Chapter 21, Problem 152AE

(a)

Interpretation Introduction

Interpretation: The equilibrium constant for the given reaction if K_a for the carboxylic group of glycine is $4.3\times {10}^{-3}$ and K_b for amino group is $6.0\times {10}^{-5}$ .

  ${}^{\text{+}}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{COO}}_{\text{2}}{}^{\text{-}}{\text{+ H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+ H}}_{\text{3}}{\text{O}}^{\text{+}}$

Concept Introduction: Amino acids are the organic compounds which have two functional groups; amino group and carboxylic acid.

The carboxylic acid group is acidic in nature and the acid dissociation constant can be written as K_a whereas amino group is basic in nature as it can accept H+ ion to form −NH₃+ ion. The base dissociation constant can be written as K_b.

Answer to Problem 152AE

  ${\text{K}}_{\text{a }}\text{ = 1}{\text{.67×10}}^{-10}$

Explanation of Solution

Given:

  ${}^{\text{+}}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{COO}}_{\text{2}}{}^{\text{-}}{\text{+ H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+ H}}_{\text{3}}{\text{O}}^{\text{+}}\mathrm{...}(a)$

K_a for the carboxylic group of glycine = $4.3\times {10}^{-3}$

K_b for amino group = $6.0\times {10}^{-5}$

The base dissociation of glycine can be written as:

  ${\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+ H}}_{\text{3}}{\text{O}}^{\text{+}}{\rightleftharpoons }^{\text{+}}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{COO}}_{\text{2}}{}^{\text{-}}{\text{+ H}}_{\text{2}}{\text{O K}}_{\text{b}}\text{ = }6.0\times {10}^{-5}$

This reaction is reverse of the equation (a). Thus the equilibrium constant for the reaction (a) can be written as:

  $\begin{array}{l}{}^{\text{+}}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{COO}}_{\text{2}}{}^{\text{-}}{\text{+ H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+ H}}_{\text{3}}{\text{O}}^{\text{+}}\mathrm{...}\text{(a)}\\ {\text{K}}_{\text{a }}\text{ =}\frac{{\text{10}}^{\text{-14}}}{{\text{ K}}_{\text{b}}}\text{=}\frac{{\text{10}}^{-14}}{\text{ 6}{\text{.0 ×10}}^{\text{-5}}}\text{= 1}{\text{.67×10}}^{-10}\end{array}$

(b)

Interpretation Introduction

Interpretation: The equilibrium constant for the given reaction if K_a for the carboxylic group of glycine is $4.3\times {10}^{-3}$ and K_b for amino group is $6.0\times {10}^{-5}$ should be calculated.

  ${\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COOH+ OH}}^{\text{-}}$

Concept Introduction: Amino acids are the organic compounds which have two functional groups; amino group and carboxylic acid.

The carboxylic acid group is acidic in nature and the acid dissociation constant can be written as K_a whereas amino group is basic in nature as it can accept H+ ion to form −NH₃+ ion. The base dissociation constant can be written as K_b.

Answer to Problem 152AE

  ${\text{K}}_{\text{b }}=\text{2}{\text{.3 ×10}}^{-12}$

Explanation of Solution

Given:

  ${\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COOH+ OH}}^{\text{-}}\mathrm{...}\text{(b)}$

K_a for the carboxylic group of glycine = $4.3\times {10}^{-3}$

K_b for amino group = $6.0\times {10}^{-5}$

The acid dissociation of glycine can be written as:

  ${\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COOH + OH}}^{\text{-}}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+ H}}_{\text{2}}{\text{O K}}_{\text{a}}\text{ = 4}{\text{.3×10}}^{\text{-3}}$

This reaction is reverse of the equation (b). Thus the equilibrium constant for the reaction (b) can be written as:

  $\begin{array}{l}{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+H}}_{\text{2}}\text{O}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COOH+ OH}}^{\text{-}}\mathrm{...}\text{(b)}\\ {\text{K}}_{\text{b }}\text{ =}\frac{{\text{10}}^{\text{-14}}}{{\text{ K}}_a}\text{=}\frac{{\text{10}}^{\text{-14}}}{\text{ 4}\text{.3}\times {\text{10}}^{\text{-3}}}\text{= 2}{\text{.3 ×10}}^{-12}\end{array}$

(c)

Interpretation Introduction

Interpretation: Calculate the equilibrium constant for the given reaction if K_a for the carboxylic group of glycine is $4.3\times {10}^{-3}$ and K_b for amino group is $6.0\times {10}^{-5}$ .

  ${}^{\text{+}}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}\text{COOH}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{ + 2H}}^{\text{+}}$

Concept Introduction: Amino acids are the organic compounds which have two functional groups; amino group and carboxylic acid.

The carboxylic acid group is acidic in nature and the acid dissociation constant can be written as K_a whereas amino group is basic in nature as it can accept H+ ion to form −NH₃+ ion. The base dissociation constant can be written as K_b.

Answer to Problem 152AE

  ${\text{K}}_a\text{=1}{\text{.6 ×10}}^{-10}$

Explanation of Solution

Given:

  ${}^{\text{+}}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}\text{COOH}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{ + 2H}}^{\text{+}}\mathrm{...}\text{(c)}$

K_a for the carboxylic group of glycine = $4.3\times {10}^{-3}$

K_b for amino group = $6.0\times {10}^{-5}$

The acid dissociation of glycine can be written as:

  $\begin{array}{l}{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COOH + OH}}^{\text{-}}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{+ H}}_{\text{2}}{\text{O K}}_{\text{a}}\text{ = 4}{\text{.3×10}}^{\text{-3}}\mathrm{...}\text{(a)}\\ {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COOH+ H}}_{\text{3}}{\text{O}}^{\text{+}}{\rightleftharpoons }^{\text{+}}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{COOH+ H}}_{\text{2}}{\text{O K}}_{\text{b}}\text{ = }6.0\times {10}^{-5}\mathrm{...}\text{(b)}\end{array}$

In the given reaction dissociation of acid takes place.

Thus,

  $\begin{array}{l}{}^{\text{+}}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}\text{COOH}\rightleftharpoons {\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{\text{-}}{\text{ + 2H}}^{\text{+}}\mathrm{...}\text{(c)}\\ {\text{K}}_{\text{a }}\text{ =}\frac{{\text{10}}^{\text{-14}}}{{\text{ K}}_b}\text{=}\frac{{\text{10}}^{\text{-14}}}{\text{ 6}\text{.0}\times {\text{10}}^{\text{-3}}}\text{= 1}{\text{.6 ×10}}^{-10}\end{array}$