Chapter 21, Problem 136AE

(a)

Interpretation Introduction

Interpretation: The equilibrium constants for the following reactions are to be calculated.

Concept introduction: Equilibrium constant is defined as the ratio of the concentration of the products to the concentration of the reactants present at equilibrium in a reversible reaction at a given temperature. It is denoted by K and is a dimensionless quantity. Its value decide the direction of the chemical reaction, if the value of K is less than one the reaction will proceed towards left but when its value is greater than one the reaction will proceed in the right direction.

To determine: The equilibrium constant for the reaction.

Answer to Problem 136AE

Answer

The equilibrium constant for the reaction is, $\underset{\_}{K_a=1.6\times 10^{-10}}$.

Explanation of Solution

Explanation

The equilibrium constant for the reaction is $\underset{\_}{1.6\times 10^{-10}}$.

The given reaction is,

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

In the given reaction, dissociation of an acid occurs.

The equilibrium constant expression for the above reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}^{-}\right]\left[{\text{H}}_{\text{3}}{\text{O}}^{+}\right]}{\left[{}^{+}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}{\text{COO}}^{-}\right]}$

Where,

• $K_{\text{a}}$ is the acid dissociation constant.

Given

$\text{Basicity of amino group}\left(K_{\text{b}}\right)\text{=}\text{6}{\text{.0×10}}^{-\text{5}}$

The given reaction occurs in an aqueous medium. In an aqueous solution, the acidity and basicity of an acid and its conjugate base is related by the expression,

$K_{\text{a}}\text{ × }{K}_{\text{b}}={\text{10}}^{-\text{14}}$

Substitute the value of $K_{\text{b}}$ in the above expression.

$\begin{array}{c}{K}_{\text{a}}={\frac{\text{10}}{K_{\text{b}}}}^{-\text{14}}\\ =\frac{{\text{10}}^{-\text{14}}}{\text{6}{\text{.0×10}}^{-\text{10}}}\\ =\text{1}{\text{.6×10}}^{-\text{10}}\end{array}$

Hence, the equilibrium constant for the reaction is $\underset{\_}{1.6\times 10^{-10}}$.

(b)

Interpretation Introduction

Interpretation: The equilibrium constants for the following reactions are to be calculated.

Concept introduction: Equilibrium constant is defined as the ratio of the concentration of the products to the concentration of the reactants present at equilibrium in a reversible reaction at a given temperature. It is denoted by K and is a dimensionless quantity. Its value decide the direction of the chemical reaction, if the value of K is less than one the reaction will proceed towards left but when its value is greater than one the reaction will proceed in the right direction.

To determine: The equilibrium constant for the reaction.

Answer to Problem 136AE

Answer

The equilibrium constant for the reaction is, $\underset{\_}{K_b=2.32\times 10^{-12}}$.

Explanation of Solution

Explanation

The equilibrium constant for the reaction is $\underset{\_}{2.32\times 10^{-12}}$.

The given reaction is,

${\text{H}}_2{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}^{-}+{\text{H}}_{\text{2}}\text{O }\rightleftharpoons {\text{ H}}_{\text{2}}{\text{NCH}}_{\text{2}}\text{COOH}+{\text{OH}}^{-}$

In the given reaction dissociation of a base occurs. The equilibrium constant expression for the above reaction is,

$K_{\text{b}}=\frac{\left[{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}\text{COOH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{-}\right]}$

Where,

• ${\text{K}}_{\text{b}}$ is the base ionization constant.

$K_{\text{a}}\text{of carboxylic acid=}4.3{\text{×10}}^{-\text{3}}$

The given reaction occurs in an aqueous medium. In an aqueous solution, the acidity and basicity of an acid and its conjugate base is related by the expression,

$K_{\text{a}}\text{ × }{K}_{\text{b}}={\text{10}}^{-\text{14}}$

Substitute the value of $K_{\text{a}}$ in the above expression.

$\begin{array}{c}{K}_{\text{b}}={\frac{\text{10}}{{\text{K}}_{\text{a}}}}^{-\text{14}}\\ =\frac{{\text{10}}^{-\text{14}}}{\text{4}{\text{.3×10}}^{-\text{3}}}\\ =2.32{\text{×10}}^{-\text{12}}\end{array}$

Hence, the equilibrium constant for the reaction is $\underset{\_}{2.32\times 10^{-12}}$.

(c)

Interpretation Introduction

Interpretation: The equilibrium constants for the following reactions are to be calculated.

Concept introduction: Equilibrium constant is defined as the ratio of the concentration of the products to the concentration of the reactants present at equilibrium in a reversible reaction at a given temperature. It is denoted by K and is a dimensionless quantity. Its value decide the direction of the chemical reaction, if the value of K is less than one the reaction will proceed towards left but when its value is greater than one the reaction will proceed in the right direction.

To determine: The equilibrium constant for the reaction.

Answer to Problem 136AE

Answer

The equilibrium constant for the reaction is, $\underset{\_}{K_a=1.6\times 10^{-10}}$.

Explanation of Solution

Explanation

The equilibrium constant for the reaction is $\underset{\_}{1.6\times 10^{-10}}$.

The given reaction is,

${}^{+}{\text{H}}_3{\text{NCH}}_{\text{2}}\text{COOH }\rightleftharpoons {\text{ 2H}}^{+}+{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{COO}}^{--}$

In the given reaction dissociation of an acid occurs. Therefore, the equilibrium constant expression for the above reaction is,

$K_{\text{a}}=\frac{\left[{\text{H}}_{\text{2}}{\text{NCH}}_{\text{2}}{\text{CO}}_{\text{2}}^{-}\right]\left[{\text{2H}}^{+}\right]}{\left[{}^{+}{\text{H}}_{\text{3}}{\text{NCH}}_{\text{2}}\text{COOH}\right]}$

Where,

• ${\text{K}}_{\text{a}}$ is the acid dissociation constant.

$\text{Basicity of amino group}\left(K_{\text{b}}\right)=\text{6}{\text{.0×10}}^{-\text{5}}$

The given reaction occurs in an aqueous medium. In an aqueous solution, the acidity and basicity of an acid and its conjugate base is related by the expression,

$K_{\text{a}}\text{ × }{K}_{\text{b}}={\text{10}}^{-\text{14}}$

Substitute the value of $K_{\text{b}}$ in the above expression.

$\begin{array}{c}{K}_{\text{a}}={\frac{\text{10}}{K_{\text{b}}}}^{-\text{14}}\\ =\frac{{\text{10}}^{-\text{14}}}{\text{6}{\text{.0×10}}^{-\text{10}}}\\ =\text{1}{\text{.6×10}}^{-\text{10}}\end{array}$

Hence, the equilibrium constant for the reaction is $\underset{\_}{1.6\times 10^{-10}}$.