Chapter 15, Problem 63E

Consider the titration of 40.0 mL of 0.200 M HClO₄ 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. 10.0 mL

c. 40.0 mL

d. 80.0 mL

e. 100.0 mL

Interpretation Introduction

Interpretation:

The titration concentration of $\text{KOH}$, volume and concentration of ${\text{HClO}}_{\text{4}}$ is given. The $\text{pH}$ of resulting solution after the addition of $\text{KOH}$ is to be calculated for the given volumes of $\text{KOH}$.

Concept introduction:

Titration is a process to determine the concentration of an unknown solution with the help of a solution of a known concentration value.

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]$

The relation between $\text{pH}$ and $\text{pOH}$ is,

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

To determine the $\text{pH}$ of the resulting solution, after the addition of $\text{0}\text{.0}\text{mL}$ of $\text{KOH}$.

Answer to Problem 63E

The $\text{pH}$ of the resulting solution, after the addition of $\text{0}\text{.0}\text{mL}$ of $\text{KOH}$ is $\underset{\_}{0.699}$.

Explanation of Solution

Explanation

Given

Concentration of ${\text{HClO}}_{\text{4}}$ is $\text{0}\text{.200}\text{M}$.

${\text{HClO}}_{\text{4}}$ is a strong acid; hence, it completely dissociates into ions. The major species present in the solution are ${\text{H}}^{+},{\text{ClO}}_{\text{4}}{}^{-}$ and ${\text{H}}_{\text{2}}\text{O}$.

The $\text{pH}$ is determined by $\left[{\text{H}}^{+}\right]$ present in the solution.

Since ${\text{HClO}}_{\text{4}}$ completely dissociates into ions, $\text{0}\text{.200}\text{M}$ ${\text{HClO}}_{\text{4}}$ contains $\text{0}\text{.200}\text{M}$ ${\text{H}}^{+}$.

Formula

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]$

Where,

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

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

$\begin{array}{c}\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]\\ =-{\log }_{\text{10}}\left(\text{0}\text{.200}\right)\\ =\underset{\_}{0.699}\end{array}$

Interpretation Introduction

Interpretation:

The titration concentration of $\text{KOH}$, volume and concentration of ${\text{HClO}}_{\text{4}}$ is given. The $\text{pH}$ of resulting solution after the addition of $\text{KOH}$ is to be calculated for the given volumes of $\text{KOH}$.

Concept introduction:

Titration is a process to determine the concentration of an unknown solution with the help of a solution of a known concentration value.

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]$

The relation between $\text{pH}$ and $\text{pOH}$ is,

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

To determine the concentration of ${\text{H}}^{+}$

Answer to Problem 63E

The $\text{pH}$ of the resulting solution, after the addition of $\text{10}\text{.0}\text{mL}$ of $\text{KOH}$ is $\underset{\_}{0.853}$.

Explanation of Solution

Explanation

Given

Concentration of ${\text{HClO}}_{\text{4}}$ is $\text{0}\text{.200}\text{M}$.

Concentration of $\text{KOH}$ is $\text{0}\text{.100}\text{M}$.

Volume of ${\text{HClO}}_{\text{4}}$ original solution is $\text{40}\text{mL}$.

Volume of $\text{KOH}$ added is $\text{10}\text{mL}$.

Large quantities of the ${\text{H}}^{\text{+}}$ ions and the ${\text{OH}}^{-}$ ions are present. The $\text{1}\text{.0}\text{mmol}$ $\left(10.0\text{mL}\times \text{0}\text{.100}\text{M}\right)$ of the ${\text{OH}}^{-}$ added will react with $\text{1}\text{.0}\text{mmol}$ ${\text{H}}^{+}$ to form water:

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \stackrel{}{\to }& {\text{H}}_{\text{2}}\text{O}\\ \begin{array}{l}\text{Before}\\ \text{reaction}\end{array}& \begin{array}{l}\text{40}\text{.0}\text{mL}\times \text{0}\text{.200}\text{M}\\ =\text{8}\text{.0}\text{mmol}\end{array}& & \begin{array}{l}\text{10}\text{.0}\text{mL}\times \text{0}\text{.100}\text{M}\\ =\text{1}\text{.00}\text{mmol}\end{array}& & \\ \begin{array}{l}\text{After}\\ \text{reaction}\end{array}& \begin{array}{l}\text{8}\text{.0}-\text{1}\text{.00}\\ =\text{7}\text{.0}\text{mmol}\end{array}& & \begin{array}{l}\text{1}\text{.00}-\text{1}\text{.00}\\ =\text{0}\end{array}& & \end{array}$

After the reaction, the solution contains ${\text{H}}^{+},{\text{ClO}}_{\text{4}}{}^{-},{\text{K}}^{+}$ and ${\text{H}}_{\text{2}}\text{O}$ (the ${\text{OH}}^{-}$ ions get consumed).

Formula

The ${\text{H}}^{+}$ ion concentration, remaining in a given solution, is calculated by the formula,

$\left[{\text{H}}^{+}\right]=\frac{\text{mmol}{\text{H}}^{+}\text{left}}{\text{Volume}\text{of}\text{original}\text{solution}\text{(mL})\times \text{Volume}\text{of}\text{compound}\text{added}\text{(mL)}}$

Substitute the values of $\text{mmoles}$ of ${\text{H}}^{+}$ left, volume of the original solution and the volume of the compound added in the above equation.

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{\text{mmol}{\text{H}}^{+}\text{left}}{\text{Volume}\text{of}\text{original}\text{solution}\text{(mL})\times \text{Volume}\text{of}\text{compound}\text{added}\text{(mL})}\\ =\frac{\text{7}\text{.0}\text{mmol}}{\left(\text{40}\text{.0}+\text{10}\text{.0}\right)\text{mL}}\\ =\underset{\_}{0.14M}\end{array}$

To find the $\text{pH}$ of the resulting solution, after the addition of $\text{10}\text{.0}\text{mL}$ of $\text{KOH}$

Formula

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]$

Where,

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

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

$\begin{array}{c}\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]\\ =-{\log }_{\text{10}}\left(\text{0}\text{.14}\right)\\ =\underset{\_}{0.853}\end{array}$

Interpretation Introduction

Interpretation:

The titration concentration of $\text{KOH}$, volume and concentration of ${\text{HClO}}_{\text{4}}$ is given. The $\text{pH}$ of resulting solution after the addition of $\text{KOH}$ is to be calculated for the given volumes of $\text{KOH}$.

Concept introduction:

Titration is a process to determine the concentration of an unknown solution with the help of a solution of a known concentration value.

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]$

The relation between $\text{pH}$ and $\text{pOH}$ is,

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

Answer to Problem 63E

The $\text{pH}$ of the resulting solution, after the addition of $\text{40}\text{.0}\text{mL}$ of $\text{KOH}$ is $\underset{\_}{1.30}$.

Explanation of Solution

Explanation

To determine the concentration of ${\text{H}}^{+}$

Given

Concentration of ${\text{HClO}}_{\text{4}}$ is $\text{0}\text{.200}\text{M}$.

Concentration of $\text{KOH}$ is $\text{0}\text{.100}\text{M}$.

Volume of ${\text{HClO}}_{\text{4}}$ original solution is $\text{40}\text{mL}$.

Volume of $\text{KOH}$ added is $\text{40}\text{.0}\text{mL}$.

Large quantities of the ${\text{H}}^{\text{+}}$ ions and the ${\text{OH}}^{-}$ ions are present. The $\text{4}\text{.0}\text{mmol}$ $\left(40.0\text{mL}\times \text{0}\text{.100}\text{M}\right)$ of the ${\text{OH}}^{-}$ added will react with $\text{1}\text{.0}\text{mmol}$ ${\text{H}}^{+}$ to form water:

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \stackrel{}{\to }& {\text{H}}_{\text{2}}\text{O}\\ \begin{array}{l}\text{Before}\\ \text{reaction}\end{array}& \begin{array}{l}\text{40}\text{.0}\text{mL}\times \text{0}\text{.200}\text{M}\\ =\text{8}\text{.0}\text{mmol}\end{array}& & \begin{array}{l}\text{40}\text{.0}\text{mL}\times \text{0}\text{.100}\text{M}\\ =\text{4}\text{.00}\text{mmol}\end{array}& & \\ \begin{array}{l}\text{After}\\ \text{reaction}\end{array}& \begin{array}{l}\text{8}\text{.0}-\text{4}\text{.00}\\ =\text{4}\text{.0}\text{mmol}\end{array}& & \begin{array}{l}\text{4}\text{.00}-\text{4}\text{.00}\\ =\text{0}\end{array}& & \end{array}$

After the reaction, the solution contains ${\text{H}}^{+},{\text{ClO}}_{\text{4}}{}^{-},{\text{K}}^{+}$ and ${\text{H}}_{\text{2}}\text{O}$ (the ${\text{OH}}^{-}$ ions get consumed).

Formula

The ${\text{H}}^{+}$ ion concentration, remaining in a given solution, is calculated by the formula,

$\left[{\text{H}}^{+}\right]=\frac{\text{mmol}{\text{H}}^{+}\text{left}}{\text{Volume}\text{of}\text{original}\text{solution}\text{(mL})\times \text{Volume}\text{of}\text{compound}\text{added}\text{(mL)}}$

Substitute the values of $\text{mmoles}$ of ${\text{H}}^{+}$ left, volume of original solution and volume of compound added in the above equation.

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{\text{mmol}{\text{H}}^{+}\text{left}}{\text{Volume}\text{of}\text{original}\text{solution}\text{(mL})\times \text{Volume}\text{of}\text{compound}\text{added}\text{(mL})}\\ =\frac{\text{4}\text{.0}\text{mmol}}{\left(\text{40}\text{.0}+\text{40}\text{.0}\right)\text{mL}}\\ =\underset{\_}{0.05M}\end{array}$

To find the $\text{pH}$of the resulting solution, after the addition of $\text{40}\text{.0}\text{mL}$of $\text{KOH}$

The concentration of ${\text{H}}^{+}$ is $\text{0}\text{.05}\text{M}$.

Formula

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]$

Where,

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

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

$\begin{array}{c}\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]\\ =-{\log }_{\text{10}}\left(\text{0}\text{.05}\right)\\ =\underset{\_}{1.30}\end{array}$

Interpretation Introduction

Interpretation:

The titration concentration of $\text{KOH}$, volume and concentration of ${\text{HClO}}_{\text{4}}$ is given. The $\text{pH}$ of resulting solution after the addition of $\text{KOH}$ is to be calculated for the given volumes of $\text{KOH}$.

Concept introduction:

Titration is a process to determine the concentration of an unknown solution with the help of a solution of a known concentration value.

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]$

The relation between $\text{pH}$ and $\text{pOH}$ is,

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

Answer to Problem 63E

The $\text{pH}$ of the resulting solution, after the addition of $\text{80}\text{.0}\text{mL}$ of $\text{KOH}$ is $\underset{\_}{7}$.

Explanation of Solution

Explanation

To determine the $\text{pH}$ of the resulting solution, after the addition of $\text{80}\text{.0}\text{mL}$ of $\text{KOH}$.

Given

Concentration of ${\text{HClO}}_{\text{4}}$ is $\text{0}\text{.200}\text{M}$.

Concentration of $\text{KOH}$ is $\text{0}\text{.100}\text{M}$.

Volume of ${\text{HClO}}_{\text{4}}$ original solution is $\text{40}\text{mL}$.

Volume of $\text{KOH}$ added is $\text{80}\text{.0}\text{mL}$.

Large quantities of the ${\text{H}}^{\text{+}}$ ions and the ${\text{OH}}^{-}$ ions are present. The $\text{8}\text{.0}\text{mmol}$ $\left(80.0\text{mL}\times \text{0}\text{.100}\text{M}\right)$ of the ${\text{OH}}^{-}$ added will react with $\text{1}\text{.0}\text{mmol}$ ${\text{H}}^{+}$ to form water:

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \stackrel{}{\to }& {\text{H}}_{\text{2}}\text{O}\\ \begin{array}{l}\text{Before}\\ \text{reaction}\end{array}& \begin{array}{l}\text{40}\text{.0}\text{mL}\times \text{0}\text{.200}\text{M}\\ =\text{8}\text{mmol}\end{array}& & \begin{array}{l}\text{80}\text{.0}\text{mL}\times \text{0}\text{.100}\text{M}\\ =\text{8}\text{.00}\text{mmol}\end{array}& & \\ \begin{array}{l}\text{After}\\ \text{reaction}\end{array}& \begin{array}{l}\text{8}\text{.00}-\text{8}\text{.00}\\ =\text{0}\end{array}& & \begin{array}{l}\text{8}\text{.00}-\text{8}\text{.00}\\ =\text{0}\end{array}& & \end{array}$

After the reaction, the solution contains ${\text{ClO}}_{\text{4}}{}^{-},{\text{K}}^{+}$ and ${\text{H}}_{\text{2}}\text{O}$ (the ${\text{OH}}^{-}$ ions and ${\text{H}}^{+}$ get consumed). It means that solution is completely neutralized. Therefore, the $\text{pH}$ value of the solution is $\underset{\_}{7}$.

Interpretation Introduction

Interpretation:

The titration concentration of $\text{KOH}$, volume and concentration of ${\text{HClO}}_{\text{4}}$ is given. The $\text{pH}$ of resulting solution after the addition of $\text{KOH}$ is to be calculated for the given volumes of $\text{KOH}$.

Concept introduction:

Titration is a process to determine the concentration of an unknown solution with the help of a solution of a known concentration value.

The $\text{pH}$ is calculated using the formula,

$\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]$

The relation between $\text{pH}$ and $\text{pOH}$ is,

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

Answer to Problem 63E

The $\text{pH}$ of the resulting solution, after the addition of $\text{100}\text{.0}\text{mL}$ of $\text{KOH}$ is $\underset{\_}{12.15}$.

Explanation of Solution

Explanation

To determine the concentration of ${\text{OH}}^{-}$

Given

Concentration of ${\text{HClO}}_{\text{4}}$ is $\text{0}\text{.200}\text{M}$.

Concentration of $\text{KOH}$ is $\text{0}\text{.100}\text{M}$.

Volume of ${\text{HClO}}_{\text{4}}$ original solution is $\text{40}\text{mL}$.

Volume of $\text{KOH}$ added is $\text{100}\text{.0}\text{mL}$.

Large quantities of the ${\text{H}}^{\text{+}}$ ions and the ${\text{OH}}^{-}$ ions are present. The $\text{10}\text{.0}\text{mmol}$ $\left(100.0\text{mL}\times \text{0}\text{.100}\text{M}\right)$ of the ${\text{OH}}^{-}$ added will react with $\text{1}\text{.0}\text{mmol}$ ${\text{H}}^{+}$ to form water:

$\begin{array}{cccccc}& {\text{H}}^{+}& +& {\text{OH}}^{-}& \stackrel{}{\to }& {\text{H}}_{\text{2}}\text{O}\\ \begin{array}{l}\text{Before}\\ \text{reaction}\end{array}& \begin{array}{l}\text{40}\text{.0}\text{mL}\times \text{0}\text{.200}\text{M}\\ =\text{8}\text{.0}\text{mmol}\end{array}& & \begin{array}{l}\text{100}\text{.0}\text{mL}\times \text{0}\text{.100}\text{M}\\ =\text{10}\text{.0}\text{mmol}\end{array}& & \\ \begin{array}{l}\text{After}\\ \text{reaction}\end{array}& \begin{array}{l}\text{8}\text{.0}-\text{8}\text{.00}\\ =\text{0}\end{array}& & \begin{array}{l}\text{10}\text{.00}-\text{8}\text{.00}\\ =\text{2}\end{array}& & \end{array}$

After the reaction, the solution contains ${\text{OH}}^{-},{\text{ClO}}_{\text{4}}{}^{-},{\text{K}}^{+}$ and ${\text{H}}_{\text{2}}\text{O}$ (the ${\text{H}}^{+}$ ions get consumed).

Formula

The ${\text{OH}}^{-}$ ion concentration, remaining in a given solution, is calculated by the formula,

$\left[{\text{OH}}^{-}\right]=\frac{\text{mmol}{\text{OH}}^{-}\text{left}}{\text{Volume}\text{of}\text{original}\text{solution}\text{(mL})\times \text{Volume}\text{of}\text{compound}\text{added}\text{(mL)}}$

Substitute the values of the $\text{mmoles}$ of ${\text{OH}}^{-}$ left, volume of original solution and volume of compound added in the above equation.

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{\text{mmol}{\text{OH}}^{-}\text{left}}{\text{Volume}\text{of}\text{original}\text{solution}\text{(mL})\times \text{Volume}\text{of}\text{compound}\text{added}\text{(mL})}\\ =\frac{\text{2}\text{.0}\text{mmol}}{\left(\text{40}\text{.0}+\text{100}\text{.0}\right)\text{mL}}\\ =\underset{\_}{0.0142M}\end{array}$

To determine the concentration of $\text{pOH}$

Formula

The $\text{pOH}$ is calculated using the formula,

$\text{pOH}=-{\log }_{\text{10}}\left[{\text{OH}}^{-}\right]$

Where,

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

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

$\begin{array}{c}\text{pOH}=-{\log }_{\text{10}}\left[{\text{OH}}^{-}\right]\\ =-{\log }_{\text{10}}\left(\text{0}\text{.0142}\right)\\ =\underset{\_}{1.85}\end{array}$

To determine the $\text{pH}$ of the resulting solution, after the addition of $\text{100}\text{.0}\text{mL}$ of $\text{KOH}$

Formula

The relation between $\text{pH}$ and $\text{pOH}$ is,

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

Where,

• $\text{pH}$ is a measure of concentration of ${\text{H}}^{\text{+}}$.
• $\text{pOH}$ is a measure of concentration of ${\text{OH}}^{-}$.

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

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