Chapter 5, Problem 5.30P

(a)

Interpretation Introduction

Interpretation:

The value of $pH,pOH$ and fraction of solute protonated or deprotonated in the given aqueous solution of lactic acid has to be calculated.

Concept Introduction:

• $\text{pH}$: The concentration of hydrogen ion is measured using $\text{pH}$ scale.  The $\text{pH}$ of a solution is a figure that expresses the acidity or the alkalinity of a given solution.
• It is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

• If the value of $\text{pH}$ is less than $7$, then the solution is acidic whereas if the value of $\text{pH}$ is greater than $7$, then the solution is basic.
• $\text{pOH}$ is similar to $\text{pH}$. The only difference is that in $\text{pOH}$ the concentration of hydroxide ion is used as a scale while in $\text{pH}$, the concentration of hydronium ion is used.

Explanation of Solution

  $\begin{array}{l}{\text{CH}}_{\text{3}}\text{CH}\left(\text{OH}\right){\text{COOH+H}}_{\text{2}}\text{O}\to {\text{CH}}_{\text{3}}\text{CH}\left(\text{OH}\right){\text{COO}}^{-}+H_3{O}^{+}\\ Initial0.15000\\ Change-xxx\\ ---------------------------------------\\ Final\text{0}\text{.150-x}xx\end{array}$

$\begin{array}{c}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[C{H}_{3}CH\left(OH\right)CO{O}^{-}\right]}{\left[C{H}_{3}CH\left(OH\right)COOH\right]}\\ =\frac{x.x}{0.150-x}\\ 8.4\times {10}^{-4}=\frac{x^2}{0.150}\\ x=\sqrt{8.4\times {10}^{-4}\times 0.150}\\ \left[H_3{O}^{+}\right]=0.011M\end{array}$

$\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ =-\log \left(0.011M\right)\\ =1.95\\ pOH=14-pH\\ =14-1.95\\ =\mathrm{12.05.}\end{array}$

The fraction deprotonated from given acid concentration is determined as shown below,

$\begin{array}{l}=\frac{0.011}{0.150}\\ =\mathrm{0.073.}\end{array}$

(b)

Interpretation Introduction

Interpretation:

The value of $pH,pOH$ and fraction of solute protonated or deprotonated in the given aqueous solution has to be calculated.

Concept Introduction:

• $\text{pH}$: The concentration of hydrogen ion is measured using $\text{pH}$ scale. The $\text{pH}$ of a solution is a figure that expresses the acidity or the alkalinity of a given solution.
• It is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

• If the value of $\text{pH}$ is less than $7$, then the solution is acidic whereas if the value of $\text{pH}$ is greater than $7$, then the solution is basic.
• $\text{pOH}$ is similar to $\text{pH}$. The only difference is that in $\text{pOH}$ the concentration of hydroxide ion is used as a scale while in $\text{pH}$, the concentration of hydronium ion is used.

Explanation of Solution

$\begin{array}{l}{\text{CH}}_{\text{3}}\text{CH}\left(\text{OH}\right){\text{COOH+H}}_{\text{2}}\text{O}\to {\text{CH}}_{\text{3}}\text{CH}\left(\text{OH}\right){\text{COO}}^{-}+H_3{O}^{+}\\ Initial2.4\times {10}^{-4}00\\ Change-xxx\\ ---------------------------------------\\ Final2.4\times {10}^{-4}\text{-x}xx\end{array}$

$\begin{array}{c}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[C{H}_{3}CH\left(OH\right)CO{O}^{-}\right]}{\left[C{H}_{3}CH\left(OH\right)COOH\right]}\\ =\frac{x.x}{\left(2.4\times {10}^{-4}\right)-x}\\ 8.4\times {10}^{-4}=\frac{x^2}{2.4\times {10}^{-4}}\\ x=\sqrt{8.4\times {10}^{-4}\times 2.4\times {10}^{-4}}\\ \left[H_3{O}^{+}\right]=4.49\times {10}^{-4}M\end{array}$

$\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ =-\log \left(4.49\times {10}^{-4}M\right)\\ =3.35\\ pOH=14-pH\\ =14-3.35\\ =10.65\end{array}$

The fraction deprotonated from given acid concentration is determined as shown below,

$\begin{array}{l}=\frac{4.49\times {10}^{-4}}{2.4\times {10}^{-4}}\\ =\mathrm{1.9.}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The value of $pH,pOH$ and fraction of solute protonated or deprotonated in the given aqueous solution has to be calculated.

Concept Introduction:

• $\text{pH}$: The concentration of hydrogen ion is measured using $\text{pH}$ scale. The $\text{pH}$ of a solution is a figure that expresses the acidity or the alkalinity of a given solution.
• It is defined as the negative base-10 logarithm of the hydrogen or hydronium ion concentration.

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

• If the value of $\text{pH}$ is less than $7$, then the solution is acidic whereas if the value of $\text{pH}$ is greater than $7$, then the solution is basic.
• $\text{pOH}$ is similar to $\text{pH}$. The only difference is that in $\text{pOH}$ the concentration of hydroxide ion is used as a scale while in $\text{pH}$, the concentration of hydronium ion is used.

Explanation of Solution

  $\begin{array}{l}{\text{C}}_6{\text{H}}_{\text{5}}{\text{SO}}_{\text{3}}{\text{H+H}}_{\text{2}}\text{O}\to {\text{C}}_6{\text{H}}_{\text{5}}{\text{SO}}_{\text{3}}{}^{-}+H_3{O}^{+}\\ Initial0.2500\\ Change-xxx\\ ---------------------------------------\\ Final0.25\text{-x}xx\end{array}$

$\begin{array}{c}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[C_6{H}_{5}S{O}_3{}^{-}\right]}{\left[C_6{H}_{5}S{O}_{3}H\right]}\\ =\frac{x.x}{\left(0.25\right)-x}\\ 0.20=\frac{x^2}{0.25-x}\\ x^2+0.20x-0.050=0\\ x=0.145\\ \left[H_3{O}^{+}\right]=0.145M\end{array}$

$\begin{array}{c}pH=-\log \left[H_3{O}^{+}\right]\\ =-\log \left(0.145M\right)\\ =0.8\\ pOH=14-pH\\ =14-0.8\\ =13.2\end{array}$

The fraction deprotonated from given acid concentration is determined as shown below,

$\begin{array}{l}=\frac{0.145}{0.25}\\ =\mathrm{0.58.}\end{array}$