Chapter 25, Problem 25.41QP

Interpretation Introduction

Interpretation:

The predominant form of glycine at $pH$ $1,7$ and $12$ has to be predicted using the given equation.

Concept Introduction:

According to equation given in chapter 16 in the text book (equation 16.4),

$pH$ and $p{K}_a$ are related as,

$pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}$

Explanation of Solution

Given data:

$\begin{array}{l}p{K}_{a}ofcarboxylgroup,CO{O}^{-}=2.3\\ p{K}_{a}ofammoniumgroup,N{H}_3^{+}=9.6\end{array}$

At $pH=1,$

Concentration of $-COOH$

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ 1=2.3+\log \frac{\left[-CO{O}^{-}\right]}{\left[-COOH\right]}\\ \frac{\left[-COO{H}^{-}\right]}{\left[-CO{O}^{-}\right]}=\text{20 }\end{array}$

Concentration of $-N{H}_3^{+}$

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ 1=\text{9}\text{.6}+\log \frac{\left[-N{H}_2\right]}{\left[-N{H}_3^{+}\right]}\\ \frac{\left[-N{H}_3^{+}\right]}{\left[-N{H}_2\right]}=\text{4}\times {\text{10}}^8\end{array}$

Concentration of $-N{H}_3^{+}$ is greater than $-COOH$.  Hence, the predominant species is ${}^{+}N{H}_3-C{H}_2-COOH.$

At $pH=\text{7},$

Concentration of $-COOH$

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ 7=2.3+\log \frac{\left[-CO{O}^{-}\right]}{\left[-COOH\right]}\\ \frac{\left[-CO{O}^{-}\right]}{\left[-COOH\right]}=\text{5}\times {\text{10}}^4\end{array}$

Concentration of $-N{H}_3^{+}$

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ 7=\text{9}\text{.6}+\log \frac{\left[-N{H}_2\right]}{\left[-N{H}_3^{+}\right]}\\ \frac{\left[-N{H}_3^{+}\right]}{\left[-N{H}_2\right]}=\text{4}\times {\text{10}}^2\end{array}$

Concentration of $-COOH$ is greater than $-N{H}_3^{+}$.  Hence, the predominant species is ${}^{+}N{H}_3-C{H}_2-CO{O}^{-}.$

At $pH=\text{12},$

Concentration of $-COOH$

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ 12=2.3+\log \frac{\left[-CO{O}^{-}\right]}{\left[-COOH\right]}\\ \frac{\left[-CO{O}^{-}\right]}{\left[-COOH\right]}=\text{5}\times {\text{10}}^9\end{array}$

Concentration of $-N{H}_3^{+}$

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ 12=\text{9}\text{.6}+\log \frac{\left[-N{H}_2\right]}{\left[-N{H}_3^{+}\right]}\\ \frac{\left[-N{H}_2\right]}{\left[-N{H}_3^{+}\right]}=\text{2}\text{.5}\times {\text{10}}^2\end{array}$

Concentration of $-COOH$ is greater than $-N{H}_3^{+}$.  Hence, the predominant species is $N{H}_2-C{H}_2-CO{O}^{-}.$