Chapter 5, Problem 5D.4E

(a)

Interpretation Introduction

Interpretation:

The molar concentration of $H_3{O}^{+}$ ions and the pH of the given $\text{NaOH}$ solutions has to be calculated.

Concept Introduction:

$\text{pH:}$

The concentration of hydrogen ion in given solution is given as $\text{pH}$.

The negative logarithm of base ten hydrogen ion concentrations is known as $\text{pH}$.

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

Explanation of Solution

Given:

Volume and molarity of $\text{HCl}\text{=}\text{25}{\text{.0 cm}}^{\text{3}}\text{(0}\text{.025}L)\text{ and 0}\text{.144 M}$

Volume and molarity of $\text{NaOH}\text{=}\text{25}{\text{.0 cm}}^{\text{3}}\text{(0}\text{.025}L)\text{ and 0}\text{.125 M}$

Total volume of solution is $\text{0}\text{.025+0}\text{.025}\text{=}\text{0}\text{.050}\text{L}$

Mole of $\text{HCl}$ is,

  $\begin{array}{l}\text{=}\text{0}\text{.025L×0}\text{.144 M}\\ \text{=}\text{0}\text{.0036}\text{mol}\end{array}$

Mole of $\text{NaOH}$ is,

  $\begin{array}{l}\text{=}\text{0}\text{.025L×0}\text{.125 M}\\ \text{=}\text{0}\text{.003125}\text{mol}\end{array}$

The reaction between given acid and base is,

    $\begin{array}{l}{\text{HCl}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(aq)}}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+Cl}}_{\text{(aq)}}\\ {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}+{\text{HO}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }2H_2{O}_{(l)}\end{array}$

The IEC table for the above reaction is,

    $\begin{array}{l}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}+{\text{HO}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }2H_2{O}_{(l)}\\ Initial0.00360.0031250\\ Change0.0036-0.003125-0.0031250.003125\\ Equlibrium0.00047500.003125\end{array}$

From the above table the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$ remains in solution so

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$ in the solution is,

    ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}=\frac{0.000475mol}{0.050L}=0.0095M$

$pH$ of the solution is,

    $\begin{array}{l}pH=-\log [H_3{O}^{+}]\\ pH=-\log [0.0095M]\\ =2.022\end{array}$

(b)

Interpretation Introduction

Interpretation:

The molar concentration of $O{H}^{-}$ ions and the pH of the given $\text{HCl and KOH}$ solutions has to be calculated.

Concept Introduction:

$\text{pH:}$

The concentration of hydrogen ion in given solution is given as $\text{pH}$.

The negative logarithm of base ten hydrogen ion concentration is known as $\text{pH}$.

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

Explanation of Solution

Given:

Volume and molarity of $\text{HCl}\text{=}\text{25}{\text{.0 cm}}^{\text{3}}\text{(0}\text{.025}L)\text{ and 0}\text{.15 M}$

Volume and molarity of $\text{KOH}\text{=}35.{\text{0 cm}}^{\text{3}}\text{(0}\text{.035}L)\text{ and 0}\text{.15 M}$

Total volume of solution is $\text{0}\text{.025+0}\text{.035}\text{=}\text{0}\text{.060}\text{L}$

Mole of $\text{HCl}$ is,

  $\begin{array}{l}\text{=}\text{0}\text{.025}\text{L×0}\text{.15 M}\\ \text{=}\text{0}\text{.00375}\text{mol}\end{array}$

Mole of $\text{KOH}$ is,

  $\begin{array}{l}\text{=}\text{0}\text{.035L×0}\text{.15 M}\\ \text{=}\text{0}\text{.00525}\text{mol}\end{array}$

The reaction between given acid and base is,

    $\begin{array}{l}{\text{HCl}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(aq)}}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+Cl}}_{\text{(aq)}}\\ {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}+{\text{HO}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }2H_2{O}_{(l)}\end{array}$

The IEC table for the above reaction is,

    $\begin{array}{l}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}+{\text{HO}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }2H_2{O}_{(l)}\\ Initial0.003750.005250\\ Change-0.003750.00525-0.00375-0.00375\\ Equlibrium00.001560.00375\end{array}$

From the above table the concentration of ${\text{OH}}^{-}{}_{\text{(aq)}}$ remains in solution so

The concentration of ${\text{OH}}^{-}{}_{\text{(aq)}}$ in the solution is,

    ${\text{OH}}^{-}{}_{\text{(aq)}}=\frac{0.0015mol}{0.060L}=0.025M$

$pH$ of the solution is,

    $\begin{array}{l}pH=14-\log [O{H}^{-}]\\ pH=-\log [0.025]\\ =14-1.6020\\ =12.4\end{array}$

(c)

Interpretation Introduction

Interpretation:

The molar concentration of $H_3{O}^{+}$ ions and the pH of the given ${\text{HNO}}_{\text{3}}\text{ and NaOH}$ solutions has to be calculated.

Concept Introduction:

$\text{pH:}$

The concentration of hydrogen ion in given solution is given as $\text{pH}$.

The negative logarithm of base ten hydrogen ion concentration is known as $\text{pH}$.

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

Explanation of Solution

Given:

Volume and molarity of $\text{HCl}\text{=}21.2{\text{ cm}}^{\text{3}}\text{(0}\text{.0212}L)\text{ and 0}\text{.22 M}$

Volume and molarity of $\text{NaOH}\text{=}10.{\text{0 cm}}^{\text{3}}\text{(0}\text{.010}L)\text{ and 0}\text{.30 M}$

Total volume of solution is $\text{0}\text{.0212+0}\text{.010}\text{=}\text{0}\text{.0312}\text{L}$

Mole of $\text{HCl}$ is,

  $\begin{array}{l}\text{=}\text{0}\text{.0212L×0}\text{.22 M}\\ \text{=}0.00467\text{mol}\end{array}$

Mole of $\text{NaOH}$ is,

  $\begin{array}{l}\text{=}\text{0}\text{.010L×0}\text{.30 M}\\ \text{=}\text{0}\text{.003}\text{mol}\end{array}$

The reaction between given acid and base is,

    $\begin{array}{l}{\text{HCl}}_{\text{(aq)}}{\text{+H}}_{\text{2}}{\text{O}}_{\text{(aq)}}\stackrel{}{\to }{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}{\text{+Cl}}_{\text{(aq)}}\\ {\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}+{\text{HO}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }2H_2{O}_{(l)}\end{array}$

The IEC table for the above reaction is,

    $\begin{array}{l}{\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}+{\text{HO}}^{-}{}_{\text{(aq)}}\stackrel{}{\to }2H_2{O}_{(l)}\\ Initial0.004670.0030\\ Change0.00467-0.003-0.0030.003\\ Equlibrium0.0016700.003\end{array}$

From the above table the concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$ remains in solution so

The concentration of ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}$ in the solution is,

    ${\text{H}}_{\text{3}}{\text{O}}^{\text{+}}{}_{\text{(aq)}}=\frac{0.00137mol}{0.0312L}=0.05M$

$pH$ of the solution is,

    $\begin{array}{l}pH=-\log [H_3{O}^{+}]\\ pH=-\log [0.05M]\\ =1.30\end{array}$