Chapter 15, Problem 59E

Consider the titration of 100.0 mL of 0.200 M acetic acid (K_a = 1.8 × 10⁻⁵) by 0.100 M KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added.

a. 0.0 mL

b. 50.0 mL

c. 100.0 mL

d. 150.0 mL

e. 200.0 mL

f. 250.0 mL

Interpretation Introduction

Interpretation: The titration of Acetic acid with different volumes of $\text{KOH}$ is given. The $\text{pH}$ of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of $\text{pH}$ of solution when $0.0\text{}\text{mL}$ $\text{KOH}$ has been added to it.

Answer to Problem 59E

The value of $\text{pH}$ of solution when $0.0\text{}\text{mL}$ $\text{KOH}$ has been added is $\underset{\_}{2.72}$.

Explanation of Solution

Explanation

The concentration of ${\text{H}}^{+}$.is $\underset{\_}{0.00189\text{}M}$.

Given:

The concentration of Acetic acid is $0.200\text{M}$.

The concentration of $\text{KOH}$ is $0.100\text{M}$.

The volume of Acetic acid is $100.0\text{mL}$.

The volume of $\text{KOH}$ is $0.0\text{mL}$.

The value of $K_{\text{a}}$ of Acetic acid is $1.8\times {10}^{-5}$

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $100\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{100}\text{mL}=100\times 0.001\text{L}\\ =\text{0}\text{.100}\text{L}\end{array}$

The concentration of any species is given as,

$\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}$ (1)

Rearrange the above equation to obtain the value of number of moles.

$\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}$ (2)

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

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.200}\text{M}\times \text{0}\text{.1}\text{L}\\ =\text{0}\text{.02}\text{moles}\end{array}$

Make the ICE table for the reaction between ${\text{CH}}_3\text{COOH}$ and $\text{KOH}$.

$\begin{array}{cccccccc}& {\text{CH}}_3\text{COOH}& +& \text{KOH}& \stackrel{}{\rightleftharpoons }& {\text{CH}}_3{\text{COO}}^{-}{\text{K}}^{+}& +& {\text{H}}_2\text{O}\\ \text{Initial moles:}& 0.02& & 0& & 0& & \\ \text{Change}:& -0& & -0& & +0.02& & \\ \text{Final}\text{moles}:& 0.02& & 0& & 0.02& & \end{array}$

The above equation shows the presence of equilibrium condition in the solution.

Make the ICE table for the dissociation reaction of ${\text{CH}}_{\text{3}}\text{COOH}$.

$\begin{array}{ccccc}& {\text{CH}}_3\text{COOH}& \stackrel{}{\rightleftharpoons }& {\text{CH}}_3{\text{COO}}^{-}& {\text{H}}^{+}\\ \text{Initial}\text{(M)}:& 0.2& & 0& 0\\ \text{Change}\left(\text{M}\right):& -x& & x& x\\ \text{Equilibrium}\left(\text{M}\right):& 0.2-x& & x& x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(x\right)\left(x\right)}{\left(0.2-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{a}}$ is very small, hence, $\left(0.2-x\right)$ is taken as $\left(0.2\right)$.

Simplify the above equation,

$\begin{array}{c}1.8\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{\left(0.2-x\right)\text{M}}\\ 1.8\times {10}^{-5}=\frac{\left(x\right)\left(x\right)}{0.2\text{M}}\\ x=\underset{\_}{0.00189\text{}M}\end{array}$

It is the concentration of ${\text{H}}^{+}$.

Explanation

The value of $\text{pH}$ of solution when $0.0\text{}\text{mL}$ $\text{KOH}$ has been added is $\underset{\_}{2.72}$.

The $\text{pH}$ of the solution is calculated by the formula,

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

Where,

• $\left[{\text{H}}^{+}\right]$ is the concentration of Hydrogen ions.

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above equation.

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(0.00189\right)\\ =\underset{\_}{2.72}\end{array}$

The value of $\text{pH}$ of solution when $0.0\text{}\text{mL}$ $\text{KOH}$ has been added is. $\underset{\_}{2.72}$.

Interpretation Introduction

Interpretation: The titration of Acetic acid with different volumes of $\text{KOH}$ is given. The $\text{pH}$ of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of $\text{pH}$ of solution when $50.0\text{}\text{mL}$ $\text{KOH}$ has been added to it.

Answer to Problem 59E

The value of $\text{pH}$ of solution when $50.0\text{}\text{mL}$ $\text{KOH}$ has been added is. $\underset{\_}{4.26}$.

Explanation of Solution

Explanation

The concentration of $\left[{\text{H}}^{\text{+}}\right]$ is. $\underset{\_}{5.5\times 10^{-5}\text{}M}$.

Given

The volume of $\text{KOH}$ is $50.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $50\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{50}\text{mL}=50\times 0.001\text{L}\\ =\text{0}\text{.05}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{KOH}$ in equation (2) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.05\text{L}\\ =\text{0}\text{.005}\text{moles}\end{array}$

Make the ICE table for the reaction between ${\text{CH}}_3\text{COOH}$ and $\text{KOH}$.

$\begin{array}{cccccccc}& {\text{CH}}_3\text{COOH}& +& \text{KOH}& \stackrel{}{\rightleftharpoons }& {\text{CH}}_3{\text{COO}}^{-}{\text{K}}^{+}& +& {\text{H}}_2\text{O}\\ \text{Initial moles:}& 0.02& & 0.005& & 0& & \\ \text{Change}:& -0.005& & -0.005& & +0.005& & \\ \text{Final}\text{moles}:& 0.015& & 0& & 0.005& & \end{array}$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{CH}}_3\text{COOH}+\text{Volume}\text{of}\text{KOH}\\ =\text{0}\text{.1}\text{L}+0.05\text{L}\\ =\text{0}\text{.15}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{CH}}_3\text{COOH}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.015\text{moles}}{0.15\text{}\text{L}}\\ =0.1\text{M}\end{array}$

Substitute the value of number of moles of ${\text{CH}}_3{\text{COO}}^{-}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.005\text{moles}}{0.15\text{}\text{L}}\\ =0.033\text{M}\end{array}$

Make the ICE table for the dissociation reaction of ${\text{CH}}_{\text{3}}\text{COOH}$.

$\begin{array}{ccccc}& {\text{CH}}_3\text{COOH}& \stackrel{}{\rightleftharpoons }& {\text{CH}}_3{\text{COO}}^{-}& {\text{H}}^{+}\\ \text{Initial}\text{(M)}:& 0.1& & 0.033& 0\\ \text{Chang}\left(\text{M}\right):& -x& & +x& x\\ \text{Equilibrium}\left(\text{M}\right):& 0.1-x& & 0.033+x& x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.033+x\right)\left(x\right)}{\left(0.1-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{a}}$ is very small, hence, $\left(0.1-x\right)$ is taken as $\left(0.1\right)$ and $\left(0.033+x\right)$ is taken as $0.033$.

Simplify the above equation,

$\begin{array}{c}1.8\times {10}^{-5}=\frac{\left(0.033\right)\left(x\right)}{\left(0.1\right)\text{M}}\\ 1.8\times {10}^{-6}=\left(0.033\right)\left(x\right)\\ x=\underset{\_}{5.5\times 10^{-5}\text{}M}\end{array}$

It is the concentration of ${\text{H}}^{+}$.

Explanation

The value of $\text{pH}$ of solution when $50.0\text{}\text{mL}$ $\text{KOH}$ has been added is. $\underset{\_}{4.26}$.

The $\text{pH}$ of the solution is calculated by the formula,

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

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above equation.

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(5.5\times {10}^{-5}\right)\\ =\underset{\_}{4.26}\end{array}$

The value of $\text{pH}$ of solution when $50.0\text{}\text{mL}$ $\text{KOH}$ has been added is. $\underset{\_}{4.26}$.

Interpretation Introduction

Interpretation: The titration of Acetic acid with different volumes of $\text{KOH}$ is given. The $\text{pH}$ of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of $\text{pH}$ of solution when $100.0\text{}\text{mL}$ $\text{KOH}$ has been added to it.

Answer to Problem 59E

The value of $\text{pH}$ of solution when $100.0\text{}\text{mL}$ $\text{KOH}$ has been added is. $\underset{\_}{4.74}$.

Explanation of Solution

Explanation

The $\left[{\text{H}}^{\text{+}}\right]$ is. $\underset{\_}{1.8\times 10^{-5}\text{}M}$.

Given

The volume of $\text{KOH}$ is $100.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $100\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{100}\text{mL}=100\times 0.001\text{L}\\ =\text{0}\text{.1}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{KOH}$ in equation (2) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.1\text{L}\\ =\text{0}\text{.01}\text{moles}\end{array}$

Make the ICE table for the reaction between ${\text{CH}}_3\text{COOH}$ and $\text{KOH}$.

$\begin{array}{cccccccc}& {\text{CH}}_3\text{COOH}& +& \text{KOH}& \stackrel{}{\rightleftharpoons }& {\text{CH}}_3{\text{COO}}^{-}{\text{K}}^{+}& +& {\text{H}}_2\text{O}\\ \text{Initial moles:}& 0.02& & 0.01& & 0& & \\ \text{Change}:& -0.01& & -0.01& & +0.01& & \\ \text{Final}\text{moles}:& 0.01& & 0& & 0.01& & \end{array}$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{CH}}_3\text{COOH}+\text{Volume}\text{of}\text{KOH}\\ =\text{0}\text{.1}\text{L}+0.1\text{L}\\ =\text{0}\text{.2}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{CH}}_3\text{COOH}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.01\text{moles}}{0.2\text{}\text{L}}\\ =0.05\text{M}\end{array}$

Substitute the value of number of moles of ${\text{CH}}_3{\text{COO}}^{-}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.01\text{moles}}{0.2\text{}\text{L}}\\ =0.05\text{M}\end{array}$

Make the ICE table for the dissociation reaction of ${\text{CH}}_{\text{3}}\text{COOH}$.

$\begin{array}{ccccc}& {\text{CH}}_3\text{COOH}& \stackrel{}{\rightleftharpoons }& {\text{CH}}_3{\text{COO}}^{-}& {\text{H}}^{+}\\ \text{Initial}\text{(M)}:& 0.05& & 0.05& 0\\ \text{Change}\left(\text{M}\right):& -x& & +x& x\\ \text{Equilibrium}\left(\text{M}\right):& 0.05-x& & 0.05+x& x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.05+x\right)\left(x\right)}{\left(0.05-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{a}}$ is very small, hence, $\left(0.05-x\right)$ is taken as $\left(0.05\right)$ and $\left(0.05+x\right)$ is taken as $0.05$.

Simplify the above equation,

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

It is the concentration of ${\text{H}}^{+}$.

Explanation

The value of $\text{pH}$ of solution when $100.0\text{}\text{mL}$ $\text{KOH}$ has been added is. $\underset{\_}{4.74}$.

The $\text{pH}$ of the solution is calculated by the formula,

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

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above equation.

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

The value of $\text{pH}$ of solution when $100.0\text{}\text{mL}$ $\text{KOH}$ has been added is. $\underset{\_}{4.74}$.

Interpretation Introduction

Interpretation: The titration of Acetic acid with different volumes of $\text{KOH}$ is given. The $\text{pH}$ of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of $\text{pH}$ of solution when $150.0\text{}\text{mL}$ $\text{KOH}$ has been added to it.

Answer to Problem 59E

The value of $\text{pH}$ of solution when $150.0\text{}\text{mL}$ $\text{KOH}$ has been added is $\underset{\_}{5.22}$.

Explanation of Solution

Explanation

The $\left[{\text{H}}^{\text{+}}\right]$ is $\underset{\_}{0.6\times 10^{-5}\text{}M}$.

Given

The volume of $\text{KOH}$ is $150.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $150\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{150}\text{mL}=150\times 0.001\text{L}\\ =\text{0}\text{.15}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{KOH}$ in equation (2) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.15\text{L}\\ =\text{0}\text{.015}\text{moles}\end{array}$

Make the ICE table for the reaction between ${\text{CH}}_3\text{COOH}$ and $\text{KOH}$.

$\begin{array}{cccccccc}& {\text{CH}}_3\text{COOH}& +& \text{KOH}& \stackrel{}{\rightleftharpoons }& {\text{CH}}_3{\text{COO}}^{-}{\text{K}}^{+}& +& {\text{H}}_2\text{O}\\ \text{Initial moles:}& 0.02& & 0.015& & 0& & \\ \text{Change}:& -0.015& & -0.015& & +0.015& & \\ \text{Final}\text{moles}:& 0.005& & 0& & 0.015& & \end{array}$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{CH}}_3\text{COOH}+\text{Volume}\text{of}\text{KOH}\\ =\text{0}\text{.1}\text{L}+0.15\text{L}\\ =\text{0}\text{.25}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{CH}}_3\text{COOH}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.005\text{moles}}{0.25\text{}\text{L}}\\ =0.02\text{M}\end{array}$

Substitute the value of number of moles of ${\text{CH}}_3{\text{COO}}^{-}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.015\text{moles}}{0.25\text{}\text{L}}\\ =0.06\text{M}\end{array}$

Make the ICE table for the dissociation reaction of ${\text{CH}}_{\text{3}}\text{COOH}$.

$\begin{array}{ccccc}& {\text{CH}}_3\text{COOH}& \stackrel{}{\rightleftharpoons }& {\text{CH}}_3{\text{COO}}^{-}& {\text{H}}^{+}\\ \text{Initial}\text{(M)}:& 0.02& & 0.06& 0\\ \text{Change}\left(\text{M}\right):& -x& & +x& x\\ \text{Equilibrium}\left(\text{M}\right):& 0.02-x& & 0.06+x& x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{a}}=\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}$

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{a}}=\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}\\ \text{1}\text{.8}\times {\text{10}}^{-5}=\frac{\left(0.06+x\right)\left(x\right)}{\left(0.02-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{a}}$ is very small, hence, $\left(0.02-x\right)$ is taken as $\left(0.02\right)$ and $\left(0.06+x\right)$ is taken as $0.06$.

Simplify the above equation,

$\begin{array}{c}1.8\times {10}^{-5}=\frac{\left(0.06\right)\left(x\right)}{\left(0.02\right)\text{M}}\\ x=\underset{\_}{0.6\times 10^{-5}\text{}M}\end{array}$

It is the concentration of ${\text{H}}^{+}$.

Explanation

The value of $\text{pH}$ of solution when $150.0\text{}\text{mL}$ $\text{KOH}$ has been added is $\underset{\_}{5.22}$.

The $\text{pH}$ of the solution is calculated by the formula,

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

Substitute the value of $\left[{\text{H}}^{+}\right]$ in the above equation.

$\begin{array}{c}\text{pH}=-\log \left[{\text{H}}^{+}\right]\\ =-\log \left(0.6\times {10}^{-5}\right)\\ =\underset{\_}{5.22}\end{array}$

The value of $\text{pH}$ of solution when $150.0\text{}\text{mL}$ $\text{KOH}$ has been added is $\underset{\_}{5.22}$.

Interpretation Introduction

Interpretation: The titration of Acetic acid with different volumes of $\text{KOH}$ is given. The $\text{pH}$ of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of $\text{pH}$ of solution when $200.0\text{}\text{mL}$ $\text{KOH}$ has been added to it.

Answer to Problem 59E

The value of $\text{pH}$ of solution when $200.0\text{}\text{mL}$ $\text{KOH}$ has been added is $\underset{\_}{8.8}$.

Explanation of Solution

Explanation

The $\left[{\text{H}}^{\text{+}}\right]$ is $\underset{\_}{0.61\times 10^{-5}\text{}M}$.

Given

The volume of $\text{KOH}$ is $200.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $200\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{200}\text{mL}=200\times 0.001\text{L}\\ =\text{0}\text{.2}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{KOH}$ in equation (2) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.2\text{L}\\ =\text{0}\text{.02}\text{moles}\end{array}$

Make the ICE table for the reaction between ${\text{CH}}_3\text{COOH}$ and $\text{KOH}$.

$\begin{array}{cccccccc}& {\text{CH}}_3\text{COOH}& +& \text{KOH}& \stackrel{}{\rightleftharpoons }& {\text{CH}}_3{\text{COO}}^{-}{\text{K}}^{+}& +& {\text{H}}_2\text{O}\\ \text{Initial moles:}& 0.02& & 0.02& & 0& & \\ \text{Change}:& -0.02& & -0.02& & +0.02& & \\ \text{Final}\text{moles}:& 0& & 0& & 0.02& & \end{array}$

The above reaction shows the presence of equilibrium in the solution.

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{CH}}_3\text{COOH}+\text{Volume}\text{of}\text{KOH}\\ =\text{0}\text{.1}\text{L}+0.2\text{L}\\ =\text{0}\text{.3}\text{L}\end{array}$

Substitute the value of number of moles of ${\text{CH}}_3{\text{COO}}^{-}{\text{K}}^{+}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.02\text{moles}}{0.3\text{}\text{L}}\\ =0.067\text{M}\end{array}$

The reaction is now written as,

${\text{CH}}_3{\text{COO}}^{-}{\text{K}}^{+}+{\text{H}}_2\text{O}\rightleftharpoons {\text{CH}}_3\text{COOH}+\text{KOH}$

Make the ICE table for the above reaction.

$\begin{array}{cccccccc}& {\text{CH}}_3{\text{COO}}^{-}{\text{K}}^{+}& +& {\text{H}}_2\text{O}& \stackrel{}{\rightleftharpoons }& {\text{CH}}_3\text{COOH}& +& \text{KOH}\\ \text{Initial moles:}& 0.067& & 0& & 0& & 0\\ \text{Change}:& -x& & 0& & +x& & x\\ \text{Final}\text{moles}:& 0.067-x& & 0& & x& & x\end{array}$

The equilibrium ratio for the given reaction is,

$K_{\text{b}}=\frac{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]\left[\text{KOH}\right]}{\left[{\text{CH}}_3{\text{COO}}^{-}{\text{K}}^{+}\right]}$

The value of $K_{\text{b}}$ for acetate ion is $5.6\times {10}^{-10}$.

Substitute the calculated concentration values in the above expression.

$\begin{array}{c}{K}_{\text{b}}=\frac{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]\left[\text{KOH}\right]}{\left[{\text{CH}}_3{\text{COO}}^{-}{\text{K}}^{+}\right]}\\ \text{5}\text{.6}\times {\text{10}}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.067-x\right)\text{M}}\end{array}$

Since, value of $K_{\text{b}}$ is very small, hence, $\left(0.067-x\right)$ is taken as $\left(0.067\right)$.

Simplify the above equation,

$\begin{array}{c}\text{5}\text{.6}\times {\text{10}}^{-10}=\frac{\left(x\right)\left(x\right)}{\left(0.067\right)\text{M}}\\ x=\underset{\_}{0.61\times 10^{-5}\text{}M}\end{array}$

It is the concentration of ${\text{H}}^{+}$.

Explanation

The value of $\text{pH}$ of solution when $200.0\text{}\text{mL}$ $\text{KOH}$ has been added is $\underset{\_}{8.8}$.

The $\text{pOH}$ of the solution is calculated by the formula,

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

Where,

• $\left[{\text{OH}}^{-}\right]$ is the concentration of Hydroxide ions.

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(0.61\times {10}^{-5}\right)\\ =5.21\end{array}$

The relationship between $\text{pOH}$ is given as,

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

Substitute the value of $\text{pOH}$ in the above equation as,

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+\text{5}\text{.21}=14\\ \text{pH}=\underset{\_}{8.8}\end{array}$

The value of $\text{pH}$ of solution when $200.0\text{}\text{mL}$ $\text{KOH}$ has been added is $\underset{\_}{8.8}$.

Interpretation Introduction

Interpretation: The titration of Acetic acid with different volumes of $\text{KOH}$ is given. The $\text{pH}$ of each solution is to be calculated.

Concept introduction: Titration is a quantitative chemical analysis method that is used for the determination of concentration of an unknown solution. In acid base titration, the neutralization of either acid or base is done with a base or acid respectively of known concentration. This helps to determine the unknown concentration of acid or base.

When the amount of the titrant added is just sufficient for the neutralization of analyte is called equivalence point. At this point equal equivalents of both the acid and base are added.

To determine: The value of $\text{pH}$ of solution when $250.0\text{}\text{mL}$ $\text{KOH}$ has been added to it.

Answer to Problem 59E

The value of $\text{pH}$ of solution when $250.0\text{}\text{mL}$ $\text{KOH}$ has been added is $\underset{\_}{12.15}$.

Explanation of Solution

Explanation

The $\left[{\text{OH}}^{-}\right]$ is $\underset{\_}{0.014M}$.

Given

The volume of $\text{KOH}$ is $250.0\text{mL}$.

The conversion of $\text{mL}$ into $\text{L}$ is done as,

$1\text{mL}=0.001\text{L}$

Hence the conversion of $250\text{mL}$ into $\text{L}$ is done as,

$\begin{array}{c}\text{250}\text{mL}=250\times 0.001\text{L}\\ =\text{0}\text{.25}\text{L}\end{array}$

Substitute the value of concentration and volume of $\text{KOH}$ in equation (2) as,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\times \text{Volume}\text{of}\text{solution}\text{in}\text{litres}\\ =\text{0}\text{.100}\text{M}\times 0.25\text{L}\\ =\text{0}\text{.025}\text{moles}\end{array}$

Make the ICE table for the reaction between ${\text{CH}}_3\text{COOH}$ and $\text{KOH}$.

$\begin{array}{cccccccc}& {\text{CH}}_3\text{COOH}& +& \text{KOH}& \stackrel{}{\rightleftharpoons }& {\text{CH}}_3{\text{COO}}^{-}{\text{K}}^{+}& +& {\text{H}}_2\text{O}\\ \text{Initial moles:}& 0.02& & 0.025& & 0& & \\ \text{Change}:& -0.02& & -0.02& & +0.02& & \\ \text{Final}\text{moles}:& 0& & 0.005& & 0.02& & \end{array}$

There is presence of excess Hydroxide ions in solution.

Total volume of solution $\begin{array}{c}=\text{Volume}\text{of}{\text{CH}}_3\text{COOH}+\text{Volume}\text{of}\text{KOH}\\ =\text{0}\text{.1}\text{L}+0.25\text{L}\\ =\text{0}\text{.35}\text{L}\end{array}$

Substitute the value of number of moles of $\text{KOH}$ and final volume of solution in equation (1).

$\begin{array}{c}\text{Concentration}=\frac{\text{Number}\text{of}\text{moles}}{\text{Volume}\text{of}\text{solution}\text{in}\text{litres}}\\ =\frac{0.005\text{moles}}{0.35\text{}\text{L}}\\ =\underset{\_}{0.014M}\end{array}$

Explanation

The value of $\text{pH}$ of solution when $250.0\text{}\text{mL}$ $\text{KOH}$ has been added is $\underset{\_}{12.15}$.

The $\text{pOH}$ of the solution is calculated by the formula,

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

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

$\begin{array}{c}\text{pOH}=-\log \left[{\text{OH}}^{-}\right]\\ =-\log \left(0.014\right)\\ =1.85\end{array}$

The relationship between $\text{pOH}$ is given as,

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

Substitute the value of $\text{pOH}$ in the above equation.

$\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}+\text{1}\text{.85}=14\\ \text{pH}=\underset{\_}{12.15}\end{array}$

The value of $\text{pH}$ of solution when $250.0\text{}\text{mL}$ $\text{KOH}$ has been added is $\underset{\_}{12.15}$.