Chapter 16, Problem 4KSP

Determine pH at the equivalence point in the titration of 26.0 mL 1.12 M pyridine with $0.93M HC\text{l at 25°C}\text{.}$

(a) 7.00

(b) 2.76

(c) 11.24

(d) 1.73

(e) 12.27

Interpretation Introduction

Interpretation:

The pH at the equivalence point in the titration of pyridine with hydrochloric acid is to be determined.

Concept introduction:

When a weak base is titrated against a strong acid, the conjugate acid of the weak base is formed in the reaction, as shown:

$\text{BOH}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\to {\text{B}}^{+}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

This conjugate acid now acts as a Bronsted acid and reacts with water to form weak base and hydronium ions according to the reaction:

${\text{B}}^{+}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons \text{BOH}\left(aq\right)+{\text{H}}^{+}\left(aq\right)$

Here, ${\text{B}}^{+}$ is the ion that forms the weak base $\text{BOH}$. The pH equivalence of this solution is now determined by the $\left[{\text{H}}^{+}\right]$ ion.

The relationship between ${\text{K}}_{\text{b}}$, ${\text{K}}_{\text{a}}$, and ${\text{K}}_{\text{w}}$ indicates the quantitative basis of the reciprocal relationship between the strength of an acid and its conjugate base, or vice-versa.

${\text{K}}_{\text{a}}\times {\text{K}}_{\text{b}}={\text{K}}_{\text{w}}$ …… (1)

Where, ${\text{K}}_{\text{a}}$ is dissociation constant of acid, ${\text{K}}_{\text{b}}$ is dissociation constant of base, and ${\text{K}}_{\text{w}}$ is dissociation constant of water.

${\text{K}}_{\text{a}}$ is the measure of dissociation of an acid, known as acid-dissociation constant and is specific at particular temperatures.

${\text{K}}_{\text{a}}=\frac{\left[{\text{H}}^{+}\right]\left[\text{BOH}\right]}{\left[{\text{B}}^{+}\right]}$ …… (2)

The formula to calculate the pH of the solution from the concentration of hydronium ions is expressed as

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$ …… (3)

Molarity $\left(M\right)$ is the concentration of any substance given by number of moles $\left(n\right)$ of the substance divided by the volume of solution, $\left(V\right)$ in liters.

$M=\frac{n}{V}$

Rearrange this equation in terms of moles as shown

$n=M\times V$

When volume is given in $\text{mL}$ instead of liter, then millimoles are calculated as

$mn=M\times V\left(\text{mL}\right)$ …… (4)

Answer to Problem 4KSP

Correct answer: Option (b).

Explanation of Solution

Given information:

The concentration of pyridine $\left({\text{C}}_5{\text{H}}_5\text{N}\right)$ at 25°Cis $1.12M$ and the volume is $26.0\text{ mL}$. The concentration of $\text{HCl}$ used in the titration is $0.93M$.

Reason for correct option:

From the given values of concentration and volume, calculate the number of millimoles of pyridineusing equation (4)

$\begin{array}{c}{\left(nm\right)}_{{\text{C}}_5{\text{H}}_5\text{N}}=1.12\frac{\text{mol}}{\cancel{\text{L}}}\times 26.0\text{ m}\cancel{\text{L}}\\ =29.12\text{ mmol}\end{array}$

Being a strong acid, $\text{HCl}$ ionizes completely into ${\text{H}}^{+}{\text{ and Cl}}^{-}$ ions. The concentration of ${\text{H}}^{+}$ is equal to the concentration of $\text{HCl}$. The ${\text{H}}^{+}$ ions thus reacts with the weak base to produce its conjugate acid and water, according to the reaction shown given below:

${\text{C}}_5{\text{H}}_5{\text{N}}_{\left(\text{aq}\right)}+{\text{H}}^{+}{}_{\left(\text{aq}\right)}\to {\text{C}}_5{\text{H}}_5{\text{NH}}^{+}{}_{\left(\text{aq}\right)}$

At equivalence point, during the titration process, millimoles of weak base must be equal to the millimoles of the strong acid. Thus,

${\left(nm\right)}_{\text{HCl}}=29.12\text{ mmol}$

As the concentration of $\text{HCl}$ is given, the volume of $\text{HCl}$ used is calculated using equation (4):

$\begin{array}{c}{\left(nm\right)}_{\text{HCl}}={\left(M\right)}_{\text{HCl}}\times V\left(\text{mL}\right)\\ V\left(\text{mL}\right)=\frac{{\left(nm\right)}_{\text{HCl}}}{{\left(M\right)}_{\text{HCl}}}\\ =\frac{\text{29}\text{.12 m}\cancel{\text{mol}}}{0.93\frac{\cancel{\text{mol}}}{\text{L}}}\\ =31.3\text{ mL}\end{array}$

Thus, the total volume of the solution containing ${\text{C}}_5{\text{H}}_5\text{N and HCl}$ is $57.3\text{ mL}$.

During titration, the weak base completelyneutralizes. Thus, the moles of weak base reacted is equal to the moles of its conjugate acid formed. Therefore,

${\left(nm\right)}_{{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}}=29.12\text{ mmol}$

Thus, the concentration of the conjugate acid $\left({\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\right)$ is calculated as

$\begin{array}{c}{\left(M\right)}_{{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}}=\frac{{\left(nm\right)}_{{\text{C}}_5{\text{H}}_5{\text{NH}}^{+}}}{V\left(\text{mL}\right)}\\ =\frac{29.12\cancel{\text{m}}\text{mol}}{57.3\cancel{\text{m}}\text{L}}\\ =0.51M\end{array}$

The anion ${\text{C}}_5{\text{H}}_5{\text{NH}}^{+}$ comes from the weak base pyridine $\left({\text{C}}_5{\text{H}}_5\text{N}\right)$, thus the cation recombines with water to produce the base and hydronium ions according to the reaction:

${\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{C}}_5{\text{H}}_5\text{N}\left(aq\right)+{\text{H}}_3{\text{O}}^{+}\left(aq\right)$

From table $16.7$, ${\text{K}}_{\text{b}}$ of pyridine is $1.7\times {10}^{-9}$.

${\text{K}}_{\text{a}}$ of ${\text{C}}_5{\text{H}}_5{\text{NH}}^{+}$, which is the conjugate acid of pyridine, is calculated using equation (1), also substitute the values of ${\text{K}}_{\text{b}}$ and ${\text{K}}_{\text{w}}$ as shown

$\begin{array}{c}\left(1.7\times {10}^{-9}\right)\times {\text{K}}_{\text{a}}=1.0\times {10}^{-14}\\ {\text{K}}_{\text{a}}=\frac{1.0\times {10}^{-14}}{1.7\times {10}^{-9}}\\ {\text{K}}_{\text{a}}=5.9\times {10}^{-6}\end{array}$

Now, prepare an equilibrium table and represent each of the species in terms of $x$ as

$\begin{array}{ccccc}& {\text{C}}_5{\text{H}}_5{\text{NH}}^{+}\left(aq\right)& {\text{H}}_2\text{O}\left(aq\right)& {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\left(aq\right)& {\text{C}}_5{\text{H}}_5\text{N}\left(aq\right)\\ \text{Initial concentration}\left(M\right)& 0.51& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(M\right)& 0.51-x& -& x& x\end{array}$

Now, substitute these concentrations in equation (2)

${\text{K}}_{\text{a}}=\frac{\left(x\right)\left(x\right)}{\left(0.51-x\right)}$

Since the value of ${\text{K}}_{\text{a}}$ is very small, the amount of acid that recombines to form the base is less. Therefore, $\left(0.51-x\right)$ can be approximated as $0.51$. Now, substitute the value of ${\text{K}}_{\text{a}}$ in the equation above

$\begin{array}{c}5.9\times {10}^{-6}=\frac{\left(x\right)\left(x\right)}{\left(0.51\right)}\\ x^2=3.0\times {10}^{-6}\\ x=\sqrt{3.0\times {10}^{-6}}\\ x=1.7\times {10}^{-3}\end{array}$

Thus,

$\begin{array}{c}\left[{\text{C}}_5{\text{H}}_5\text{N}\right]=1.7\times {10}^{-3}M\\ \left[{\text{H}}_3{\text{O}}^{+}\right]=1.7\times {10}^{-3}M\end{array}$

Now, substitute the value of $\left[{\text{H}}_3{\text{O}}^{+}\right]$ in equation (3) as follows:

$\begin{array}{c}\text{pH}=-\log \left[1.7\times {10}^{-3}\right]\\ =2.76\end{array}$

Therefore, the equivalence pH of the solution is $2.76$. Thus, option (b) is correct.

Reason for incorrect options:

Since $7.00$ is not obtained as the equivalence pH of the solution, option (a) is incorrect.

Since $11.24$ is not obtained as the equivalence pH of the solution, option (c) is incorrect.

Since $1.73$ is not obtained as the equivalence pH of the solution, option (d) is incorrect.

Since $12.27$ is not obtained as the equivalence pH of the solution, option (e) is incorrect.

Therefore, options (a), (c), (d), and (e) are incorrect.