Chapter 9, Problem 9.8P

(a)

Interpretation Introduction

Interpretation:

The ratio of $\frac{\left[{\text{CH}}_{\text{3}}{\text{NH}}_2\right]}{\left[{\text{CH}}_{\text{3}}{\text{NH}}_3^{+}\right]}$ at $\text{pH}$ equals to 4 has to be calculated.

Concept Introduction:

The equation for buffer is described by Henderson-Hasselbach equation and it is a rearranged form of equilibrium constant, $K_{\text{a}}$. It relates $\text{pH}$ of buffer solution and $\text{p}{K}_{\text{a}}$ value. It also helps to find equilibrium $\text{pH}$ in acid-base reactions. Consider an equation of dissociation of an acid as follows:

  $\text{HA}\rightleftharpoons {\text{H}}^{+}+{\text{A}}^{-}$

Formula to calculate $\text{pH}$ is as follows:

  $\text{pH}=\text{p}{K}_{\text{a}}+\log \left(\frac{\left[{\text{A}}^{-}\right]}{\left[\text{HA}\right]}\right)$

Here,

$\text{p}{K}_{\text{a}}$ is acid dissociation constant of weak acid $\text{HA}$.

$\left[{\text{A}}^{-}\right]$ is concentration of conjugate base ${\text{A}}^{-}$

$\left[\text{HA}\right]$ is concentration of acid $\text{HA}$.

(b)

Interpretation Introduction

Interpretation:

The ratio of $\frac{\left[{\text{CH}}_{\text{3}}{\text{NH}}_2\right]}{\left[{\text{CH}}_{\text{3}}{\text{NH}}_3^{+}\right]}$ at $\text{pH}$ equals to 10.632 has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Interpretation Introduction

Interpretation:

The ratio of $\frac{\left[{\text{CH}}_{\text{3}}{\text{NH}}_2\right]}{\left[{\text{CH}}_{\text{3}}{\text{NH}}_3^{+}\right]}$ at $\text{pH}$ equals to 4 has to be calculated.

Concept Introduction:

Refer to part (a).