Chapter 2, Problem 2.44P

What is $K_{\text{a}}$ for each compound? Use a calculator when necessary.

a. $\begin{array}{l}{\text{H}}_{\text{2}}\text{S}\\ \text{p}{K}_{\text{a}}=7.0\end{array}$

b. $\begin{array}{l}{\text{ClCH}}_{\text{2}}\text{COOH}\\ \text{p}{K}_{\text{a}}=2.8\end{array}$

c. $\begin{array}{l}\text{HCN}\\ \text{p}{K}_{\text{a}}=9.1\end{array}$

Interpretation Introduction

(a)

Interpretation: The value of $K_{\text{a}}$ for ${\text{H}}_{\text{2}}\text{S}$ is to be calculated.

Concept introduction: The equilibrium constant of an acid dissociation reaction depends on the strength of an acid involved in the reaction. The $K_{\text{a}}$ value is large for strong acids and small for weak acids.

Acid dissociation constant $K_{\text{a}}$ represents the acidity of an acid in a solution.

For an acid dissociation reaction,

$HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

$K_a=\frac{\left[H_3{O}^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$

$p{K}_a=-\log K_a$

The strength of an acid decreases as the value of $\text{p}{K}_{\text{a}}$ increases.

Answer to Problem 2.44P

The value of $K_{\text{a}}$ for ${\text{H}}_{\text{2}}\text{S}$ is ${10}^{-7}$.

Explanation of Solution

The given value of $\text{p}{K}_{\text{a}}$ is $7.0$.

The value of $K_{\text{a}}$ is calculated by the formula,

$K_{\text{a}}={10}^{-\text{p}{K}_{\text{a}}}$

Substitute the value of $\text{p}{K}_{\text{a}}$ in the above formula to calculate the value of $K_{\text{a}}$.

$\begin{array}{l}{K}_{\text{a}}={10}^{-(7)}\\ K_{\text{a}}={10}^{-7}\end{array}$

Hence, the value of ${\text{K}}_{\text{a}}$ is ${10}^{-7}$.

Conclusion

The value of $K_{\text{a}}$ for ${\text{H}}_{\text{2}}\text{S}$ is ${10}^{-7}$.

Interpretation Introduction

(b)

Interpretation: The value of $K_{\text{a}}$ for ${\text{ClCH}}_{\text{2}}\text{COOH}$ is to be calculated.

Concept introduction: The equilibrium constant of an acid dissociation reaction depends on the strength of an acid involved in the reaction. The $K_{\text{a}}$ value is large for strong acids and small for weak acids.

Acid dissociation constant $K_{\text{a}}$ represents the acidity of an acid in a solution.

For an acid dissociation reaction,

$HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

${\text{pK}}_{\text{a}}=\frac{\left[H_3{O}^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$

$p{K}_a=-\log K_a$

The strength of an acid decreases as the value of $\text{p}{K}_{\text{a}}$ increases.

Answer to Problem 2.44P

The value of $K_{\text{a}}$ for ${\text{ClCH}}_{\text{2}}\text{COOH}$ is $1.585\times {10}^{-3}$.

Explanation of Solution

The given value of $\text{p}{K}_{\text{a}}$ is $2.8$.

The value of $K_{\text{a}}$ is calculated by the formula,

$K_{\text{a}}={10}^{-\text{p}{K}_{\text{a}}}$

Substitute the value of $\text{p}{K}_{\text{a}}$ in the above formula to calculate the value of $K_{\text{a}}$.

$\begin{array}{l}{K}_{\text{a}}={10}^{-(2.8)}\\ K_{\text{a}}=0.001585\\ K_{\text{a}}=1.585\times {10}^{-3}\end{array}$

Hence, the value of $K_{\text{a}}$ is $1.585\times {10}^{-3}$.

Conclusion

The value of $K_{\text{a}}$ for ${\text{ClCH}}_{\text{2}}\text{COOH}$ is $1.585\times {10}^{-3}$.

Interpretation Introduction

(c)

Interpretation: The value of $K_{\text{a}}$ for $\text{HCN}$ is to be calculated.

Concept introduction: The equilibrium constant of an acid dissociation reaction depends on the strength of an acid involved in the reaction. The $K_{\text{a}}$ value is large for strong acids and small for weak acids.

Acid dissociation constant $K_{\text{a}}$ represents the acidity of an acid in a solution.

For an acid dissociation reaction,

$HA+H_{2}O\underset{}{\overset{}{\rightleftharpoons }}{H}_3{O}^{+}+A^{-}$

$K_a=\frac{\left[H_3{O}^{+}\right]\left[A^{-}\right]}{\left[HA\right]}$

$\text{p}{K}_{\text{a}}=-\log K_a$

The strength of an acid decreases as the value of $\text{p}{K}_{\text{a}}$ increases.

Answer to Problem 2.44P

The value of $K_{\text{a}}$ for $\text{HCN}$ is $7.943\times {10}^{-10}$.

Explanation of Solution

The given value of $\text{p}{K}_{\text{a}}$ is $9.1$.

The value of $K_{\text{a}}$ is calculated by the formula,

$K_{\text{a}}={10}^{-\text{p}{K}_{\text{a}}}$

Substitute the value of $\text{p}{K}_{\text{a}}$ in the above formula to calculate the value of $K_{\text{a}}$.

$\begin{array}{l}{K}_{\text{a}}={10}^{-(9.1)}\\ K_{\text{a}}=8\times {10}^{-10}\end{array}$

Hence, the value of $K_{\text{a}}$ is $8\times {10}^{-10}$.

Conclusion

The value of $K_{\text{a}}$ for $\text{HCN}$ is $8\times {10}^{-10}$.