Chapter 9, Problem 9.12P

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of solution formed from solution of chloroacetic acid and sodium chloroacetate 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 $\text{HA}$.

Concentration of a solution is expressed by molarity and is denoted by $M$. It is defined as the ratio of number of moles of solute to volume of solution in $1\text{ L}$.

The formula to calculate moles is as follows:

  $\text{Moles}=\frac{\text{mass of solute}}{\text{molar mass of solute}}$

The formula used to calculate concentration is as follows:

  $\text{Concentration}=\frac{\text{moles}}{\text{volume of solution}\left(\text{L}\right)}$

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ of solution formed from solution of chloroacetic acid, sodium chloroacetate, ${\text{HNO}}_3$ and $\text{Ca}{\left(\text{OH}\right)}_2$ has to be calculated.

Concept Introduction:

Refer to part (a).