Chapter 1, Problem 11P

Interpretation Introduction

Interpretation:

The values of ratio of acetate ion to acetic acid and ethylamine to ethylammonium is to be determined.

Concept introduction:

The expression to calculate ${\text{H}}^{+}$ ion concentration is as follows,

$\text{pH}=-\log \left[{\text{H}}^{+}\right]$ ...... (1)

The expression for the ratio of acetate to acetic acid, which can be done with help of expression of $\text{p}{K}_{\text{a}}$ given below,

$K_{\text{a}}=\frac{\left[{\text{CH}}_3{\text{COO}}^{-}\right]\left[{\text{H}}^{+}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}$ ...... (2)

Where,

• $\left[{\text{CH}}_3{\text{COO}}^{-}\right]$ is the concentration of acetate ion.
• $\left[{\text{CH}}_3\text{COOH}\right]$ is the concentration of acetic acid.
• $\left[{\text{H}}^{+}\right]$ is the concentration of hydrogen ions.
• $K_{\text{a}}$ is equilibrium constant of the acid.

The ratio of ethylamine to ethylammonium is calculated as follows:

$K_{\text{b}}\text{=}\frac{\left[{\text{C}}_2{\text{H}}_5{\text{NH}}_3{}^{+}\right]\left[{\text{OH}}^{-}\right]}{\left[{\text{C}}_2{\text{H}}_5{\text{NH}}_2\right]}$ ...... (3)

Where,

• $K_{\text{b}}$ is the dissociation constant of the base.
• $\left[{\text{C}}_2{\text{H}}_5{\text{NH}}_3{}^{+}\right]$ is the concentration of ethylammonium ion.
• $\left[{\text{OH}}^{-}\right]$ is the concentration of hydroxyl ion.
• $\left[{\text{C}}_2{\text{H}}_5{\text{NH}}_2\right]$ is the concentration of ethylamine.

Answer to Problem 11P

The ratio of acetate to acetic acid at $\text{pH}$

$7.4$ is $446.67$ and that of ethylamine to ethylammonium is $5.0119\times {10}^{-4}$.

Explanation of Solution

Rearrange equation (1) for $\left[{\text{H}}^{+}\right]$ ,

$\left[{\text{H}}^{+}\right]={10}^{-\text{pH}}$ ...... (4)

The value of $\text{pH}$ the solution is $7.4$.

Substitute the value of $\text{pH}$ in equation (4),

$\begin{array}{c}\left[{\text{H}}^{+}\right]={10}^{-7.4}\\ =3.9810\times {10}^{-8}\text{M}\end{array}$

The value of concentration of ${\text{H}}^{\text{+}}$ ion is $3.9810\times {10}^{-8}\text{M}$.

The expression to calculate $\text{p}{K}_{\text{a}}$ is as follows:

$\text{p}{K}_{\text{a}}=-\log \left(K_{\text{a}}\right)$ ...... (5)

Rearrange equation (5) for $K_{\text{a}}$.

$K_{\text{a}}={10}^{-\text{p}{K}_{\text{a}}}$ ...... (6)

The $\text{p}{K}_{\text{a}}$ value of acetic acid from problem 7 is 4.75.

Substitute the value of $\text{p}{K}_{\text{a}}$ in equation (6).

$\begin{array}{c}{K}_{\text{a}}={10}^{-4.75}\\ =1.7782\times {10}^{-5}\end{array}$

Rearrange equation (2) for $\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}$ ,

$\frac{\left[{\text{CH}}_3{\text{COO}}^{-}\right]}{\left[{\text{CH}}_3\text{COOH}\right]}=\frac{K_{\text{a}}}{\left[{\text{H}}^{+}\right]}$ ...... (7)

The value of $K_{\text{a}}$ is $1.7782\times {10}^{-5}$.

The value of $\left[{\text{H}}^{\text{+}}\right]$ is $3.9810\times {10}^{-8}\text{M}$.

Substitute the values of $\left[{\text{H}}^{\text{+}}\right]$ and $K_{\text{a}}$ in equation (7),

$\begin{array}{c}\frac{\left[{\text{CH}}_{\text{3}}{\text{COO}}^{-}\right]}{\left[{\text{CH}}_{\text{3}}\text{COOH}\right]}=\frac{1.7782\times {10}^{-5}}{3.9810\times {10}^{-8}}\\ =446.67\end{array}$

The value of $\left[{\text{H}}^{\text{+}}\right]$ is available, to find the value of $\left[{\text{OH}}^{-}\right]$ the expression for dissociation of water at room temperature is used.

$1\times {10}^{-14}=\left[{\text{H}}^{+}\right]\left[{\text{OH}}^{-}\right]$ ...... (8)

Rearrange equation for (8) for $\left[{\text{OH}}^{-}\right]$ ,

$\begin{array}{l}\left[{\text{OH}}^{-}\right]=\frac{1\times {10}^{-14}}{3.9810\times {10}^{-8}}\\ \left[{\text{OH}}^{-}\right]=2.5119\times {10}^{-7}\end{array}$

The value of $\text{p}{K}_{\text{a}}$ is given as 10.70 in problem 10. The expression that relates $\text{p}{K}_{\text{b}}$ with $\text{p}{K}_{\text{a}}$ is as follows:

$\text{p}{K}_{\text{a}}+\text{p}{K}_{\text{b}}=14$ ...... (9)

Rearrange equation (9) for $\text{p}{K}_{\text{b}}$ ,

$\text{p}{K}_{\text{b}}=14-\text{p}{K}_{\text{a}}$ ...... (10)

The value of $\text{p}{K}_{\text{a}}$ is 10.70.

Substitute the value of $\text{p}{K}_{\text{a}}$ in equation (10),

$\begin{array}{c}\text{p}{K}_{\text{b}}=14-10.70\\ =3.3\end{array}$

To find the value of $K_{\text{b}}$ the following expression is used,

$\text{p}{K}_{\text{b}}=-\text{log}\left(K_{\text{b}}\right)$

Rearrange for $K_{\text{b}}$ ,

$K_{\text{b}}={10}^{-\text{p}{K}_{\text{b}}}$

Substitute the value of $\text{p}{K}_{\text{b}}$ in the above equation,

$\begin{array}{c}{K}_{\text{b}}={10}^{-3.3}\\ =5.0118\times {10}^{-4}\end{array}$

Rearrange the above equation (3) for $\frac{\left[{\text{C}}_2{\text{H}}_5{\text{NH}}_2\right]}{\left[{\text{C}}_2{\text{H}}_5{\text{NH}}_3{}^{+}\right]}$ ,

$\frac{\left[{\text{C}}_2{\text{H}}_5{\text{NH}}_3{}^{+}\right]}{\left[{\text{C}}_2{\text{H}}_5{\text{NH}}_2\right]}=\frac{K_{\text{b}}}{\left[{\text{OH}}^{-}\right]}$

Substitute the values of $K_{\text{b}}$, $\left[{\text{OH}}^{-}\right]$ in the above equation,

$\begin{array}{c}\frac{\left[{\text{C}}_2{\text{H}}_5{\text{NH}}_2\right]}{\left[{\text{C}}_2{\text{H}}_5{\text{NH}}_3{}^{+}\right]}=\frac{2.5119\times {10}^{-7}}{5.0118\times {10}^{-4}}\\ =5.0119\times {10}^{-4}\end{array}$

Thus, the value of ethylamine to ethyl ammonium is $5.0119\times {10}^{-4}$.

Conclusion

Thus, the ratio of acetate to acetic acid at $\text{pH}$ 7.4 is $446.67$ and that of ethylamine to ethylammonium is $5.0119\times {10}^{-4}$.