Chapter 17, Problem 35QP

A 25.0-mL solution of 0.100 M ${\text{CH}}_{\text{3}}\text{COOH}$ is titrated with a 0.200 M KOH solution. Calculate the pH after the following additions of the KOH solution: (a) 0.0 mL, (b) 5.0 mL (c) 10.0 mL, (d) 12.5 mL, (e) 15.0 mL.

Interpretation Introduction

Interpretation:

The pH values of the solution after the addition of different volumes of $\text{KOH}$ are to be determined with given volume of solution and concentration of ${\text{CH}}_3\text{COOH}$ and $\text{KOH}$ solution.

Concept introduction:

Acid–base titration is a technique used to analyze the unknown concentration of the acid or the base through the known concentration of the acid and base.

The equivalence point is the point in the acid–base titration of the chemical reaction where the number of moles of the titrant and the unknown concentration of the analyte are equal. It is used to identify the unknown concentration of the analyte.

The value of pH expresses the acidity or alkalinity of a solution on a logarithmic scale.

When the value of pH is equal to $7$, the solution is neutral; a solution with pH less than 7 is acidic and greater than 7 is basic or alkaline.

The formula to calculate the pH of a solution is:

$\text{pH}=-\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=-\text{log}\left[{\text{H}}^{+}\right]$ as $\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=\left[{\text{H}}^{+}\right]$

pOH is a measure of the basicity of a solution, which depends on the concentration of hydroxide ions and the temperature of the solution.

The formula to calculate the pOH of a solution is:

$\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$

The relation between pH and pOH is as:

$\text{pH}+\text{pOH}=14$

A base dissociation constant, K_b, (also known base-ionization constant) is a quantitative measure of the strength of an base in solution.

The ionization of a weak base can be represented by the equation: ${\text{B + H}}_2\text{O }\rightleftharpoons {\text{ HB}}^{+}{\text{ + OH}}^{-}$

Where B is weak base and ${\text{HB}}^{+}$ is conjugated acid.

The base ionization constant for a weak base is:

$K_b=\frac{\left[{\text{HB}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{B}\right]}$

Where $K_b$ is base ionization constant or equilibrium constant.

An acid dissociation constant, K_a, (also known as acidity constant, or acid-ionization constant) is a quantitative measure of the strength of an acid in solution

The dissociation of a weak acid in an aqueous solution:

$HA\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons H_3{O}^{+}\left(aq\right)+A^{-}\left(aq\right)$

The acid ionization constant for a weak acid is:

$K_a=\frac{[{\text{H}}_3{\text{O}}^{+}][A^{-}]}{[\text{HA}]}$

The number of moles of compound initially present in the solution is calculated as:

$\text{Mole}=\text{Concentration}\times \text{Volume}$

The molarity of a compound is given by the expression as:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}$

The pH of the buffer solution by the Henderson–Hasselbalch equation can be calculated as: $\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{[\text{Conjugate base}]}{[\text{weak acid}]}$

Answer to Problem 35QP

Solution:

a)

$\text{pH}=2.87$

b)

$\text{pH}=4.56$

c)

$\text{pH}=5.34$

d)

$\text{pH}=8.78$

e)

$\text{pH}=12.10$

Explanation of Solution

Given information: The concentration of ${\text{CH}}_3\text{COOH}$ is $0.1\text{ M}$, the concentration of $\text{KOH}$ is $0.2\text{ M}$ and the volume of the solution is $\text{25}\text{mL}$.

Explanation:

a) $\text{0}\text{.0}\text{mL}$

The neutralization reaction between ${\text{CH}}_{\text{3}}\text{COOH}$ and $\text{KOH}$ is as follows:

${\text{CH}}_3\text{COOH}\left(aq\right)+\text{KOH}\left(aq\right)\rightleftharpoons {\text{CH}}_3\text{COOK}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)$

From the above equation, 1 mole of ${\text{CH}}_3\text{COOH}$ reacts with 1 mole of $\text{KOH}$.

Let $x$ be the degree of dissociation.

The initial ionization change table for the ionization of ${\text{CH}}_{\text{3}}\text{COOH}$ is as follows:

$\begin{array}{l}{\text{CH}}_3\text{COOH}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{H}}_3{\text{O}}^{+}\left(\text{aq}\right)+{\text{CH}}_3{\text{COO}}^{-}\left(aq\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{M}\right)& 0.100& -& \text{}\text{}0& 0\\ \text{Change}\left(\text{M}\right)& -x& -& +x& +x\\ \text{Equilibrium}\left(\text{M}\right)& 0.100-x& -& x& x\end{array}\end{array}$

The equilibrium expression for the reaction is represented as follows:

$K_a=\frac{[{\text{H}}_3{\text{O}}^{+}][A^{-}]}{[\text{HA}]}$

$K_a\text{=}\frac{\left[{\text{CH}}_3{\text{COO}}^{-}\right]\left[{\text{H}}_3{\text{O}}^{+}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}$

Here, $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is the concentration of the acetate ion, $\left[{\text{H}}_3{\text{O}}^{+}\right]$ is the concentration of the hydronium ion, $\left[{\text{CH}}_3\text{COOH}\right]$ is the concentration of acetic acid and $K_{\text{a}}$ is the acid dissociation constant.

Substitutes the values of $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$, $\left[{\text{H}}_3{\text{O}}^{+}\right]$, $\left[{\text{CH}}_3\text{COOH}\right]$ and $K_{\text{a}}$ in the above expression.

$1.8\times {10}^{-5}=\frac{\left[x\right]\left[x\right]}{\left[\text{0}\text{.100}-x\right]}$

The value of $x$ is very small as compared to $0.100$ ; hence, it can be neglected.

$\begin{array}{c}1.8\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{\left(0.100\right)}\\ \left(1.8\times {10}^{-5}\times 0.300\right)=x^2\\ x=\sqrt{1.8\times {10}^{-5}\times 0.100}\\ x=1.34\times {10}^{-3}\end{array}$

Concentration of $\left[{\text{H}}_3{\text{O}}^{+}\right]=\text{1}.34\times {10}^{-3}$

The value of $\text{pH}$ is calculated by the expression as follows:

$\text{pH}=-\text{log}\left[{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}\right]=-\text{log}\left[{\text{H}}^{+}\right]$

$\text{pH}=-\log \left(\left[{\text{H}}_3{\text{O}}^{+}\right]\right)$

Substitute the value of $\left[{\text{H}}_3{\text{O}}^{+}\right]$.

$\begin{array}{l}\text{pH}=-\log \left(1.34\times {10}^{-3}\right)\\ =2.87\end{array}$

Hence, the $\text{pH}$ value at the equivalence point is $2.87$.

Explanation:

b) $\text{5}\text{.0}\text{mL}$

The number of moles of ${\text{CH}}_3\text{COOH}$ initially present in $\text{10}\text{mL}$ of the solution is calculated as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$

Substitute the values of the concentration and the volume of ${\text{CH}}_3\text{COOH}$ in the above expression.

$\begin{array}{c}\text{Moles of}{\text{CH}}_3\text{COOH}=\left(\text{25mL}\right)\left(\frac{\text{0}\text{.100}{\text{mol CH}}_3\text{COOH}}{{\text{1000 mL CH}}_{\text{3}}\text{COOH solution}}\right)\\ =2.50\times {10}^{-3}\text{mol}\\ \end{array}$

The number of moles of $\text{KOH}$ initially present in $5\text{mL}$ of the solution is calculated as:

$\text{Mole}=\text{Concentration}\times \text{Volume}$

Substitute the values of the concentration and the volume of $\text{KOH}$ in the above expression.

$\begin{array}{l}\text{Moles}\text{of}\text{KOH}=\text{5}\text{mL}\times \left(\frac{0.200\text{mol KOH}}{1000\text{mL KOH solution}}\right)\\ =1.00\times {10}^{-3}\text{mol}\end{array}$

On mixing the two solutions, the molarity of the solution changes, but the number of moles will remain the same.

The changes in the number of moles of ${\text{CH}}_3\text{COOH}$ and $\text{KOH}$ at equilibrium are as follows:

$\begin{array}{l}{\text{CH}}_3\text{COOH}\left(aq\right)+\text{KOH}\left(aq\right)\rightleftharpoons {\text{CH}}_3\text{COOK}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{mole}\right)& 2.50\times {10}^{-3}& \text{}1.00\times {10}^{-3}& 0& -\\ \text{Change}\left(\text{mole}\right)& -1.00\times {10}^{-3}& -1.00\times {10}^{-3}& +1.00\times {10}^{-3}& -\\ \text{Equilibrium}\left(\text{mole}\right)& 1.50\times {10}^{-3}& 0& 1.00\times {10}^{-3}& -\end{array}\end{array}$

Determine the pH of the buffer solution by the Henderson–Hasselbalch equation. This is a basic buffer.

$K_{\text{a}}\text{of}{\text{CH}}_{\text{3}}\text{COOH}=1.8\times {10}^{-5}$

The pH of the buffer is calculated as follows:

$\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{[\text{Conjugate base}]}{[\text{weak acid}]}$

$\begin{array}{l}\text{pH}=\text{p}{K}_{\text{a}}+\log \frac{[{\text{CH}}_3\text{COOK}]}{[{\text{CH}}_3\text{COOH}]}\\ =-\log \left(1.8\times {10}^{-5}\right)+\log \frac{\left[1.0\times {10}^{-3}\right]}{\left[1.50\times {10}^{-3}\right]}\\ =4.56\end{array}$

Hence, the $\text{pH}$ value at the equivalence point is $4.56$.

Explanation:

c) $\text{10}\text{.0}\text{mL}$

The number of moles of ${\text{CH}}_3\text{COOH}$ initially present in $\text{10}\text{mL}$ of the solution is calculated as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$

Substitute the values of the concentration and the volume of ${\text{CH}}_3\text{COOH}$ in the above expression.

$\begin{array}{c}\text{Moles of}{\text{CH}}_3\text{COOH}=\left(\text{25 mL×}\frac{\text{0}\text{.100}{\text{mol CH}}_3\text{COOH}}{{\text{1000 mL CH}}_{\text{3}}\text{COOH solution}}\right)\\ =2.50\times {10}^{-3}\text{mol}\\ \end{array}$

The number of moles of $\text{KOH}$ initially present in $10\text{mL}$ of the solution is calculated as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$

Substitute the values of the concentration and the volume of $\text{KOH}$ in the above expression.

$\begin{array}{c}\text{Moles}\text{of}\text{KOH}=\text{10}\text{mL}\times \left(\frac{0.200\text{mol KOH}}{1000\text{mL KOH solution}}\right)\\ =\text{2}.00\times {10}^{-3}\text{mol}\end{array}$

On mixing the two solutions, the molarity of the solution changes, but the number of moles will remain the same.

The changes in the number of moles of ${\text{CH}}_3\text{COOH}$ and $\text{KOH}$ at equilibrium are as follows:

$\begin{array}{l}{\text{CH}}_3\text{COOH}\left(aq\right)+\text{KOH}\left(aq\right)\rightleftharpoons {\text{CH}}_3\text{COOK}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{mole}\right)& 2.50\times {10}^{-3}& \text{}2.00\times {10}^{-3}& 0& -\\ \text{Change}\left(\text{mole}\right)& -2.00\times {10}^{-3}& -2.00\times {10}^{-3}& +2.00\times {10}^{-3}& -\\ \text{Equilibrium}\left(\text{mole}\right)& 0.50\times {10}^{-3}& 0& 2.00\times {10}^{-3}& -\end{array}\end{array}$

Determine the pH of the buffer solution by the Henderson–Hasselbalch equation.

The pH of the buffer is calculated as follows:

$\text{pH}={\text{pK}}_{\text{a}}+\log \frac{[\text{Conjugate base}]}{[\text{weak acid}]}$

$\begin{array}{l}\text{pH}={\text{pK}}_{\text{a}}+\log \frac{[{\text{CH}}_3\text{COOK}]}{[{\text{CH}}_3\text{COOH}]}\\ =-\log \left(1.8\times {10}^{-5}\right)+\log \frac{\left[2.0\times {10}^{-3}\right]}{\left[0.50\times {10}^{-3}\right]}\\ =5.34\end{array}$

Hence, the $\text{pH}$ value at the equivalence point is $5.34$.

Explanation:

d) $\text{15}\text{.0}\text{mL}$

The number of moles of ${\text{CH}}_3\text{COOH}$ initially present in $\text{10}\text{mL}$ of the solution is calculated as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$

Substitute the values of the concentration and the volume of ${\text{CH}}_3\text{COOH}$ in the above expression.

$\begin{array}{c}\text{Moles of}{\text{CH}}_3\text{COOH}=\left(\text{25 mL}\right)\text{×}\left(\frac{\text{0}\text{.100}{\text{mol CH}}_3\text{COOH}}{{\text{1000mL CH}}_{\text{3}}\text{COOH solution}}\right)\\ =2.50\times {10}^{-3}\text{mol}\\ \end{array}$

The number of moles of $\text{KOH}$ initially present in $12.5\text{mL}$ of the solution is calculated as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$

Substitute the values of the concentration and the volume of $\text{KOH}$ in the above expression.

$\begin{array}{c}\text{Moles}\text{of}\text{KOH}=\text{12}\text{.5}\text{mL}\times \left(\frac{0.200\text{mol KOH}}{1000\text{mL KOH solution}}\right)\\ =\text{2}.50\times {10}^{-3}\text{mol}\end{array}$

On mixing the two solutions, the molarity of the solution changes, but the number of moles will remain the same.

The changes in the number of moles of ${\text{CH}}_3\text{COOH}$ and $\text{KOH}$ at equilibrium are as:

$\begin{array}{l}{\text{CH}}_3\text{COOH}\left(aq\right)+\text{KOH}\left(aq\right)\rightleftharpoons {\text{CH}}_3\text{COOK}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{mole}\right)& 2.50\times {10}^{-3}& \text{}2.5\times {10}^{-3}& 0& -\\ \text{Change}\left(\text{mole}\right)& -2.50\times {10}^{-3}& -2.5\times {10}^{-3}& +2.5\times {10}^{-3}& -\\ \text{Equilibrium}\left(\text{mole}\right)& 0& 0& 2.5\times {10}^{-3}& -\end{array}\end{array}$

At this stage, an equivalence point of the titration is reached. $2.50\times {10}^{-3}$ moles of ${\text{CH}}_3\text{COOH}$ and $2.50\times {10}^{-3}$ moles of $\text{KOH}$ produce $2.50\times {10}^{-3}$ moles of ${\text{CH}}_3\text{COOH}$. At the equivalence point, only salt is present, which is the conjugate base of ${\text{CH}}_3\text{COOH}$.

The molarity of ${\text{CH}}_3{\text{COO}}^{-}$ is given by the expression as follows:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}$

$\begin{array}{c}\text{Total}\text{Volume}=\left(25\text{mL}+12.5\text{mL}\right)\\ =\text{37}\text{.5 mL}\\ =0.0375\text{L}\end{array}$

Substitute the values of the number of moles and the volume of ${\text{CH}}_3\text{COOH}$ in the above expression,

$\begin{array}{l}\text{Molarity}=\left(\frac{2.50\times {10}^{-3}\text{mol}}{0.0375\text{L}}\right)\\ =0.0667\text{M}\end{array}$

Let $x$ be the degree of dissociation.

The initial concentration change for the ionization of ${\text{CH}}_3{\text{COO}}^{-}$ is as follows:

$\begin{array}{l}{\text{CH}}_3{\text{COO}}^{-}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\rightleftharpoons {\text{CH}}_3\text{COOH}\left(aq\right)+{\text{OH}}^{-}\left(aq\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{M}\right)& 0.0667& -& 0& 0\\ \text{Change}\left(\text{M}\right)& -x& -& +x& +x\\ \text{Equilibrium}\left(\text{M}\right)& 0.0667-x& -& x& x\end{array}\end{array}$

The equilibrium expression for the reaction is written as follows:

$K_b=\frac{\left[{\text{HB}}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[\text{B}\right]}$

$K_{\text{b}}=\frac{\left[{\text{OH}}^{-}\right]\left[{\text{CH}}_3\text{COOH}\right]}{\left[{\text{CH}}_3{\text{COO}}^{-}\right]}$

Here, $\left[{\text{OH}}^{-}\right]$ is the concentration of the hydroxide ion, $\left[{\text{CH}}_3\text{COOH}\right]$ is the concentration of acetic acid, $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is the concentration of the acetate ion, and $K_{\text{b}}$ is the base dissociation constant.

Substitute the values of $\left[{\text{OH}}^{-}\right]$, $\left[{\text{CH}}_3\text{COOH}\right]$, $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$, and $K_{\text{b}}$ in the expression,

$\begin{array}{c}5.56\times {10}^{-10}=\frac{\left[x\right]\left[x\right]}{\left[\text{0}\text{.0667}-x\right]}\\ 5.56\times {10}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.0667\right)}\\ \left(5.56\times {10}^{-10}\times 0.0667\right)=x^2\end{array}$

The value of x is very small as compared to $0.0667$; hence, it can be neglected.

$\begin{array}{l}x=\sqrt{5.56\times {10}^{-10}\times 0.0667}\\ x=6.09\times {10}^{-6}\end{array}$

Concentration of $\left[{\text{OH}}^{-}\right]=x=\text{ 6}.09\times {10}^{-6}$

The $\text{pOH}$ is calculated by using the expression as follows:

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

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{c}\text{pOH}=-\log \left(6.09\times {10}^{-6}\right)\\ =5.22\end{array}$

The value of $\text{pH}$ is calculated using the expression as follows:

$\begin{array}{l}\text{pH}+\text{pOH}=14\\ \text{pH}=14-\text{pOH}\\ \text{}\text{}\text{}\text{}\text{}\text{}=14-5.22\\ =8.78\end{array}$

Hence, the $\text{pH}$ value at the equivalence point is $8.78$.

e)

Given information: The concentration of ${\text{CH}}_3\text{COOH}$ is $0.1\text{ M}$, the concentration of $\text{KOH}$ is $0.2\text{ M}$, the volume of the solution is $\text{25}\text{mL}$, and $15\text{mL of KOH}$ is added to the solution.

Explanation:

The number of moles of ${\text{CH}}_3\text{COOH}$ initially present in $\text{10}\text{mL}$ of the solution is calculated as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$

Substitute the values of the concentration and the volume of ${\text{CH}}_3\text{COOH}$ in the above expression,

$\begin{array}{c}\text{Moles of}{\text{CH}}_3\text{COOH}=\left(\text{25mL×}\frac{\text{0}\text{.100}{\text{mol CH}}_3\text{COOH}}{{\text{1000mL CH}}_{\text{3}}\text{COOH solution}}\right)\\ =2.50\times {10}^{-3}\text{mol}\\ \end{array}$

The number of moles of $\text{KOH}$ initially present in $15\text{mL}$ of the solution is calculated as follows:

$\text{Mole}=\text{Concentration}\times \text{Volume}$

Substitute the values of the concentration and the volume of $\text{KOH}$ in the above expression.

$\begin{array}{l}\text{Moles}\text{of}\text{KOH}=\text{15}\text{mL}\times \left(\frac{0.200\text{mol KOH}}{1000\text{mL KOH solution}}\right)\\ =\text{3}.00\times {10}^{-3}\text{mol}\end{array}$

On mixing the two solutions, the molarity of the solution changes, but the number of moles will remain the same.

The changes in the number of moles of ${\text{CH}}_3\text{COOH}$ and $\text{KOH}$ at equilibrium are as follows:

$\begin{array}{l}{\text{CH}}_3\text{COOH}\left(aq\right)+\text{KOH}\left(aq\right)\rightleftharpoons {\text{CH}}_3\text{COOK}\left(aq\right)+{\text{H}}_2\text{O}\left(l\right)\\ \begin{array}{ccccc}\text{Initial}\left(\text{mole}\right)& 2.50\times {10}^{-3}& \text{}3.00\times {10}^{-3}& 0& -\\ \text{Change}\left(\text{mole}\right)& -2.50\times {10}^{-3}& -2.50\times {10}^{-3}& +2.50\times {10}^{-3}& -\\ \text{Equilibrium}\left(\text{mole}\right)& 0& 0.5\times {10}^{-3}& 2.50\times {10}^{-3}& -\end{array}\end{array}$

The molarity of $\text{KOH}$ is as follows:

$\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}$

$\begin{array}{l}\text{Total}\text{Volume}=40\text{mL}\\ \text{=}\text{0}\text{.0400}\text{L}\end{array}$

Substitute the values of the number of moles and the volume ${\text{NH}}_4^{+}$ in the above expression.

$\begin{array}{c}\text{Molarity}=\left(\frac{0.50\times {10}^{-3}\text{mol}}{0.0400\text{L}}\right)\\ =0.0125\text{M}\end{array}$

$\text{KOH}$ is a strong base. It dissociates completely in water.

The concentration of $\left[{\text{OH}}^{-}\right]=0.0125\text{M}$

The $\text{pOH}$ is calculated using the expression as follows:

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

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{l}\text{pOH}=-\log \left(0.0125\right)\\ =1.90\end{array}$

The value of $\text{pH}$ is calculated using the expression as follows:

$\text{pH}+\text{pOH}=14$

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in the above expression.

$\begin{array}{c}\text{pH}=\left(14-\text{1}\text{.90}\right)\\ =12.10\end{array}$

Hence, the $\text{pH}$ value at the equivalence point is $12.10$.