Chapter 6, Problem 6B.6E

(a)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of $\text{0}\text{.0356}\text{M}\text{HI}(aq)$ 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.6E

The $\text{pH}$ and $\text{pOH}$ of $\text{0}\text{.0356}\text{M}\text{HI}(aq)$ is $1.44$ and $12.56$ respectively.

Explanation of Solution

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

    $\text{HI}(aq)\to {\text{H}}^{+}(aq)+{\text{I}}^{-}(aq)$

One mole of hydroiodic acid dissociates to give one mole of hydrogen ion and one mole of iodide ion.  Therefore, $\text{0}\text{.0356}\text{M}$ of hydroiodic acid gives $\text{0}\text{.0356}\text{M}$ of hydronium ions.  Hence, the concentration of hydronium ion is $\text{0}\text{.0356}\text{M}$.

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

    $\begin{array}{c}\text{pH}=-\log [{\text{H}}_{\text{3}}{\text{O}}^{+}]\\ =-\log (\text{0}\text{.0356}\text{M})\\ =1.44\end{array}$

Hence, the $\text{pH}$ of hydroiodic acid solution is $1.44$.

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

    $\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pOH}=14-\text{pH}\\ \text{=14}-1.44\\ =12.56\end{array}$

Hence, the $\text{pOH}$ of hydroiodic acid solution is $12.56$.

(b)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of $\text{0}\text{.0725}\text{M}\text{HCl}(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer to Problem 6B.6E

The $\text{pH}$ and $\text{pOH}$ of $\text{0}\text{.0725}\text{M}\text{HCl}(aq)$ is $1.14$ and $12.86$ 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, $\text{0}\text{.0725}\text{M}$ of hydrochloric acid gives $\text{0}\text{.0725}\text{M}$ of hydronium ions.  Hence, the concentration of hydronium ion is $\text{0}\text{.0725}\text{M}$.

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

    $\begin{array}{c}\text{pH}=-\log [{\text{H}}_{\text{3}}{\text{O}}^{+}]\\ =-\log (\text{0}\text{.0725})\\ =1.14\end{array}$

Hence, the $\text{pH}$ of hydrochloric acid solution is $1.14$.

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

    $\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pOH}=14-\text{pH}\\ \text{=14}-1.14\\ =12.86\end{array}$

Hence, the $\text{pOH}$ of hydrochloric acid solution is $12.86$.

(c)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of $\text{3}\text{.46}\times {\text{10}}^{\text{-3}}\text{M}{\text{Ba(OH)}}_{\text{2}}(aq)$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer to Problem 6B.6E

The $\text{pH}$ and $\text{pOH}$ of $\text{3}\text{.46}\times {\text{10}}^{\text{-3}}{\text{Ba(OH)}}_{\text{2}}(aq)$ is $11.84$ and $2.16$ 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, $\text{3}\text{.46}\times {\text{10}}^{\text{-3}}\text{M}$ of barium hydroxide gives $\text{6}\text{.92}\times {\text{10}}^{\text{-3}}\text{M}$ of hydroxide ions.  Hence, the concentration of hydroxide ion is $\text{6}\text{.92}\times {\text{10}}^{\text{-3}}\text{M}$.

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

    $\begin{array}{c}\text{pOH}=-\log [{\text{OH}}^{-}]\\ =-\log (6.92\times {\text{10}}^{\text{-3}})\\ =2.16\end{array}$

Hence, the $\text{pOH}$ of barium hydroxide solution is $2.16$.

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

    $\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}=14-\text{pOH}\\ \text{=14}-2.16\\ =11.84\end{array}$

Hence, the $\text{pH}$ of barium hydroxide solution is $11.84$.

(d)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of $\text{10}\text{.9}\text{mg}$ of $\text{KOH}$ dissolved in $\text{10}\text{.0}\text{mL}$ of solution has to be calculated.

Concept Introduction:

Refer part (a).

Answer to Problem 6B.6E

The $\text{pH}$ and $\text{pOH}$ of $\text{KOH}(aq)$ is $9.23$ and $4.71$ respectively.

Explanation of Solution

Molarity of potassium hydroxide solution is calculated as shown below;

    $\begin{array}{c}\text{Molarity}=\frac{\text{number}\text{of}\text{moles}\text{of}\text{solute}}{\text{Volume}\text{of}\text{solution}}\\ =\frac{10.9\text{mg}\times \frac{1\text{g}}{1000\text{mg}}}{56.1056\text{g}\cdot {\text{mol}}^{-1}\times 10.0\text{mL}}\\ =1.94\times {10}^{-5}\text{M}\end{array}$

Therefore, $1.94\times {10}^{-5}\text{M}$ of potassium hydroxide gives $1.94\times {10}^{-5}\text{M}$ of hydroxide ions.  Hence, the concentration of hydroxide ion is $1.94\times {10}^{-5}\text{M}$.

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

    $\begin{array}{c}\text{pOH}=-\log [{\text{OH}}^{-}]\\ =-\log (1.94\times {10}^{-5})\\ =4.71\end{array}$

Hence, the $\text{pOH}$ of potassium hydroxide solution is $4.71$.

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

    $\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}=14-\text{pOH}\\ \text{=14}-4.71\\ =9.23\end{array}$

Hence, the $\text{pH}$ of potassium hydroxide solution is $9.23$.

(e)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of $\text{5}\text{.00}\text{M}\text{NaOH}(aq)$ after dilution from $\text{10}\text{.00}\text{mL}$ to $\text{2}\text{.50}\text{L}$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer to Problem 6B.6E

The $\text{pH}$ and $\text{pOH}$ of $\text{5}\text{.00}\text{M}\text{NaOH}(aq)$ is $12.30$ and $1.70$ respectively.

Explanation of Solution

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

    $\begin{array}{c}{C}_{initial}{V}_{initial}=C_{final}{V}_{final}\\ (5.00\text{M)(10}\text{.00}\text{mL)=}{C}_{final}(2500\text{mL})\\ C_{final}=\frac{(5.00\text{M)(10}\text{.00}\text{mL)}}{2500.0\text{mL}}\\ =0.02\text{M}\end{array}$

Therefore, $\text{0}\text{.02}\text{M}$ of sodium hydroxide gives $\text{0}\text{.02}\text{M}$ of hydroxide ions.  Hence, the concentration of hydroxide ion is $\text{0}\text{.02}\text{M}$.

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

    $\begin{array}{c}\text{pOH}=-\log [{\text{OH}}^{-}]\\ =-\log (0.02)\\ =1.70\end{array}$

Hence, the $\text{pOH}$ of sodium hydroxide solution is $1.70$.

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

    $\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pH}=14-\text{pOH}\\ \text{=14}-1.70\\ =12.30\end{array}$

Hence, the $\text{pH}$ of sodium hydroxide solution is $12.30$.

(f)

Interpretation Introduction

Interpretation:

The $\text{pH}$ and $\text{pOH}$ of $\text{3}\text{.5}\times {\text{10}}^{\text{-4}}\text{M}{\text{HClO}}_{\text{4}}(aq)$ after dilution from $\text{5}\text{.00}\text{mL}$ to $\text{25}\text{.0}\text{mL}$ has to be calculated.

Concept Introduction:

Refer part (a).

Answer to Problem 6B.6E

The $\text{pH}$ and $\text{pOH}$ of $\text{3}\text{.5}\times {\text{10}}^{\text{-4}}\text{M}{\text{HClO}}_{\text{4}}(aq)$ is $4.15$ and $9.85$ respectively.

Explanation of Solution

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

    $\begin{array}{c}{C}_{initial}{V}_{initial}=C_{final}{V}_{final}\\ (\text{3}\text{.5}\times {\text{10}}^{\text{-4}}\text{M)(5}\text{.00}\text{mL)=}{C}_{final}(25.0\text{mL})\\ C_{final}=\frac{(\text{3}\text{.5}\times {\text{10}}^{\text{-4}}\text{M)(5}\text{.00}\text{mL)}}{25.0\text{mL}}\\ =\text{0}\text{.7}\times {\text{10}}^{\text{-4}}\text{M}\end{array}$

Therefore, $\text{0}\text{.7}\times {\text{10}}^{\text{-4}}\text{M}$ of perchloric gives $\text{0}\text{.7}\times {\text{10}}^{\text{-4}}\text{M}$ of hydronium ions.  Hence, the concentration of hydronium ion is $\text{0}\text{.7}\times {\text{10}}^{\text{-4}}\text{M}$.

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

    $\begin{array}{c}\text{pH}=-\log [{\text{H}}_{\text{3}}{\text{O}}^{+}]\\ =-\log (\text{0}\text{.7}\times {\text{10}}^{\text{-4}})\\ =4.15\end{array}$

Hence, the $\text{pH}$ of perchloric acid solution is $4.15$.

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

    $\begin{array}{c}\text{pH}+\text{pOH}=14\\ \text{pOH}=14-\text{pH}\\ \text{=14}-4.15\\ =9.85\end{array}$

Hence, the $\text{pOH}$ of perchloric acid solution is $9.85$.