Chapter 16, Problem 16.77QP

Interpretation Introduction

Interpretation:

The $K_b$ for each of the given ions has to be calculated.

Concept Information:

Acid ionization constant ${\text{K}}_{\text{a}}$:

The equilibrium expression for the reaction ${\text{HA}}_{\text{(aq)}}\to {\text{ H}}^{\text{+}}{}_{\text{(aq)}}{\text{+A}}^{\text{-}}{}_{\text{(aq)}}$ is given below.

$K_a=\frac{[{\text{H}}^{\text{+}}{\text{][A}}^{\text{-}}\text{]}}{[\text{HA]}}$

Where $K_a$ is acid ionization constant, $[{\text{H}}^{\text{+}}\text{]}$  is concentration of hydrogen ion, $[{\text{A}}^{\text{-}}\text{]}$  is concentration of acid anion, $[\text{HA]}$ is concentration of the acid

Base ionization constant $K_b$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{B}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(l)}}\to {\text{HB}}^{\text{+}}{}_{\text{(aq)}}{\text{+OH}}^{\text{-}}{}_{\text{(aq)}}$

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where $K_b$ is base ionization constant, $[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion, $[{\text{HB}}^{\text{+}}\text{]}$ is concentration of conjugate acid, $[\text{B]}$ is concentration of the base

Relationship between $K_a$ and $K_b$

$K_a\times K_b=K_w$

Where,

$K_w$ is autoionization of water.

$\begin{array}{l}\text{The autoionization of water}:\\ K_w=[{\text{H}}^{\text{+}}\text{]}[{\text{OH}}^{\text{-}}]\\ =1.0\times {10}^{-14}\end{array}$

Answer to Problem 16.77QP

Answer

• The $K_b$ for the given ion ${\text{CN}}^{\text{-}}$  is $2.0\times {10}^{-5}$
• The $K_b$ for the given ion ${\text{F}}^{\text{-}}$ is $1.4\times {10}^{-11}$
• The $K_b$ for the given ion ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}$ is $5.6\times {10}^{-10}$
• The $K_b$ for the given ion ${\text{HCO}}_{\text{3}}{}^{-}$ is $2.4\times {10}^{-8}$  

Explanation of Solution

${\text{CN}}^{\text{-}}$ is the conjugate base of weak acid $\text{HCN}$,

The acid ionization constant $K_a$ of acid $\text{HCN}$ is $4.9\times {10}^{-10}$

The ionization constant $K_w$ is $1.0\times {10}^{-14}$

$\begin{array}{l}{K}_a\times K_b=K_w\\ K_b({\text{CN}}^{\text{-}})=\frac{K_w}{K_a(\text{HCN})}\\ =\frac{1.0\times {10}^{-14}}{4.9\times {10}^{-10}}\\ =2.0\times {10}^{-5}\end{array}$

Therefore, $K_b$ of ${\text{CN}}^{\text{-}}$ ion is $2.0\times {10}^{-5}$

The $K_b$ for the given ion ${\text{F}}^{\text{-}}$

${\text{F}}^{\text{-}}$ is the conjugate base of weak acid $\text{HF}$

The acid ionization constant $K_a$ of acid $\text{HF}$ is $7.1\times {10}^{-4}$

The ionization constant $K_w$ is $1.0\times {10}^{-14}$

$\begin{array}{l}{K}_a\times K_b=K_w\\ K_b({\text{F}}^{\text{-}})=\frac{K_w}{K_a(\text{HF})}\\ =\frac{1.0\times {10}^{-14}}{7.1\times {10}^{-4}}\\ =1.4\times {10}^{-11}\end{array}$

Therefore, $K_b$ of ${\text{F}}^{\text{-}}$ ion is $1.4\times {10}^{-11}$

The $K_b$ for the given ion ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}$

${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}$ is the conjugate base of weak acid ${\text{CH}}_{\text{3}}\text{COOH}$,

The acid ionization constant $K_a$ of acid ${\text{CH}}_{\text{3}}\text{COOH}$ is $1.8\times {10}^{-5}$

The ionization constant $K_w$ is $1.0\times {10}^{-14}$

$\begin{array}{l}{K}_a\times K_b=K_w\\ K_b({\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}})=\frac{K_w}{K_a({\text{CH}}_{\text{3}}\text{COOH})}\\ =\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}\\ =5.6\times {10}^{-10}\end{array}$

Therefore, $K_b$ of ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}$ ion is $5.6\times {10}^{-10}$

The $K_b$ for the given ion ${\text{HCO}}_{\text{3}}{}^{\text{-}}$

${\text{HCO}}_{\text{3}}{}^{\text{-}}$ is the conjugate base of weak acid ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$

The acid ionization constant $K_a$ of acid ${\text{H}}_{\text{2}}{\text{CO}}_{\text{3}}$ is $4.2\times {10}^{-7}$

The ionization constant $K_w$ is $1.0\times {10}^{-14}$

$\begin{array}{l}{K}_a\times K_b=K_w\\ K_b({\text{HCO}}_{\text{3}}{}^{\text{-}})=\frac{K_w}{K_a({\text{H}}_{\text{2}}{\text{CO}}_{\text{3}})}\\ =\frac{1.0\times {10}^{-14}}{4.2\times {10}^{-7}}\\ =2.4\times {10}^{-8}\end{array}$

Therefore, $K_b$ of ${\text{HCO}}_{\text{3}}{}^{\text{-}}$ ion is $2.4\times {10}^{-8}$

Conclusion

The $K_b$ for each of the given ions was calculated.