Chapter 15, Problem 7RQ

Sketch the titration curve for a weak acid titrated by a strong base. When performing calculations concerning weak acid–strong base titrations, the general two-slep procedure is to solve a stoichiometry problem first, then to solve an equilibrium problem to determine the pH. What reaction takes place in the stoichiometry part of the problem? What is assumed about this reaction?

At the various points in your titration curve, list the major species present after the strong base (NaOH, for example) reacts to completion with the weak acid, HA. What equilibrium problem would you solve at the various points in your titration curve to calculate the pH? Why is pH > 7.0 at the equivalence point of a weak acid-strong base titration? Does the pH at the halfway point to equivalence have to be less than 7.0? What does the pH at the halfway point equal? Compare and contrast the titration curves for a strong acid–strong base titration and a weak acid–strong base titration.

Interpretation Introduction

Interpretation: The titration curve for a weak acid titrated by a weak base is to be sketched. The reaction takes place in the stoichiometric part of the problem is to be stated. The equilibrium problem, at the various points in the titration curve to calculate the $\text{pH}$ is to be stated. The reason for the value of $\text{pH}$ is greater than $7$ at the equilibrium point for the titration of weak acid-strong base is to be stated. Whether the $\text{pH}$ at the halfway point to equivalence have to be less than $7$ is to be stated. The titration curves for a strong acid-strong base titration are to be compared and to be stated.

Concept introduction: The reaction between an acid and a base takes place with the formation of a salt and a water molecule.

The hydrogen ion concentration of the solution is known as $\text{pH}$ of the solution.

It is the negative logarithm of Hydrogen ion concentration.

At the equivalence point the moles of the acid and base in the solution are same.

At the half equivalence point the concentration of titrant is just half of the initial amount.

For the given value of dissociation constant the $\text{pH}=\text{p}{K}_{\text{a}}$.

To determine: The sketch of titration curve for a weak acid-strong base titration; The reaction takes place in stoichiometric part of the problem; The assumption about this reaction; The list of the major species present after the completion reaction between strong base with weak acid; The equilibrium problem which involve at the various points in the calculation of $\text{pH}$; The reason for the value of $\text{pH}>7$ at the equivalence point in the weak acid-strong base titration. If the $\text{pH}$ at the halfway point to equivalence point is less than $7$; The $\text{pH}$ at the halfway point; The comparison between the titration curves of a strong acid-strong base titration and a weak acid-strong base titration.

Explanation of Solution

Explanation

A titration curve is plot of $\text{pH}$ Vs volume of titrant. In such plots the $\text{pH}$ is taken as independent variable while volume of titrant is taken as independent variable. Such plots are also known as $\text{pH}$ curves.

In the calculation of weak acid-strong base titration, the two step procedure is involved. First one is stoichiometric problem and second is equilibrium problem. The step wise procedure is stated as follows,

• Stoichiometric problem: The completion reaction is to be assumed in the weak acid-strong base titration.
• Equilibrium problem: The $\text{pH}$ at the equivalence point and the position of weak acid equilibrium is determined.

The titration of $50.0\text{ml}\text{of}\text{0}\text{.10}\text{M}$ acetic acid ($K_{\text{a}}=1.8\times {10}^{-5}$) with $0.10\text{M}\text{NaOH}$ is to be assumed. The $\text{pH}$ curve for this titration is given as,

Figure 1

The equivalence point occurs on the addition of $50\text{ml}\text{of}\text{NaOH}$. Here the amount of hydroxide ion is exactly equal to the original amount of acid. At the equivalence point the $\text{pH}$ is greater than $7$ because the acetate ion present at this point is a base and it reacts with water to produce hydroxide ion.

The reaction involves in the part of stoichiometric problem is given as,

$\begin{array}{cccccccc}& {\text{OH}}^{\text{-}}& +& {\text{CH}}_{\text{3}}\text{COOH}& \to & {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}& +& {\text{H}}_{\text{2}}\text{O}\\ \text{Before}\text{reaction}& 10\times \text{0}\text{.10}=\text{1}\text{.0}\text{mmol}& & \begin{array}{l}50.0\times \text{0}\text{.10}\\ =\text{5}\text{.0}\text{mmol}\end{array}& & 0& & -\\ \text{After}\text{recation}& 0& & \begin{array}{l}5.0-1.0\\ =4.0\text{mmol}\end{array}& & 1.0\text{mmol}& & -\end{array}$

Here the amount of Hydroxide ion is lesser than the acid. Therefore, it consumed completely and called as limiting reactant.

The $\text{pH}$ at various points is calculated by representing the volumes of added $\text{NaOH}$ as,

When $0.0\text{ml}0.10\text{M}\text{NaOH}$ has been added.

In this situation only weak acid that is acetic acid id present in the solution.

Therefore the equilibrium of ${\text{CH}}_3\text{COOH}$ is represented by the equation,

${\text{CH}}_3\text{COOH}\rightleftharpoons {\text{CH}}_3{\text{COO}}^{-}+{\text{H}}^{+}$

Molarity of ${\text{CH}}_3\text{COOH}$ is $0.10\text{M}$. The number of moles is calculated by the formula,

$n=C\times V$

Where,

• $n$ is the number of moles.
• $C$ is the concentration of the solution.
• $V$ is the volume of the solution.

Substitute the values of concentrations and volume in the above equation.

$\begin{array}{c}n=C\times V\\ =0.100\text{M}\times \frac{50}{1000}\text{L}\\ =\text{0}\text{.05}\text{mol}\end{array}$

Now $x$ is supposed to be the change in moles. The equilibrium reaction with the calculated moles is expressed in ICE (initial, change, equilibrium) table as,

$\begin{array}{cccccc}& {\text{CH}}_3\text{COOH}& \rightleftharpoons & {\text{H}}^{+}& +& {\text{CH}}_3{\text{COO}}^{-}\\ \text{Initial}\text{mol}& 0.05& & 0& & 0\\ \text{Change}\text{in}\text{mol}& -x& & +x& & +x\\ \text{Equilibrium}\text{mol}& 0.05-x& & +x& & +x\end{array}$

The acid dissociation constant $\left(K_{\text{a}}\right)$ for this reaction is written as,

$K_{\text{a}}=\frac{\left(x\right)\left(x\right)}{\left(0.05-x\right)}$

Substitute the value of $K_{\text{b}}$ in the above equation,

$\begin{array}{c}{K}_{\text{b}}=\frac{\left(x\right)\left(x\right)}{\left(0.05-x\right)}\\ 1.8\times {10}^{-5}=\frac{x^2}{\left(0.05-x\right)}\end{array}$

Simplify the above equation.

$x=2.9\times {10}^{-4}$

Therefore, the concentration of ${\text{H}}^{+}$ is $2.9\times {10}^{-4}$.

The $\text{pH}$ of the above given hydride ion is calculated as,

$\begin{array}{c}\text{pH}=-\log \left({\text{H}}^{\text{+}}\right)\\ =-\log \left(2.9\times {10}^{-4}\right)\\ =-\log \left(2.9\right)+4\\ =\underset{\_}{3.53}\end{array}$

When $10.0\text{ml}0.10\text{M}\text{NaOH}$ has been added.

The species present in the mixture before the reaction are ${\text{CH}}_{\text{3}}\text{COOH,}{\text{OH}}^{-},{\text{Na}}^{+}$ and ${\text{H}}_{\text{2}}\text{O}$.

The stoichiometric problem part is represented as,

$\begin{array}{cccccccc}& {\text{OH}}^{-}& +& {\text{CH}}_{\text{3}}\text{COOH}& \to & {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}& +& {\text{H}}_{\text{2}}\text{O}\\ \text{Before}\text{reaction}& \begin{array}{l}10\text{ml}\times \text{0}\text{.10}\\ M=\text{1}\text{.0}\text{mmol}\end{array}& & \begin{array}{l}50.0\text{ml}\times \text{0}\text{.10}\text{M}\\ =\text{5}\text{.0}\text{mmol}\end{array}& & 0& & -\\ \text{After}\text{recation}& 0& & 5.0-1.0=4.0\text{mmol}& & 1.0\text{mmol}& & -\end{array}$

The equilibrium problem part is given below.

Acetic acid is much stronger acid than water and ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}$ is the conjugate base of ${\text{CH}}_{\text{3}}\text{COOH}$. The $\text{pH}$ of the acetic acid is determined as,

$\begin{array}{l}{\text{CH}}_3\text{COOH}\left(aq\right)\rightleftharpoons {\text{CH}}_3{\text{COO}}^{-}\left(aq\right)+{\text{H}}^{+}\left(aq\right)\\ K_{\text{a}}=\frac{\left[H^{+}\right]\left[{\text{CH}}_3{\text{COO}}^{-}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}\end{array}$ (1)

Where,

• $K_{\text{a}}$ is the acid dissociation constant.
• $\left[{\text{H}}^{\text{+}}\right]$ is the concentration of Hydrogen ion.
• $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is the concentration of ${\text{CH}}_3{\text{COO}}^{-}$.
• $\left[{\text{CH}}_3\text{COOH}\right]$ is the concentration of ${\text{CH}}_3\text{COOH}$.

The concentration of $\left[{\text{CH}}_3\text{COOH}\right]$ is calculated as,

$\begin{array}{c}{\left[{\text{CH}}_3\text{COOH}\right]}_0=\frac{4.0\text{mmol}}{\left(50.0+10.0\right)\text{ml}}\\ {\left[{\text{CH}}_3\text{COOH}\right]}_0=\frac{4.0}{60.0}\\ {\left[{\text{CH}}_3\text{COOH}\right]}_0=0.067\end{array}$

The concentration of $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is calculated as,

$\begin{array}{c}{\left[{\text{CH}}_3{\text{COO}}^{-}\right]}_0=\frac{1.0\text{mmol}}{\left(50.0+10.0\right)\text{ml}}\\ {\left[{\text{CH}}_3{\text{COO}}^{-}\right]}_0=\frac{1.0}{60.0}\\ {\left[{\text{CH}}_3{\text{COO}}^{-}\right]}_0=0.0167\end{array}$

The concentration of $\left[{\text{H}}^{+}\right]$ is approximately zero.

The ICE-table for the equilibrium reaction is given as,

$\begin{array}{cccccc}& {\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)& \rightleftharpoons & {\text{H}}^{\text{+}}& +& {\text{CH}}_{\text{3}}{\text{COO}}^{-}\\ \text{Initial}\text{concentration}& 0.067& & 0& & 0.0167\\ \text{Change}\text{concentration}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.067-x& & x& & 0.0167+x\end{array}$

Substitute the values of concentrations in the equation (1).

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[H^{+}\right]\left[{\text{CH}}_3{\text{COO}}^{-}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}\\ 1.8\times {10}^{-5}=\frac{x\left(0.0167+x\right)}{\left(0.067-x\right)}\end{array}$

Neglect the values of $x$, the above equation becomes,

$\begin{array}{c}1.8\times {10}^{-5}=\frac{x\left(0.0167\right)}{\left(0.067\right)}\\ x=\frac{0.12\times {10}^{-5}}{0.0167}\\ x=7.2\times {10}^{-5}\end{array}$

Therefore, the concentration of ${\text{H}}^{+}$ is $7.2\times {10}^{-5}$.

The $\text{pH}$ of the above given hydride ion is calculated as,

$\begin{array}{c}\text{pH}=-\log \left({\text{H}}^{\text{+}}\right)\\ =-\log \left(7.2\times {10}^{-5}\right)\\ =-\log \left(7.2\right)+5\\ =\underset{\_}{4.14}\end{array}$

When $25.0\text{ml}0.10\text{M}\text{NaOH}$ has been added.

The process is exactly similar as in part (b). Only volume is changed here that is $50.0+25.0=75.0\text{ml}$.

The stoichiometric problem part is represented as,

$\begin{array}{cccccccc}& {\text{OH}}^{\text{-}}& +& {\text{CH}}_{\text{3}}\text{COOH}& \to & {\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}& +& {\text{H}}_{\text{2}}\text{O}\\ \text{Before}\text{reaction}& \begin{array}{l}25\text{ml}\times \text{0}\text{.10}\text{M}\\ =\text{2}\text{.5}\text{mmol}\end{array}& & \begin{array}{l}50.0\text{ml}\times \text{0}\text{.10}\text{M}\\ =\text{2}\text{.5}\text{mmol}\end{array}& & 0& & -\\ \text{After}\text{recation}& 0& & 5.0-2.5=2.5\text{mmol}& & 2.5\text{mmol}& & -\end{array}$

The equilibrium problem part is given below.

Acetic acid is much stronger acid than water and ${\text{CH}}_{\text{3}}{\text{COO}}^{\text{-}}$ is the conjugate base of ${\text{CH}}_{\text{3}}\text{COOH}$. The $\text{pH}$ of the acetic acid is determined as,

The concentration of $\left[{\text{CH}}_3\text{COOH}\right]$ is calculated as,

$\begin{array}{c}{\left[{\text{CH}}_3\text{COOH}\right]}_0=\frac{2.5\text{mmol}}{\left(50.0+25.0\right)\text{ml}}\\ {\left[{\text{CH}}_3\text{COOH}\right]}_0=\frac{2.5}{75.0}\\ {\left[{\text{CH}}_3\text{COOH}\right]}_0=0.033\end{array}$

The concentration of $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is calculated as,

$\begin{array}{c}{\left[{\text{CH}}_3{\text{COO}}^{-}\right]}_0=\frac{2.5\text{mmol}}{\left(50.0+25.0\right)\text{ml}}\\ {\left[{\text{CH}}_3{\text{COO}}^{-}\right]}_0=\frac{2.5}{75.0}\\ {\left[{\text{CH}}_3{\text{COO}}^{-}\right]}_0=0.33\end{array}$

The concentration of $\left[{\text{H}}^{+}\right]$ is approximately zero.

The ICE-table for the equilibrium reaction is given as,

$\begin{array}{cccccc}& {\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)& \rightleftharpoons & {\text{H}}^{\text{+}}& +& {\text{CH}}_{\text{3}}{\text{COO}}^{-}\\ \text{Initial}\text{concentration}& 0.033& & 0& & 0.033\\ \text{Change}\text{concentration}& -x& & +x& & +x\\ \text{Equilibrium}\text{concentration}& 0.033-x& & x& & 0.033+x\end{array}$

Substitute the values of concentrations in the equation (1).

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[H^{+}\right]\left[{\text{CH}}_3{\text{COO}}^{-}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}\\ 1.8\times {10}^{-5}=\frac{x\left(0.033+x\right)}{\left(0.033-x\right)}\end{array}$

Neglect the values of $x$, the above equation becomes,

$\begin{array}{c}1.8\times {10}^{-5}=\frac{x\left(0.033\right)}{\left(0.033\right)}\\ x=1.8\times {10}^{-5}\end{array}$

Therefore, the concentration of ${\text{H}}^{+}$ is $1.8\times {10}^{-5}$.

The $\text{pH}$ of the above given hydride ion is calculated as,

$\begin{array}{c}\text{pH}=-\log \left({\text{H}}^{\text{+}}\right)\\ =-\log \left(1.8\times {10}^{-5}\right)\\ =-\log \left(1.8\right)+5\\ =\underset{\_}{4.74}\end{array}$

This is the special point in the titration because it is the halfway to the equivalence point.

Here the  $\left[{\text{H}}^{+}\right]=K_{\text{a}}$

Therefore, $\text{pH}=\text{p}{K}_{\text{a}}$.

When $50.0\text{ml}$ $\text{0}\text{.10}\text{M}\text{NaOH}$ has been added.

This is the equivalence point of the titration. $5.0\text{mmol}{\text{OH}}^{-}$ has been added, which will just react with the $5.0\text{mmol}{\text{CH}}_3\text{COOH}$ originally present. At this point the major species present are, ${\text{Na}}^{+},{\text{CH}}_{\text{3}}{\text{COO}}^{-}$ and ${\text{H}}_{\text{2}}\text{O}$.

This solution contains ${\text{CH}}_{\text{3}}{\text{COO}}^{-}$ which is a base. Therefore, this base combines with proton that is Hydrogen ion. The only source of protons us water. So, the reaction is as follows,

${\text{CH}}_{\text{3}}{\text{COO}}^{-}\left(aq\right)+{\text{H}}_{\text{2}}\text{O}\left(l\right)\to {\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)$.

The $K_{\text{b}}$ for this reaction is written as,

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]}\\ K_{\text{b}}=\frac{K_{\text{w}}}{K_{\text{a}}}\\ K_{\text{b}}=\frac{1.0\times {10}^{-14}}{1.8\times {10}^{-5}}\\ K_{\text{b}}=5.6\times {10}^{-10}\end{array}$

The concentration of $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is calculated as,

$\begin{array}{c}{\left[{\text{CH}}_3{\text{COO}}^{-}\right]}_0=\frac{5.0\text{mmol}}{\left(50.0+50.0\right)\text{ml}}\\ {\left[{\text{CH}}_3{\text{COO}}^{-}\right]}_0=\frac{5.0}{100.0}\\ {\left[{\text{CH}}_3{\text{COO}}^{-}\right]}_0=0.050\end{array}$

The concentration of $\left[{\text{H}}^{+}\right]$ is approximately zero.

The concentration of $\left[{\text{CH}}_3\text{COOH}\right]$ is also approximately is zero.

The corresponding ICE-table is given as,

$\begin{array}{cccccccc}& {\text{CH}}_{\text{3}}{\text{COO}}^{-}\left(aq\right)& +& {\text{H}}_{\text{2}}\text{O}\left(l\right)& \to & {\text{CH}}_{\text{3}}\text{COOH}\left(aq\right)& +& {\text{OH}}^{-}\left(aq\right)\\ \text{Initial}& 0.050& & -& & 0& & 0\\ \text{Change}& -x& & -& & +x& & +x\\ \text{Equilibrium}& 0.050-x& & -& & x& & x\end{array}$

Therefore, the equilibrium constant is written as,

$\begin{array}{l}{K}_{\text{b}}=\frac{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]}\\ 5.6\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.050-x\right)}\end{array}$

The approximation is valid by the $5\%$ rule. So,

$\begin{array}{c}5.6\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.050\right)}\\ x^2=28.0\times {10}^{-12}\\ x=\sqrt{28.0\times {10}^{-12}}\\ x=5.3\times {10}^{-6}\end{array}$

Therefore, the $\left[{\text{OH}}^{-}\right]$ concentration in terms of $K_{\text{w}}$ is written as,

$\begin{array}{c}\left[{\text{OH}}^{-}\right]\left[{\text{H}}^{+}\right]=K_{\text{w}}\\ \left(5.3\times {10}^{-6}\right)\left[{\text{H}}^{+}\right]=1.0\times {10}^{-14}\\ \left[{\text{H}}^{+}\right]=\frac{1.0\times {10}^{-14}}{5.3\times {10}^{-6}}\\ \left[{\text{H}}^{+}\right]=1.9\times {10}^{-9}\end{array}$

The $\text{pH}$ of the above given hydride ion is calculated as,

$\begin{array}{c}\text{pH}=-\log \left({\text{H}}^{\text{+}}\right)\\ =-\log \left(1.9\times {10}^{-9}\right)\\ =-\log \left(1.9\right)+9\\ =\underset{\_}{8.72}\end{array}$

The $\text{pH}$ at the equivalence point of a titration of a weak acid with a strong base is always greater than $7$ the reason is that the anion part of the acid is remains in the solution, as all of the acid has been converted to its conjugated base by the addition of strong acid.

Yes, from the titration curve it is clear that the $\text{pH}$ at the halfway point to the equivalence is less than $7$.

The value of $\text{pH}$ at the halfway point of titration is equal to $\text{p}{K}_{\text{a}}$. Mathematically it is given as,

$\begin{array}{l}\text{pH}=\text{p}{K}_{\text{a}}\\ \text{pH}=-\log \left(K_{\text{a}}\right)\end{array}$

The comparison of the titration curve for strong acid-strong base and weak acid-strong base titration is given as,

• The shape of both the curves is same after the equivalence point but before the equivalence point it is different.
• In the beginning of the titration of the weak acid-strong base, the $\text{pH}$ increases sharply than the strong acid-base titration.
• The equivalence point for the strong acid-base titration occur at the $\text{pH}$ $7$.while for the weak acid-strong base titration the equivalence point occur at the $\text{pH}>7$ because of the high basicity of conjugated base of weak acid.

Conclusion

Conclusion

All the questions based on titration and titration curve have been rightfully stated