Chapter 25, Problem 25.48QP

(a)

Interpretation Introduction

Interpretation:

From the given the $p{K}_a$ values of the two groups, ${-}^{+}N{H}_3$ and $-COOH$ of valine, the more acidic one has to be determined.

Concept Introduction:

Greater the $p{K}_a$ value of a substance lower will be its acidic property.

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

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

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

Answer to Problem 25.48QP

$-COOH$ group is more acidic.

Explanation of Solution

Given that $p{K}_a$ of ${-}^{+}N{H}_3$ and $-COOH$ groups of valine are $9.62$ and $2.32$ respectively.

Since $-COOH$ has lower $p{K}_a$ value of $2.32$, it is more acidic.

(b)

Interpretation Introduction

Interpretation:

The predominant form of valine at $pH$ $1.0,7.0$ and $12$ has to be predicted.

Concept Introduction:

Greater the $p{K}_a$ value of a substance lower will be its acidic property.

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:

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

At $pH=1,$

Concentration of $-COOH$

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ 1=2.32+\log \frac{\left[-CO{O}^{-}\right]}{\left[-COOH\right]}\\ \frac{\left[-COOH\right]}{\left[-CO{O}^{-}\right]}=\text{21 }\end{array}$

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

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ 1=\text{9}\text{.62}+\log \frac{\left[-N{H}_2\right]}{\left[-N{H}_3^{+}\right]}\\ \frac{\left[-N{H}_3^{+}\right]}{\left[-N{H}_2\right]}=\text{4}\text{.2}\times {\text{10}}^8\end{array}$

Concentration of $-N{H}_3^{+}$ is greater than $-COOH$.  Hence, the predominant species is $CH{\left(C{H}_3\right)}_2-CH\left({}^{+}N{H}_3\right)-COOH.$

At $pH=\text{7},$

Concentration of $-COOH$

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ 7=2.32+\log \frac{\left[-CO{O}^{-}\right]}{\left[-COOH\right]}\\ \frac{\left[-CO{O}^{-}\right]}{\left[-COOH\right]}=\text{4}\text{.8}\times {\text{10}}^4\end{array}$

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

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ 7=\text{9}\text{.62}+\log \frac{\left[-N{H}_2\right]}{\left[-N{H}_3^{+}\right]}\\ \frac{\left[-N{H}_3^{+}\right]}{\left[-N{H}_2\right]}=\text{4}\text{.2}\times {\text{10}}^2\end{array}$

Concentration of $-CO{O}^{-}$ is greater than $-N{H}_3^{+}$.  Hence, the predominant species is $CH{\left(C{H}_3\right)}_2-CH\left({}^{+}N{H}_3\right)-CO{O}^{-}.$

At $pH=\text{12},$

Concentration of $-COOH$

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ 12=2.32+\log \frac{\left[-CO{O}^{-}\right]}{\left[-COOH\right]}\\ \frac{\left[-CO{O}^{-}\right]}{\left[-COOH\right]}=\text{4}\text{.8}\times {\text{10}}^9\end{array}$

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

$\begin{array}{l}pH=p{K}_a+\log \frac{\left[conjugatebase\right]}{\left[acid\right]}\\ 12=\text{9}\text{.62}+\log \frac{\left[-N{H}_2\right]}{\left[-N{H}_3^{+}\right]}\\ \frac{\left[-N{H}_2\right]}{\left[-N{H}_3^{+}\right]}=\text{2}\text{.4}\times {\text{10}}^2\end{array}$

Concentration of $-CO{O}^{-}$ is greater than $-N{H}_3^{+}$.  Hence, the predominant species is $CH{\left(C{H}_3\right)}_2-CH\left(N{H}_2\right)-CO{O}^{-}.$

(c)

Interpretation Introduction

Interpretation:

Isoelectric point of valine has to be calculated.

Concept Introduction:

Isoelectric point $\left(pI\right)$ of an amino acid refers to the $pH$ at which the amino acid remains neutral and does not migrate in an electric field

It is calculated as,

$pI=\frac{p{K}_{a_1}+p{K}_{a_2}}{2}$

Where ,

$p{K}_{a_1}andp{K}_{a_2}$ are two $p{K}_a$ values of amino acid.

Answer to Problem 25.48QP

Isoelectric point of valine is calculated as $5.97$

Explanation of Solution

Given data:

$\begin{array}{l}p{K}_a{}_{1}ofcarboxylgroup,COOH,=2.32\\ p{K}_a{}_{2}ofammoniumgroup,N{H}_3^{+}=9.62\end{array}$

Therefore isoelectronic point of valine is,

$\begin{array}{l}pI=\frac{p{K}_{a_1}+p{K}_{a_2}}{2}\\ =\frac{2.32+9.62}{2}=5.97\end{array}$