Chapter 16, Problem 2KSP

Determine pH at the equivalence point in the titration of 41.0 mL 0.096 M formic acid with 0.108 M $\text{NaOH}$ at $\text{25°C}$.

(a) 12.94

(b) 7.00

(c) 5.76

(d) 8.24

(e) 1.06

Interpretation Introduction

Interpretation:

The pH at the equivalence point in the titration of formic acid with sodium hydroxide is to be determined.

Concept introduction:

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

$\text{HA}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\to {\text{A}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

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

${\text{A}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons \text{HA}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

Here, ${\text{A}}^{-}$ is the ion that forms the weak acid $\text{HA}$. The pH equivalence of this solution is now determined by the $\left[{\text{OH}}^{-}\right]$

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. …

${\text{K}}_{\text{b}}$ is the measure of dissociation of a base, known as base-dissociation constant and is specific at a particular temperature.

${\text{K}}_{\text{b}}=\frac{\left[{\text{OH}}^{-}\right]\left[\text{HA}\right]}{\left[{\text{A}}^{-}\right]}$ …… (2)

The formula to calculate the pOH of the solution from the concentration of hydroxide ions is

$\text{pOH}=-\log \left[{\text{OH}}^{-}\right]$ …… (3)

pH is the measure of acidity of a solution, which depends on the concentration of hydronium ions and the temperature of the solution. The relationship between pH and pOH is

$\text{pH}+\text{pOH}=14$ …… (4)

Molarity $\left(M\right)$ is the concentration of any substance given by the 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

$n=M\times V$

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

$mn=M\times V\left(\text{mL}\right)$ …… (5)

Answer to Problem 2KSP

Correct answer: Option (d).

Explanation of Solution

Given information:

The concentration of formic acid $\left(\text{HCOOH}\right)$ at 25°Cis $0.096M$ and $41.0\text{ mL}$ is the volume. The concentration of $\text{NaOH}$ used in the titration is $0.108M$.

Reason for correct option:

From the given values of concentration and volume of formic acid, calculate the number of millimoles of formic acid, using equation (5) as shown:

$\begin{array}{c}{\left(nm\right)}_{\text{HCOOH}}=0.096\frac{\text{mol}}{\cancel{\text{L}}}\times 41.0\text{ m}\cancel{\text{L}}\\ =3.936\text{ mmol}\end{array}$

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

${\text{HCOOH}}_{\left(\text{aq}\right)}+{\text{OH}}^{-}{}_{\left(\text{aq}\right)}\to {\text{HCOO}}^{-}{}_{\left(\text{aq}\right)}+{\text{H}}_2{\text{O}}_{\left(\text{l}\right)}$

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

${\left(nm\right)}_{\text{NaOH}}=3.936\text{ mmol}$

As, the concentration of $\text{NaOH}$ is given, volume of $\text{NaOH}$ used is calculated using equation (5) as follows:

$\begin{array}{c}{\left(nm\right)}_{\text{NaOH}}={\left(M\right)}_{\text{NaOH}}\times V\left(\text{mL}\right)\\ V\left(\text{mL}\right)=\frac{{\left(nm\right)}_{\text{NaOH}}}{{\left(M\right)}_{\text{NaOH}}}\\ =\frac{3.936\text{ m}\cancel{\text{mol}}}{0.108\frac{\cancel{\text{mol}}}{\text{L}}}\\ =36.4\text{ mL}\end{array}$

Thus, total volume of the solution containing $\text{HCOOH and NaOH}$ is $77.4\text{ mL}$.

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

${\left(nm\right)}_{{\text{HCOO}}^{-}}=3.936\text{ mmol}$

Thus, concentration of the conjugate base $\left({\text{HCOO}}^{-}\right)$ will be:

$\begin{array}{c}{\left(M\right)}_{{\text{HCOO}}^{-}}=\frac{{\left(nm\right)}_{{\text{HCOO}}^{-}}}{V\left(\text{mL}\right)}\\ =\frac{3.936\cancel{\text{m}}\text{mol}}{77.4\cancel{\text{m}}\text{L}}\\ =0.051M\end{array}$

The anion ${\text{HCOO}}^{-}$ comes from the weak formic acid $\left(\text{HCOOH}\right)$, thus the anion recombines with water to produce the acid and hydroxide ions, according to the reaction:

${\text{HCOO}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons \text{HCOOH}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$

From table $16.6$, ${\text{K}}_{\text{a}}$ of formic acid is $1.7\times {10}^{-4}$.

${\text{K}}_{\text{b}}$ of ${\text{HCOO}}^{-}$, which is the conjugate base of formic acid, is calculated using equation (1), and substitute the values of ${\text{K}}_{\text{a}}$ and ${\text{K}}_{\text{w}}$ as shown:

$\begin{array}{c}\left(1.7\times {10}^{-4}\right)\times {\text{K}}_{\text{b}}=1.0\times {10}^{-14}\\ {\text{K}}_{\text{b}}=\frac{1.0\times {10}^{-14}}{1.7\times {10}^{-4}}\\ {\text{K}}_{\text{b}}=5.9\times {10}^{-11}\end{array}$

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

$\begin{array}{ccccc}& {\text{HCOO}}^{-}\left(aq\right)& {\text{H}}_2\text{O}\left(aq\right)& {\text{OH}}^{-}\left(aq\right)& \text{HCOOH}\left(aq\right)\\ \text{Initial concentration}\left(M\right)& 0.051& -& 0& 0\\ \text{Change in concentration}\left(M\right)& -x& -& +x& +x\\ \text{Equilibrium concentration}\left(M\right)& 0.051-x& -& x& x\end{array}$

Now, substitute these concentrations in equation (2) as

${\text{K}}_{\text{b}}=\frac{\left(x\right)\left(x\right)}{\left(0.051-x\right)}$

Since the value of ${\text{K}}_{\text{b}}$ is very small, the amount of base that recombines to form acid is less. Therefore, $\left(0.051-x\right)$ can be approximated as $0.051$. Now, substitute the value of ${\text{K}}_{\text{b}}$ in above equation as shown:

$\begin{array}{c}5.9\times {10}^{-11}=\frac{\left(x\right)\left(x\right)}{\left(0.051\right)}\\ x^2=3.0\times {10}^{-12}\\ x=\sqrt{3.0\times {10}^{-12}}\\ x=1.7\times {10}^{-6}\end{array}$

Thus,

$\begin{array}{c}\left[\text{HCOOH}\right]=1.7\times {10}^{-6}M\\ \left[{\text{OH}}^{-}\right]=1.7\times {10}^{-6}M\end{array}$

Now, substitute the value of $\left[{\text{OH}}^{-}\right]$ in equation (3)

$\begin{array}{c}\text{pOH}=-\log \left[1.7\times {10}^{-6}\right]\\ =5.76\end{array}$

Substitute this value of pOH in equation (4) to calculate the pH of the solution

$\begin{array}{c}\text{pH}+5.76=14\\ \text{pH}=14-5.76\\ \text{pH}=8.24\end{array}$

Therefore, the equivalence pH of the solution is $8.24$. Thus, option (d) is correct.

Reason for incorrect options:

Since $12.94$ is not obtained as the equivalence pH of the solution, option (a) is incorrect.

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

Since $5.76$ is not obtained as the equivalence pH of the solution, option (c) is incorrect.

Since $1.06$ is not obtained as the equivalence pH of the solution, option (e) is incorrect.

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