Chapter 20, Problem 20.40P

(a)

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{a}}$ value for conjugate acid of pyridine has to be calculated.

Concept Introduction:

Ionic-product constant for water: It is the hydronium ion concentration times the $O{H}^{-}$ concentration present in the solution.

$\begin{array}{c}{H}_2{O}_{\left(l\right)}+H_2{O}_{\left(l\right)}{H}_3{O}^{+}{}_{\left(aq\right)}+O{H}^{-}{}_{\left(aq\right)}\\ K=\frac{\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]}{\left[H_{2}O\right]\left[H_{2}O\right]}\\ \left[H_3{O}^{+}\right]=\left[O{H}^{-}\right]=1.00\times {10}^{-7}M\\ \therefore K_W=K\left[H_{2}O\right]\left[H_{2}O\right]=\left[H_3{O}^{+}\right]\left[O{H}^{-}\right]\\ K_W=1.00\times {10}^{-14}M\\ K_W=K_a\cdot K_b\end{array}$

Answer to Problem 20.40P

The ${\text{K}}_{\text{a}}$ value for conjugate acid of pyridine is $\text{6}\text{.7}\times {10}^{-6}\text{ M}$.

Explanation of Solution

Given: ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}\text{N}\Rightarrow {\text{K}}_{\text{b}}\text{ = 1}\text{.5}\times {\text{10}}^{\text{-9}}\text{ M}$

    ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{N+ H}}^{\text{+}}\rightleftharpoons {\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}$

The conjugate acid of given base is ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}$.

The ${\text{K}}_{\text{a}}$ value of ${\text{C}}_{\text{5}}{\text{H}}_{\text{5}}{\text{NH}}^{\text{+}}$ is shown as follows,

    $\begin{array}{c}{K}_W=K_a\cdot K_b\\ 1.00\times {10}^{-14}\text{ M}=\text{ 1}\text{.5}\times {\text{10}}^{\text{-9}}\text{ M}\times K_a\\ K_a=\frac{1.00\times {10}^{-14}\text{ M}}{\text{1}\text{.5}\times {\text{10}}^{\text{-9}}\text{ M}}\\ =\text{ 6}\text{.7}\times {10}^{-6}\text{ M}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{a}}$ value for conjugate acid of cyanide ion has to be calculated.

Concept Introduction:

Refer to sub-part (a).

Answer to Problem 20.40P

The ${\text{K}}_{\text{a}}$ value for conjugate acid of cyanide ion is $\text{4}\text{.8}\times {10}^{-10}\text{ M}$.

Explanation of Solution

Given: ${\text{CN}}^{\text{-}}\Rightarrow {\text{K}}_{\text{b}}\text{ = 2}\text{.1}\times {\text{10}}^{\text{-5}}\text{ M}$

    ${\text{CN}}^{\text{-}}{\text{+ H}}^{\text{+}}\rightleftharpoons \text{HCN}$

The conjugate acid of given base is $\text{HCN}$.

The ${\text{K}}_{\text{a}}$ value of $\text{HCN}$ is shown as follows,

    $\begin{array}{c}{K}_W=K_a\cdot K_b\\ 1.00\times {10}^{-14}\text{ M}=\text{ 2}\text{.1}\times {\text{10}}^{\text{-5}}\text{ M}\times K_a\\ K_a=\frac{1.00\times {10}^{-14}\text{ M}}{\text{2}\text{.1}\times {\text{10}}^{\text{-5}}\text{ M}}\\ =\text{ 4}\text{.8}\times {10}^{-10}\text{ M}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{a}}$ value for conjugate acid of hydroxylamine has to be calculated.

Concept Introduction:

Refer to sub-part (a).

Answer to Problem 20.40P

The ${\text{K}}_{\text{a}}$ value for conjugate acid of hydroxylamine is $\text{1}\text{.1}\times {10}^{-6}\text{ M}$.

Explanation of Solution

Given: ${\text{NH}}_{\text{2}}\text{OH}\Rightarrow {\text{K}}_{\text{b}}\text{ = 8}\text{.7}\times {\text{10}}^{\text{-9}}\text{ M}$

    ${\text{NH}}_{\text{2}}{\text{OH+ H}}^{\text{+}}\rightleftharpoons {\text{NH}}_{\text{3}}{\text{OH}}^{\text{+}}$

The conjugate acid of given base is ${\text{NH}}_{\text{3}}{\text{OH}}^{\text{+}}$.

The ${\text{K}}_{\text{a}}$ value of ${\text{NH}}_{\text{3}}{\text{OH}}^{\text{+}}$ is shown as follows,

    $\begin{array}{c}{K}_W=K_a\cdot K_b\\ 1.00\times {10}^{-14}\text{ M}=\text{ 8}\text{.7}\times {\text{10}}^{\text{-9}}\text{ M}\times K_a\\ K_a=\frac{1.00\times {10}^{-14}\text{ M}}{\text{8}\text{.7}\times {\text{10}}^{\text{-9}}\text{ M}}\\ =\text{ 1}\text{.1}\times {10}^{-6}\text{ M}\end{array}$

(d)

Interpretation Introduction

Interpretation:

The ${\text{K}}_{\text{a}}$ value for conjugate acid of dimethylamine has to be calculated.

Concept Introduction:

Refer to sub-part (a).

Answer to Problem 20.40P

The ${\text{K}}_{\text{a}}$ value for conjugate acid of dimethylamine is $\text{1}\text{.9}\times {10}^{-11}\text{ M}$.

Explanation of Solution

Given: ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}\text{NH}\Rightarrow {\text{K}}_{\text{b}}\text{ = 5}\text{.4}\times {\text{10}}^{\text{-4}}\text{ M}$

    ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{NH+ H}}^{\text{+}}\rightleftharpoons {\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{NH}}_2{}^{\text{+}}$

The conjugate acid of given base is ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{NH}}_2{}^{\text{+}}$.

The ${\text{K}}_{\text{a}}$ value of ${\left({\text{CH}}_{\text{3}}\right)}_{\text{2}}{\text{NH}}_2{}^{\text{+}}$ is shown as follows,

    $\begin{array}{c}{K}_W=K_a\cdot K_b\\ 1.00\times {10}^{-14}\text{ M}=\text{ 1}\text{.5}\times {\text{10}}^{\text{-9}}\text{ M}\times K_a\\ K_a=\frac{1.00\times {10}^{-14}\text{ M}}{\text{5}\text{.4}\times {\text{10}}^{\text{-4}}\text{ M}}\\ =\text{ 1}\text{.9}\times {10}^{-11}\text{ M}\end{array}$