Chapter 5, Problem 5.29P

(a)

Interpretation Introduction

Interpretation:

The 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

$\begin{array}{l}{\text{C}}_{\text{6}}{\text{H}}_5\text{COOH}\text{+}{\text{H}}_{\text{2}}\text{O}\to {\text{C}}_{\text{6}}{\text{H}}_5{\text{COO}}^{-}+H_3{O}^{+}\\ Initial0.25000\\ Change-xxx\\ ---------------------------------\\ Final\text{0}\text{.250-x}xx\end{array}$

$\begin{array}{c}{K}_a=\frac{\left[H_3{O}^{+}\right]\left[C_6{H}_{5}CO{O}^{-}\right]}{\left[C_6{H}_{5}COOH\right]}\\ =\frac{x.x}{0.250-x}\\ 6.5\times {10}^{-5}=\frac{x^2}{0.250}\\ x=\sqrt{6.5\times {10}^{-5}\times 0.250}\\ \left[H_3{O}^{+}\right]=4\times {10}^{-3}M\end{array}$

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

$\begin{array}{c}=\frac{4\times {10}^{-3}}{0.250}\\ =0.016\\ Percentage=0.016\times 100\%\\ =1.6\%\end{array}$

(b)

Interpretation Introduction

Interpretation:

The 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

$\begin{array}{l}{\text{NH}}_{\text{2}}{\text{NH}}_{\text{2}}{\text{+H}}_{\text{2}}\text{O}\to {\text{NH}}_{\text{2}}{\text{NH}}_{\text{3}}{}^{+}+O{H}^{-}\\ Initial0.15000\\ Change-xxx\\ -----------------------------\\ Final0.150\text{-x}xx\end{array}$

$\begin{array}{c}{K}_a=\frac{\left[H{O}^{-}\right]\left[N{H}_{2}N{H}_3^{+}\right]}{\left[N{H}_{2}N{H}_2\right]}\\ =\frac{x.x}{\left(0.150\right)-x}\\ 1.7\times {10}^{-6}=\frac{x^2}{0.150}\\ x=\sqrt{1.7\times {10}^{-6}\times 0.150}\\ \left[H_3{O}^{+}\right]=5\times {10}^{-4}M\end{array}$

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

$\begin{array}{c}=\frac{5\times {10}^{-4}}{0.150}\\ =3.37\times {10}^{-3}\\ Percentage=0.016\times 100\%\\ =0.34\%\end{array}$

(c)

Interpretation Introduction

Interpretation:

The 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

$\begin{array}{l}{\left(C{H}_3\right)}_{\text{3}}\text{N}\text{+}{\text{H}}_{\text{2}}\text{O}\to {\left(C{H}_3\right)}_{\text{3}}\text{N}{H}^{+}+O{H}^{-}\\ Initial0.11200\\ Change-xxx\\ -----------------------------\\ Final0.112\text{-x}xx\end{array}$

$\begin{array}{c}{K}_a=\frac{\left[H{O}^{-}\right]\left[{\left(C{H}_3\right)}_{3}N{H}^{+}\right]}{\left[{\left(C{H}_3\right)}_{3}N\right]}\\ =\frac{x.x}{\left(0.112\right)-x}\\ 6.5\times {10}^{-5}=\frac{x^2}{0.112}\\ x=\sqrt{6.5\times {10}^{-5}\times 0.112}\\ \left[H_3{O}^{+}\right]=2.7\times {10}^{-3}M\end{array}$

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

$\begin{array}{c}=\frac{2.7\times {10}^{-3}}{0.112}\\ =0.024\\ Percentage=0.016\times 100\%\\ =2.41\%\end{array}$