Chapter 17.3, Problem 3RC

Interpretation Introduction

Interpretation:

The correct indicator should be identifed given the titation of ammonia $\left({\text{NH}}_{\text{3}}\right)$ and hydrochloric acid $\left(\text{HCl}\right)$ reaction, than using the figure 17.7.

Concept introduction:

Titration is a quantitative method to determine the quantity of an acid or base in a solution. This method is used to determine the concentration an acid in the solution by titrating it against a base. There are four types of acid-base titrations.

(1) Strong acid-Strong base, in this type of titration a strong acid is titrated against a strong base for example, $\text{HCl}$ is titrated against $\text{NaOH}$.

(2) Strong acid-Weak base, in this type of titration a strong acid is titrated against a weak base for example, $\text{HCl}$ is titrated against ${\text{NH}}_4\text{OH}$.

(3) Weak acid-Strong base, in this type of titration a weak acid is titrated against a strong base for example, ${\text{CH}}_3\text{COOH}$ is titrated against $\text{NaOH}$.

(4) Weak acid-Weak base, in this type of titration a weak acid is titrated against a weak base for example, ${\text{CH}}_3\text{COOH}$ is titrated against ${\text{NH}}_4\text{OH}$.

For weak base-strong acid titration the $\text{pH}$ value can be calculated at various points before and after equivalence point.

The equilibrium established during the titration of ${\text{NH}}_3$ with $\text{HCl}$. The equilibrium can be represented as,

${\text{NH}}_3\left(\text{aq}\right){\text{+ H}}_3{\text{O}}^{+}\left(\text{aq}\right)\rightleftharpoons {\text{H}}_2\text{O}\left(\text{l}\right)+{\text{NH}}_4{}^{+}\left(\text{aq}\right)$

Here, the ${\text{H}}_3{\text{O}}^{+}$ ion are generated from $\text{HCl}$, and the reaction is written as,

$\text{HCl}\left(\text{aq}\right)+{\text{H}}_2\text{O}\left(\text{l}\right)\to {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{Cl}}^{-}\left(\text{aq}\right)$

At equivalence point all the base will be neutralized, and there will be only ${\text{NH}}_4{}^{+}$ ion and ${\text{H}}_3{\text{O}}^{+}$. The ${\text{H}}_3{\text{O}}^{+}$ will be produced due to the hydrolysis of ${\text{NH}}_4{}^{+}$ at equivalence point. The hydrolysis equilibrium is represented as,

${\text{NH}}_4{}^{+}\left(\text{aq}\right){\text{+H}}_2\text{O}\left(\text{l}\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{NH}}_3\left(\text{aq}\right)$

By using the value of $K_{\text{a}}$ for the ${\text{NH}}_4{}^{+}$ concentration of ${\text{H}}_3{\text{O}}^{+}$ can be calculated.

$K_{\text{a}}=\frac{\left[{\text{H}}_3{\text{O}}^{+}\right]\left[{\text{H}}_2\text{O}\right]}{\left[{\text{NH}}_4{}^{+}\right]}$ (1)

As ${\text{H}}_3{\text{O}}^{+}$ ions are produced due to hydrolysis of ${\text{NH}}_4{}^{+}$, the value of $\text{pH}$ will be lower than $7$ at equivalence point for the weak base-strong acid titrations.