Chapter 6, Problem 6B.5E

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of $\text{0}\text{.0146}\text{M}{\text{HNO}}_{\text{3}}(aq)$ has to be calculated.

Concept Introduction:

Negative logarithm of the concentration of hydronium ion is the $\text{pH}$ of solution.  This can be given in equation form as follows;

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

Answer to Problem 6B.5E

The $\text{pH}$ and $\text{pOH}$ of $\text{0}\text{.0146}\text{M}{\text{HNO}}_{\text{3}}(aq)$ is $1.84$ and $12.16$ respectively.

Explanation of Solution

Nitric acid is a strong acid and it completely dissociates in water.  This can be represented as follows;

    ${\text{HNO}}_{\text{3}}(aq)\to {\text{H}}^{+}(aq)+{\text{NO}}_{\text{3}}{}^{-}(aq)$

One mole of nitric acid dissociates to give one mole of hydrogen ion and one mole of nitrate ion.  Therefore, $\text{0}\text{.0146}\text{M}$ of nitric acid gives $\text{0}\text{.0146}\text{M}$ of hydronium ions.  Hence, the concentration of hydronium ion is $\text{0}\text{.0146}\text{M}$.

The $\text{pH}$ of the nitric acid solution is calculated as shown below;

    $\begin{array}{c}\text{pH}=-\log [{\text{H}}_{\text{3}}{\text{O}}^{+}]\\ =-\log (0.0146)\\ =1.84\end{array}$

Hence, the $\text{pH}$ of nitric acid solution is $1.84$.

The $\text{pOH}$ of the solution can be calculated using the formula of $\text{pOH}$ as follows:

    $\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pOH}=14-\text{pH}\\ \text{=14}-1.84\\ =12.16\end{array}$

Hence, the $\text{pOH}$ of nitric acid solution is $12.16$.

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of $\text{0}\text{.11}\text{M}\text{HCl}(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer to Problem 6B.5E

The $\text{pH}$ and $\text{pOH}$ of $\text{0}\text{.11}\text{M}\text{HCl}(aq)$ is $0.96$ and $13.04$ respectively.

Explanation of Solution

Hydrochloric acid is a strong acid and it completely dissociates in water.  This can be represented as follows;

    $\text{HCl}(aq)\to {\text{H}}^{+}(aq)+{\text{Cl}}^{-}(aq)$

One mole of hydrochloric acid dissociates to give one mole of hydrogen ion and one mole of chloride ion.  Therefore, $\text{0}\text{.11}\text{M}$ of hydrochloric acid gives $\text{0}\text{.11}\text{M}$ of hydronium ions.  Hence, the concentration of hydronium ion is $\text{0}\text{.11}\text{M}$.

The $\text{pH}$ of the hydrochloric acid solution is calculated as shown below;

    $\begin{array}{c}\text{pH}=-\log [{\text{H}}_{\text{3}}{\text{O}}^{+}]\\ =-\log (0.11)\\ =0.96\end{array}$

Hence, the $\text{pH}$ of hydrochloric acid solution is $0.96$.

The $\text{pOH}$ of the solution can be calculated using the formula of $\text{pOH}$ as follows:

    $\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pOH}=14-\text{pH}\\ \text{=14}-0.96\\ =13.04\end{array}$

Hence, the $\text{pOH}$ of hydrochloric acid solution is $13.04$.

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of $\text{0}\text{.0092}\text{M}{\text{Ba(OH)}}_{\text{2}}(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer to Problem 6B.5E

The $\text{pH}$ and $\text{pOH}$ of $\text{0}\text{.0092}\text{M}{\text{Ba(OH)}}_{\text{2}}(aq)$ is $12.26$ and $1.74$ respectively.

Explanation of Solution

Barium hydroxide is a strong acid and it completely dissociates in water.  This can be represented as follows;

    ${\text{Ba(OH)}}_{\text{2}}(aq)\to {\text{Ba}}^{2+}(aq)+2{\text{OH}}^{-}(aq)$

One mole of barium hydroxide dissociates to give one mole of barium ion and two moles of hydroxide ion.  Therefore, $\text{0}\text{.0092}\text{M}$ of barium hydroxide gives $\text{0}\text{.0184}\text{M}$ of hydroxide ions.  Hence, the concentration of hydroxide ion is $\text{0}\text{.0184}\text{M}$.

The $\text{pOH}$ of the barium hydroxide solution is calculated as shown below;

    $\begin{array}{c}\text{pOH}=-\log [{\text{OH}}^{-}]\\ =-\log (0.0184)\\ =1.74\end{array}$

Hence, the $\text{pOH}$ of barium hydroxide solution is $1.74$.

The $\text{pH}$ of the solution can be calculated using the formula of $\text{pH}$ as follows:

    $\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}=14-\text{pOH}\\ \text{=14}-1.74\\ =12.26\end{array}$

Hence, the $\text{pH}$ of barium hydroxide solution is $12.26$.

(d)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of $\text{0}\text{.175}\text{M}\text{KOH}(aq)$ after dilution from $\text{2}\text{.00}\text{mL}$ to $\text{0}\text{.500}\text{L}$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer to Problem 6B.5E

The $\text{pH}$ and $\text{pOH}$ of $\text{0}\text{.175}\text{M}\text{KOH}(aq)$ is $10.85$ and $3.15$ respectively.

Explanation of Solution

Concentration of potassium hydroxide after dilution is calculated as shown below;

    $\begin{array}{c}{C}_{initial}{V}_{initial}=C_{final}{V}_{final}\\ (0.175\text{M)(2}\text{.00}\text{mL)=}{C}_{final}(50.0\text{mL})\\ C_{final}=\frac{(0.175\text{M)(2}\text{.00}\text{mL)}}{500.0\text{mL}}\\ =0.0007\text{M}\end{array}$

Therefore, $\text{0}\text{.0007}\text{M}$ of potassium hydroxide gives $\text{0}\text{.0007}\text{M}$ of hydroxide ions.  Hence, the concentration of hydroxide ion is $\text{0}\text{.0007}\text{M}$.

The $\text{pOH}$ of the potassium hydroxide solution is calculated as shown below;

    $\begin{array}{c}\text{pOH}=-\log [{\text{OH}}^{-}]\\ =-\log (0.0007)\\ =3.15\end{array}$

Hence, the $\text{pOH}$ of potassium hydroxide solution is $3.15$.

The $\text{pH}$ of the solution can be calculated using the formula of $\text{pH}$ as follows:

    $\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}=14-\text{pOH}\\ \text{=14}-3.15\\ =10.85\end{array}$

Hence, the $\text{pH}$ of potassium hydroxide solution is $10.85$.

(e)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of $\text{13}\text{.6}\text{mg}$ of $\text{NaOH}$ dissolved in $\text{0}\text{.350}\text{L}$ of solution has to be calculated.

Concept Introduction:

Refer part (a).

Answer to Problem 6B.5E

The $\text{pH}$ and $\text{pOH}$ of $\text{NaOH}(aq)$ is $10.99$ and $3.01$ respectively.

Explanation of Solution

Molarity of sodium hydroxide solution is calculated as shown below;

    $\begin{array}{c}\text{Molarity}=\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}}\\ =\frac{13.6\text{mg}\times \frac{1\text{g}}{1000\text{mg}}}{39.99\text{g}\cdot {\text{mol}}^{-1}\times 0.350\text{L}}\\ =9.72\times {10}^{-4}\text{M}\end{array}$

Therefore, $9.72\times {10}^{-4}\text{M}$ of sodium hydroxide gives $9.72\times {10}^{-4}\text{M}$ of hydroxide ions.  Hence, the concentration of hydroxide ion is $9.72\times {10}^{-4}\text{M}$.

The $\text{pOH}$ of the sodium hydroxide solution is calculated as shown below;

    $\begin{array}{c}\text{pOH}=-\log [{\text{OH}}^{-}]\\ =-\log (9.72\times {10}^{-4})\\ =3.01\end{array}$

Hence, the $\text{pOH}$ of sodium hydroxide solution is $3.01$.

The $\text{pH}$ of the solution can be calculated using the formula of $\text{pH}$ as follows:

    $\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}=14-\text{pOH}\\ \text{=14}-3.01\\ =10.99\end{array}$

Hence, the $\text{pH}$ of sodium hydroxide solution is $10.99$.

(f)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of $\text{3}\text{.5}\times {\text{10}}^{\text{-4}}\text{M}\text{HBr}(aq)$ after dilution from $\text{75}\text{.00}\text{mL}$ to $\text{0}\text{.500}\text{L}$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer to Problem 6B.5E

The $\text{pH}$ and $\text{pOH}$ of $\text{3}\text{.5}\times {\text{10}}^{\text{-4}}\text{M}\text{HBr}(aq)$ is $4.28$ and $9.72$ respectively.

Explanation of Solution

Concentration of hydrobromic acid after dilution is calculated as shown below;

    $\begin{array}{c}{C}_{initial}{V}_{initial}=C_{final}{V}_{final}\\ (\text{3}\text{.5}\times {\text{10}}^{\text{-4}}\text{M)(75}\text{.00}\text{mL)=}{C}_{final}(500.0\text{mL})\\ C_{final}=\frac{(\text{3}\text{.5}\times {\text{10}}^{\text{-4}}\text{M)(75}\text{.00}\text{mL)}}{500.0\text{mL}}\\ =\text{0}\text{.525}\times {\text{10}}^{\text{-4}}\text{M}\end{array}$

Therefore, $\text{0}\text{.525}\times {\text{10}}^{\text{-4}}\text{M}$ of hydrobromic gives $\text{0}\text{.525}\times {\text{10}}^{\text{-4}}\text{M}$ of hydronium ions.  Hence, the concentration of hydronium ion is $\text{0}\text{.525}\times {\text{10}}^{\text{-4}}\text{M}$.

The $\text{pH}$ of the hydrobromic acid solution is calculated as shown below;

    $\begin{array}{c}\text{pH}=-\log [{\text{H}}_{\text{3}}{\text{O}}^{+}]\\ =-\log (\text{0}\text{.525}\times {\text{10}}^{\text{-4}})\\ =4.28\end{array}$

Hence, the $\text{pH}$ of hydrobromic acid solution is $4.28$.

The $\text{pOH}$ of the solution can be calculated using the formula of $\text{pH}$ as follows:

    $\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pOH}=14-\text{pH}\\ \text{=14}-4.28\\ =9.72\end{array}$

Hence, the $\text{pOH}$ of hydrobromic acid solution is $9.72$.