Chapter 2, Problem 2.46P

What is the $p{K}_{\text{a}}$ for each compound?

a.

b.

c.

Interpretation Introduction

(a)

Interpretation: The value of $p{K}_a$ for the given compound 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,

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

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

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

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

Answer to Problem 2.46P

The value of $\text{p}{K}_{\text{a}}$ for the given compound is $9.3279$.

Explanation of Solution

The given value of $K_{\text{a}}$ is $4.7\times {10}^{-10}$.

The value of $\text{p}{K}_{\text{a}}$ is calculated by the formula,

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

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

$\begin{array}{rll}\text{p}{K}_{\text{a}}& =& -\log \left(4.7\times {10}^{-10}\right)\\ & =& -\left[\log \left(4.7\right)+\log {10}^{-10}\right]\\ & =& -\log \left(4.7\right)+10\\ & =& -0.6721+10\end{array}$

Simplify the above equation,

$p{K}_a=9.3279$

Hence, the value of $\text{p}{K}_{\text{a}}$ for the given compound is $9.3279$.

Conclusion

The value of $\text{p}{K}_{\text{a}}$ for the given compound is $9.3279$.

Interpretation Introduction

(b)

Interpretation: The value of $\text{p}{K}_{\text{a}}$ for the given compound 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,

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

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

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

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

Answer to Problem 2.46P

The value of $\text{p}{K}_{\text{a}}$ for the given compound is $4.6383$.

Explanation of Solution

The given value of ${\text{K}}_{\text{a}}$ is $2.3\times {10}^{-5}$.

The value of $\text{p}{K}_{\text{a}}$ is calculated by the formula,

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

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

$\begin{array}{rll}p{K}_a& =& -\log \left(2.3\times {10}^{-5}\right)\\ & =& -\left[\log \left(2.3\right)+\log {10}^{-5}\right]\\ & =& -\log \left(2.3\right)+5\\ & =& -0.3617+5\end{array}$

Simplify the above equation,

$p{K}_a=4.6383$

Hence, the value of $\text{p}{K}_{\text{a}}$ for the given compound is $4.6383$.

Conclusion

The value of $\text{p}{K}_{\text{a}}$ for the given compound is $4.6383$.

Interpretation Introduction

(c)

Interpretation: The value of $\text{p}{K}_{\text{a}}$ for the given compound 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 2.46P

The value of $\text{p}{K}_{\text{a}}$ for the given compound is $0.23$.

Explanation of Solution

The given value of $K_{\text{a}}$ is $5.9\times {10}^{-1}$.

The value of $\text{p}{K}_{\text{a}}$ is calculated by the formula,

${\text{pK}}_{\text{a}}=-\log K_a$

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

$\begin{array}{rll}\text{p}{K}_{\text{a}}& =& -\log \left(5.9\times {10}^{-1}\right)\\ & =& -\left[\log \left(5.9\right)+\log {10}^{-1}\right]\\ & =& -\log \left(5.9\right)+1\\ & =& -0.7709+1\end{array}$

Simplify the above equation.

$\text{p}{K}_{\text{a}}=0.23$

Hence, the value of $\text{p}{K}_{\text{a}}$ for the given compound is $0.23$.

Conclusion

The value of $\text{p}{K}_{\text{a}}$ for the given compound is $0.23$.