Chapter 23, Problem 29QAP

For alanine, $K_{\text{a1}}=5.1\times {10}^{-5},K_{\text{a2}}=1.8\times {10}^{-10}$ . Calculate the ratios $\left[\text{Z}\right]/\left[{\text{C}}^{+}\right]$ and $\left[\text{Z}\right]/\left[{\text{A}}^{+}\right]$ at pH
(a) 2.00.
(b) 6.00.
(c) 10.50.
What is the principal species at each pH?

Interpretation Introduction

(a)

Interpretation:

The value of $\frac{\left[\text{Z}\right]}{\left[{\text{C}}^{+}\right]}$, $\frac{\left[\text{Z}\right]}{\left[{\text{A}}^{-}\right]}$ and principal species at pH 2.00 needs to be determined.

Concept introduction:

General formula of alanine is shown as follows:

The $\text{pH}$ is calculated as negative logarithm of molar hydrogen ion concentration.

$\text{pH}=-\mathrm{l}\text{og}\left[{\text{H}}^{\text{+}}\right]$

Answer to Problem 29QAP

$\frac{\left[\text{Z}\right]}{\left[\text{C}\right]}\text{==5}{\text{.1×10}}^{-\text{1}}$

$\frac{\left[\text{Z}\right]}{\left[\text{A}\right]}\text{=5}{\text{.5×10}}^{\text{7}}$

The principal species at $\text{pH}2.00$ is cationic form.

Explanation of Solution

Under equilibrium condition, the Zwitter ionic form of alanine is shown as follows:

Under acidic condition, oxygen atom accepts the proton Zwitter ion exists in cationic form.

Under basic condition, base abstracts proton form nitrogen atom and it forms the anionic form.

Dissociation of cation is represented as follows:

${\text{K}}_{\text{a1}}\text{=}\frac{\left[\text{Z}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[\text{C}\right]}$

Dissociation of anion is represented as follows:

${\text{K}}_{\text{a2}}\text{=}\frac{\left[\text{A}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[\text{Z}\right]}$

Given value of ${\text{K}}_{\text{a1}}\text{and}{\text{K}}_{\text{a2}}$ as follows:

${\text{K}}_{\text{a1}}=5.1\times {10}^{-3},{\text{K}}_{\text{a2}}\text{=}1.8\times {10}^{-10}$

At $\text{pH}\text{=}\text{2}\text{.00}$

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

So, $\left[{\text{H}}^{\text{+}}\right]{\text{=10}}^{-\text{pH}}$

Take antilog both sides,

$\begin{array}{c}\left[{\text{H}}^{\text{+}}\right]{\text{=10}}^{-\text{2}\text{.00}}\\ \text{=0}\text{.01}\text{M}\\ \text{=}\text{1}\times {\text{10}}^{-2}\end{array}$

$\frac{\left[\text{Z}\right]}{\left[\text{C}\right]}$ can be calculated as follows:

Here, Z is Zwitter ion and C is cationic form of Zwitter ion.

$\begin{array}{c}\text{5}{\text{.1×10}}^{-\text{3}}\text{=}\frac{\left[\text{Z}\right]\text{×1}{\text{.0×10}}^{-\text{2}}}{\left[\text{C}\right]}\\ \frac{\left[\text{Z}\right]}{\left[\text{C}\right]}\text{=}\frac{\text{5}{\text{.1×10}}^{-\text{3}}}{\text{1}{\text{.0×10}}^{-2}}\\ \text{=5}{\text{.1×10}}^{-\text{1}}\end{array}$

$\frac{\left[\text{Z}\right]}{\left[\text{A}\right]}$ can be calculated as follows:

Here, Z is Zwitter ion and A is anionic form of Zwitter ion.

$\text{1}{\text{.8×10}}^{-\text{10}}\text{=}\frac{\left[\text{A}\right]\text{×1}{\text{.0×10}}^{-\text{2}}}{\left[\text{Z}\right]}$

Therefore,

$\begin{array}{c}\frac{\left[\text{Z}\right]}{\left[\text{A}\right]}\text{=}\frac{\text{1}{\text{.0×10}}^{\text{-2}}}{\text{1}{\text{.8×10}}^{\text{-10}}}\\ \text{=5}{\text{.5×10}}^{\text{7}}\end{array}$

The principal species at $\text{pH}2.00$ is cationic form.

Interpretation Introduction

(b)

Interpretation:

The value of $\frac{\left[\text{Z}\right]}{\left[{\text{C}}^{+}\right]}$, $\frac{\left[\text{Z}\right]}{\left[{\text{A}}^{-}\right]}$ and principal species at pH 6.00 needs to be determined.

Concept introduction:

General formula of alanine is shown as follows:

The $\text{pH}$ is calculated as negative logarithm of molar hydrogen ion concentration.

$\text{pH}=-\mathrm{l}\text{og}\left[{\text{H}}^{\text{+}}\right]$

Answer to Problem 29QAP

$\frac{\left[\text{Z}\right]}{\left[\text{C}\right]}\text{=5}{\text{.1×10}}^{\text{3}}$

$\frac{\left[\text{Z}\right]}{\left[\text{A}\right]}\text{=}\text{5}{\text{.5×10}}^{\text{3}}$

The principal species at $\text{pH}\text{6}\text{.00}$ is cationic and anionic form.

Explanation of Solution

Under equilibrium condition zwitter ionic form of alanine is:

Under acidic condition oxygen atom accepts the proton zwitter ion exists in cationic form.

Under basic condition base abstracts proton form nitrogen atom and it forms the anionic form.

Dissociation of cation is-

${\text{K}}_{\text{a1}}\text{=}\frac{\left[\text{Z}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[\text{C}\right]}$

Dissociation of anion is-

${\text{K}}_{\text{a2}}\text{=}\frac{\left[\text{A}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[\text{Z}\right]}$

Given value of ${\text{K}}_{\text{a1}}\text{and}{\text{K}}_{\text{a2}}$ as follows:

${\text{K}}_{\text{a1}}=5.1\times {10}^{-3},{\text{K}}_{\text{a2}}\text{=}1.8\times {10}^{-10}$

At $\text{pH}\text{=}\text{6}\text{.00}$

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

So, $\left[{\text{H}}^{\text{+}}\right]{\text{=10}}^{-\text{pH}}$

Take antilog both sides,

$\begin{array}{c}\left[{\text{H}}^{\text{+}}\right]{\text{=10}}^{-\text{6}\text{.00}}\\ \text{=1}{\text{.0×10}}^{-\text{6}}\text{M}\end{array}$

$\frac{\left[\text{Z}\right]}{\left[\text{C}\right]}$ can be calculated as:

Here, Z is zwitter ion and C is cationic form of zwitter ion.

$\text{5}{\text{.1×10}}^{-\text{3}}\text{=}\frac{\left[\text{Z}\right]\text{×1}{\text{.0×10}}^{-\text{6}}}{\left[\text{C}\right]}$

$\begin{array}{c}\frac{\left[\text{Z}\right]}{\left[\text{C}\right]}\text{=}\frac{\text{5}{\text{.1×10}}^{-\text{3}}}{\text{1}{\text{.0×10}}^{-\text{6}}}\\ \text{=5}{\text{.1×10}}^{\text{3}}\end{array}$

$\frac{\left[\text{Z}\right]}{\left[\text{A}\right]}$ can be calculated as follows:

Here, Z is zwitter ion and A is anionic form of zwitter ion.

$\text{1}{\text{.8×10}}^{-\text{10}}\text{=}\frac{\left[\text{A}\right]\text{×1}{\text{.0×10}}^{-\text{6}}}{\left[\text{Z}\right]}$

$\begin{array}{c}\frac{\left[\text{Z}\right]}{\left[\text{A}\right]}\text{=}\frac{\text{1}{\text{.0×10}}^{-\text{6}}}{\text{1}{\text{.8×10}}^{-\text{10}}}\\ \text{=}\text{5}{\text{.5×10}}^{\text{3}}\end{array}$

The principal species at $\text{pH}\text{6}\text{.00}$ is cationic and anionic form.

Interpretation Introduction

(c)

Interpretation:

The value of $\frac{\left[\text{Z}\right]}{\left[{\text{C}}^{+}\right]}$, $\frac{\left[\text{Z}\right]}{\left[{\text{A}}^{-}\right]}$ and principal species at pH 10.50 needs to be determined.

Concept introduction:

General formula of alanine is shown as follows:

The $\text{pH}$ is calculated as negative logarithm of molar hydrogen ion concentration.

$\text{pH}=-\mathrm{l}\text{og}\left[{\text{H}}^{\text{+}}\right]$

Answer to Problem 29QAP

$\frac{\left[\text{Z}\right]}{\left[\text{C}\right]}\text{=1}{\text{.6×10}}^{\text{8}}$

$\frac{\left[\text{Z}\right]}{\left[\text{A}\right]}\text{=}\text{1}{\text{.8×10}}^{-1}$

The principal species at $\text{pH}\text{=}\text{10}\text{.50}$ is anionic form.

Explanation of Solution

Under equilibrium condition zwitter ionic form of alanine is:

Under acidic condition oxygen atom accepts the proton zwitter ion exists in cationic form.

Under basic condition base abstracts proton form nitrogen atom and it forms the anionic form.

Dissociation of cation is-

${\text{K}}_{\text{a1}}\text{=}\frac{\left[\text{Z}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[\text{C}\right]}$

Dissociation of anion is-

${\text{K}}_{\text{a2}}\text{=}\frac{\left[\text{A}\right]\left[{\text{H}}^{\text{+}}\right]}{\left[\text{Z}\right]}$

Given value of ${\text{K}}_{\text{a1}}\text{and}{\text{K}}_{\text{a2}}$ as follows:

${\text{K}}_{\text{a1}}=5.1\times {10}^{-3},{\text{K}}_{\text{a2}}\text{=}1.8\times {10}^{-10}$

At $\text{pH}\text{=}\text{10}\text{.50}$

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

So, $\left[{\text{H}}^{\text{+}}\right]{\text{=10}}^{-\text{pH}}$

Take antilog both sides,

$\begin{array}{c}\left[{\text{H}}^{\text{+}}\right]{\text{=10}}^{-\text{10}\text{.50}}\\ \text{=3}{\text{.2×10}}^{-\text{11}}\text{M}\end{array}$

$\frac{\left[\text{Z}\right]}{\left[\text{C}\right]}$ can be calculated as:

Here, Z is zwitter ion and C is cationic form of zwitter ion.

$\text{5}{\text{.1×10}}^{-\text{3}}\text{=}\frac{\left[\text{Z}\right]\text{×3}{\text{.2×10}}^{-\text{11}}}{\left[\text{C}\right]}$

$\begin{array}{c}\text{So,}\frac{\left[\text{Z}\right]}{\left[\text{C}\right]}\text{=}\frac{\text{5}{\text{.1×10}}^{-\text{3}}}{\text{3}{\text{.2×10}}^{-\text{11}}}\\ \text{=1}{\text{.6×10}}^{\text{8}}\end{array}$

$\frac{\left[\text{Z}\right]}{\left[\text{A}\right]}$ can be calculated as follows:

Here, Z is zwitter ion and A is anionic form of zwitter ion.

$\text{1}{\text{.8×10}}^{-\text{10}}\text{=}\frac{\left[\text{A}\right]\text{×}3.2\times {10}^{-11}}{\left[\text{Z}\right]}$

$\begin{array}{c}\frac{\left[\text{Z}\right]}{\left[\text{A}\right]}\text{=}\frac{\text{3}{\text{.2×10}}^{-11}}{\text{1}{\text{.8×10}}^{-\text{10}}}\\ \text{=}\text{1}{\text{.8×10}}^{-1}\end{array}$

The principal species at $\text{pH}\text{=}\text{10}\text{.50}$ is anionic form.