Chapter 14, Problem 58E

Consider the titration of 80.0 mL of 0.100 M Ba(OH)₂ by 0.400 M HCl. Calculate the pH of the resulting solution after the following volumes of HCl have been added.

a. 0.0 mL

b. 20.0 mL

c. 30.0 mL

d. 40.0 mL

e. 80.0 mL

Interpretation Introduction

Interpretation:

The concentration of $\text{HCl}$ used in titration of a stated volume and concentration of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is given. The $\text{pH}$ of the resulting solution after the addition of a given volume of $\text{HCl}$ is to be calculated.

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

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

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

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

Answer to Problem 58E

The $\text{pH}$ of the resulting solution, after the addition of $\text{0}\text{mL}$ of $\text{HCl}$, is $\underset{\_}{13.301}$.

Explanation of Solution

Explanation

Given

Concentration of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is $\text{0}\text{.100}\text{M}$.

The $\text{pOH}$ is determined by the $\left[{\text{OH}}^{-}\right]$ present in the solution.

In the given titration,

$0.0\text{mL}$ of $\text{0}\text{.4}\text{M}$ $\text{HCl}$ is added to the given $\text{80}\text{mL}$ of $\text{0}\text{.100}\text{M}$ $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$.

Since, $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is a strong base, it completely dissociates into its ions. The major species present are ${\text{OH}}^{-},{\text{Ba}}^{\text{2}+}$ and ${\text{H}}_{\text{2}}\text{O}$.

The complete dissociation of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ leads to the formation of $2{\text{OH}}^{-}$ ions.

Therefore, the concentration of ${\text{OH}}^{-}$ in $\text{0}\text{.100}\text{M}$ $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is,

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\text{2}\times \text{0}\text{.100}\text{M}\\ =\underset{\_}{0.200M}\end{array}$

The concentration of ${\text{OH}}^{-}$ is $\text{0}\text{.200}\text{M}$.

Formula

The $\text{pOH}$ is calculated by 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 $\left[{\text{OH}}^{-}\right]$ in the above equation.

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

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.

$\begin{array}{c}\text{pH}+\text{pOH}=\text{14}\\ \text{pH}+\text{0}\text{.699}=\text{14}\\ \text{pH}=\underset{\_}{13.301}\end{array}$

Interpretation Introduction

Interpretation:

The concentration of $\text{HCl}$ used in titration of a stated volume and concentration of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is given. The $\text{pH}$ of the resulting solution after the addition of a given volume of $\text{HCl}$ is to be calculated.

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

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

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

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

Answer to Problem 58E

The $\text{pH}$ of the resulting solution, after the addition of $\text{20}\text{.0}\text{mL}$ of $\text{HCl}$, is $\underset{\_}{12.9}$.

Explanation of Solution

Explanation

Given

Concentration of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is $\text{0}\text{.100}\text{M}$.

Concentration of $\text{HCl}$ is $\text{0}\text{.400}\text{M}$.

Volume of original $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ solution is $\text{80}\text{mL}$.

Volume of $\text{HCl}$ added is $\text{20}\text{.0}\text{mL}$.

Formula

The number of moles of compound in a solution is calculated using the formula,

$\text{Moles}\text{of}\text{compound}=\text{Concentration}\text{of}\text{compound}\times \text{Volume}\text{of}\text{solution}$

Substitute the values of concentration and volume of $\text{HCl}$ in the above equation.

$\begin{array}{c}\text{Moles}\text{of}\text{compound}=\text{Concentration}\text{of}\text{compound}\times \text{Volume}\text{of}\text{solution}\\ =\text{0}\text{.4}\text{M}\times \text{20}\text{.0}\text{mL}\\ =\underset{\_}{8.0mmol}\end{array}$

In the given titration,

$\text{20}\text{.0}\text{mL}$ of $\text{HCl}$ is added to $\text{80}\text{mL}$ of $\text{0}\text{.100}\text{M}$ $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$.

The number of moles of $\text{HCl}$ added is $\text{8}\text{.0}\text{mmol}$.

$\text{8}\text{.0}\text{mmol}$ of ${\text{H}}^{+}$ will react with $\text{8}\text{.0}\text{mmol}$ of ${\text{OH}}^{-}$ in the solution.

The number of moles of ${\text{OH}}^{-}$ originally present is,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\text{of}\left[{\text{OH}}^{-}\right]\times \text{Volume}\text{of}\text{solution}\\ =\text{0}\text{.200}\text{M}\times \text{80}\text{.0}\text{mL}\\ =\text{16}\text{mmol}\end{array}$

The number of moles of ${\text{OH}}^{-}$ remaining in the solution is calculated by the formula,

$\text{Number}\text{of}\text{moles}\text{of}{\text{OH}}^{-}\text{left}=\text{Total}\text{moles}\text{of}{\text{OH}}^{-}-\text{Number}\text{of}\text{moles}\text{that}\text{reacted}$

Substitute the value of the total number of moles of ${\text{OH}}^{-}$ and the moles of ${\text{OH}}^{-}$ that reacted in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{OH}}^{-}\text{left}=\text{16}\text{.0}\text{mmol}-\text{8}\text{.0}\text{mmol}\\ =\text{8}\text{.0}\text{mmol}\end{array}$

Formula

The $\text{pOH}$ will be determined by ${\text{OH}}^{-}$ remaining in the solution,

The remaining ${\text{OH}}^{-}$ is calculated by the formula,

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

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

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{\text{mmol}{\text{OH}}^{-}\text{left}}{\text{Volume}\text{of}\text{original}\text{solution}\text{(mL})\times \text{Volume}\text{of}\text{compound}\text{added}}\\ =\frac{\text{8}\text{.0}\text{mmol}}{\left(\text{20}\text{.0}+\text{80}\text{.0}\right)\text{mL}}\\ =\underset{\_}{0.08M}\end{array}$

The concentration of ${\text{OH}}^{-}$ is $\text{0}\text{.08}\text{M}$.

Formula

The $\text{pOH}$ is calculated by 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 concentration of $\left[{\text{OH}}^{-}\right]$ in the above equation.

$\begin{array}{c}\text{pOH}=-{\log }_{\text{10}}\left[{\text{OH}}^{-}\right]\\ =-{\log }_{\text{10}}\left(\text{0}\text{.08}\right)\\ =\underset{\_}{1.1}\end{array}$

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.

$\begin{array}{c}\text{pH}+\text{pOH}=\text{14}\\ \text{pH}+\text{1}\text{.1}=\text{14}\\ \text{pH}=\underset{\_}{12.9}\end{array}$

Interpretation Introduction

Interpretation:

The concentration of $\text{HCl}$ used in titration of a stated volume and concentration of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is given. The $\text{pH}$ of the resulting solution after the addition of a given volume of $\text{HCl}$ is to be calculated.

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

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

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

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

Answer to Problem 58E

The $\text{pH}$ of the resulting solution, after the addition of $\text{30}\text{.0}\text{mL}$ of $\text{HCl}$, is $\underset{\_}{12.557}$.

Explanation of Solution

Explanation

Given

Concentration of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is $\text{0}\text{.100}\text{M}$.

Concentration of $\text{HCl}$ is $\text{0}\text{.400}\text{M}$.

Volume of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ original solution is $\text{80}\text{mL}$.

Volume of $\text{HCl}$ added is $\text{30}\text{.0}\text{mL}$.

Formula

The number of moles of compound in a solution is calculated using the formula,

$\text{Moles}\text{of}\text{compound}=\text{Concentration}\text{of}\text{compound}\times \text{Volume}\text{of}\text{solution}$

Substitute the values of concentration and volume of $\text{HCl}$ in the above equation.

$\begin{array}{c}\text{Moles}\text{of}\text{compound}=\text{Concentration}\text{of}\text{compound}\times \text{Volume}\text{of}\text{solution}\\ =\text{0}\text{.4}\text{M}\times \text{30}\text{mL}\\ =\underset{\_}{12.0mmol}\end{array}$

In the given titration, $\text{30}\text{.0}\text{mL}$ of $\text{HCl}$ is added to $\text{80}\text{mL}$ of $\text{0}\text{.100}\text{M}$ $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$. It means that $\text{12}\text{.0}\text{mmol}$ of ${\text{H}}^{+}$ will react with $\text{12}\text{.0}\text{mmol}$ of ${\text{OH}}^{-}$.

The number of moles of ${\text{OH}}^{-}$ originally present is,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\text{of}\left[{\text{OH}}^{-}\right]\times \text{Volume}\text{of}\text{solution}\\ =\text{0}\text{.200}\text{M}\times \text{80}\text{.0}\text{mL}\\ =\text{16}\text{mmol}\end{array}$

The number of moles of ${\text{OH}}^{-}$ remaining in the solution is calculated by the formula,

$\text{Number}\text{of}\text{moles}\text{of}{\text{OH}}^{-}\text{left}=\text{Total}\text{moles}\text{of}{\text{OH}}^{-}-\text{Number}\text{of}\text{moles}\text{that}\text{reacted}$

Substitute the value of the total number of moles of ${\text{OH}}^{-}$ and the moles of ${\text{OH}}^{-}$ that reacted in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{OH}}^{-}\text{left}=\text{16}\text{.0}\text{mmol}-\text{12}\text{.0}\text{mmol}\\ =\text{4}\text{.0}\text{mmol}\end{array}$

Formula

The $\text{pOH}$ will be determined by ${\text{OH}}^{-}$ remaining in the solution,

The remaining ${\text{OH}}^{-}$ is calculated by the formula,

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

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

$\begin{array}{c}\left[{\text{OH}}^{-}\right]=\frac{\text{mmol}{\text{OH}}^{-}\text{left}}{\text{Volume}\text{of}\text{original}\text{solution}\text{(mL})\times \text{Volume}\text{of}\text{compound}\text{added}}\\ =\frac{\text{4}\text{.0}\text{mmol}}{\left(\text{30}\text{.0}+\text{80}\text{.0}\right)\text{mL}}\\ =\underset{\_}{0.036M}\end{array}$

Formula

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

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

Where,

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

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

$\begin{array}{c}\text{pOH}=-{\log }_{\text{10}}\left[{\text{OH}}^{-}\right]\\ =-{\log }_{\text{10}}\text{0}\text{.036}\text{M}\\ =\underset{\_}{1.443}\end{array}$

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.

$\begin{array}{c}\text{pH}+\text{pOH}=\text{14}\\ \text{pH}+\text{1}\text{.443}=\text{14}\\ \text{pH}=\underset{\_}{12.557}\end{array}$

Interpretation Introduction

Interpretation:

The concentration of $\text{HCl}$ used in titration of a stated volume and concentration of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is given. The $\text{pH}$ of the resulting solution after the addition of a given volume of $\text{HCl}$ is to be calculated.

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

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

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

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

Answer to Problem 58E

The $\text{pH}$ of the resulting solution, after the addition of $\text{40}\text{.0}\text{mL}$ of $\text{HCl}$, is $\underset{\_}{7}$.

Explanation of Solution

Explanation

Given

Concentration of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is $\text{0}\text{.100}\text{M}$.

Concentration of $\text{HCl}$ is $\text{0}\text{.400}\text{M}$.

Volume of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ original solution is $\text{80}\text{mL}$.

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

Formula

The number of moles of compound in a solution is calculated using the formula,

$\text{Moles}\text{of}\text{compound}=\text{Concentration}\text{of}\text{compound}\times \text{Volume}\text{of}\text{solution}$

Substitute the values of concentration and volume of $\text{HCl}$ in the above equation.

$\begin{array}{c}\text{Moles}\text{of}\text{compound}=\text{Concentration}\text{of}\text{compound}\times \text{Volume}\text{of}\text{solution}\\ =\text{0}\text{.4}\text{M}\times \text{40}\text{mL}\\ =\underset{\_}{16.0mmol}\end{array}$

In the given titration, $\text{40}\text{.0}\text{mL}$ of $\text{HCl}$ is added to $\text{80}\text{mL}$ of $\text{0}\text{.100}\text{M}$ $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$. It means that $\text{16}\text{.0}\text{mmol}$ of ${\text{H}}^{+}$ will react with $\text{16}\text{.0}\text{mmol}$ of ${\text{OH}}^{-}$ in $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ (all 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 concentration of $\text{HCl}$ used in titration of a stated volume and concentration of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is given. The $\text{pH}$ of the resulting solution after the addition of a given volume of $\text{HCl}$ is to be calculated.

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

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

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

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

Answer to Problem 58E

The $\text{pH}$ of the resulting solution, after the addition of $\text{80}\text{.0}\text{mL}$ of $\text{HCl}$, is $\underset{\_}{1}$.

Explanation of Solution

Explanation

Given

Concentration of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ is $\text{0}\text{.100}\text{M}$.

Concentration of $\text{HCl}$ is $\text{0}\text{.400}\text{M}$.

Volume of $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$ original solution is $\text{80}\text{mL}$.

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

Formula

The number of moles of compound in a solution is calculated using the formula,

$\text{Moles}\text{of}\text{compound}=\text{Concentration}\text{of}\text{compound}\times \text{Volume}\text{of}\text{solution}$

Substitute the values of concentration and volume of $\text{HCl}$ in the above equation.

$\begin{array}{c}\text{Moles}\text{of}\text{compound}=\text{Concentration}\text{of}\text{compound}\times \text{Volume}\text{of}\text{solution}\\ =\text{0}\text{.4}\text{M}\times \text{80}\text{mL}\\ =\underset{\_}{32.0mmol}\end{array}$

In the given titration, $\text{80}\text{.0}\text{mL}$ of $\text{HCl}$ is added to $\text{80}\text{mL}$ of $\text{0}\text{.100}\text{M}$ $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$. If $\text{32}\text{.0}\text{mmol}$ of ${\text{H}}^{+}$ will react with $\text{16}\text{.0}\text{mmol}$ of ${\text{OH}}^{-}$ in $\text{Ba}{\left(\text{OH}\right)}_{\text{2}}$.

The number of moles of ${\text{OH}}^{-}$ originally present is,

$\begin{array}{c}\text{Number}\text{of}\text{moles}=\text{Concentration}\text{of}\left[{\text{OH}}^{-}\right]\times \text{Volume}\text{of}\text{solution}\\ =\text{0}\text{.200}\text{M}\times \text{80}\text{.0}\text{mL}\\ =\text{16}\text{mmol}\end{array}$

The number of moles of ${\text{OH}}^{-}$ remaining in the solution is calculated by the formula,

$\text{Number}\text{of}\text{moles}\text{of}{\text{OH}}^{-}\text{left}=\text{Total}\text{moles}\text{of}{\text{OH}}^{-}-\text{Number}\text{of}\text{moles}\text{that}\text{reacted}$

Substitute the value of the total number of moles of ${\text{OH}}^{-}$ and the moles of ${\text{OH}}^{-}$ that reacted in the above expression.

$\begin{array}{c}\text{Number}\text{of}\text{moles}\text{of}{\text{OH}}^{-}\text{left}=\text{32}\text{.0}\text{mmol}-\text{16}\text{.0}\text{mmol}\\ =\text{16}\text{.0}\text{mmol}\end{array}$

Formula

The concentration of ${\text{H}}^{+}$ ions remaining in the solution is calculated by the formula,

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

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

$\begin{array}{c}\left[{\text{H}}^{+}\right]=\frac{\text{mmol}{\text{H}}^{+}\text{left}}{\text{Volume}\text{of}\text{original}\text{solution}\text{(mL})\times \text{Volume}\text{of}\text{compound}\text{added}}\\ =\frac{\text{16}\text{.0}\text{mmol}}{\left(\text{40}\text{.0}+\text{10}\text{.0}\right)\text{mL}}\\ =\underset{\_}{0.1M}\end{array}$

The concentration of ${\text{H}}^{+}$ is $\text{0}\text{.1}\text{M}$.

Formula

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

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

Where,

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

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

$\begin{array}{c}\text{pH}=-{\log }_{\text{10}}\left[{\text{H}}^{+}\right]\\ =-{\log }_{\text{10}}\left(\text{0}\text{.1}\right)\\ =\underset{\_}{1}\end{array}$