{O}^{+}\right]}{H}\\ \\ =\frac{0.0066\times 1.1220\times {10}^{-13}}{{10}^{-12}+0.0066\times {10}^{-8}+1.1220\times {10}^{-13}+2.1854\times {10}^{-22}}\\ \\ =\mathrm{0.00001.}\end{array}$

At $pH=6$:

$\begin{array}{l}f\left(H{A}^{-}\right)=\frac{K_{a_1}{K}_{a_2}\left[H_3{O}^{+}\right]}{H}\\ \\ =\frac{0.0066\times 1.1220\times {10}^{-15}}{{10}^{-18}+0.0066\times {10}^{-12}+1.1220\times {10}^{-15}+2.1854\times {10}^{-22}}\\ \\ =\mathrm{0.0009.}\end{array}$

Similarly, the fraction of $H{A}^{-}$ can be calculated for rest of $pH$.

The fraction of $A^{-}$ can also be calculated as given above.

All the results have been plotted in the below graph.

Figure 1