Chapter 15, Problem 15.57QP

(a)

Interpretation Introduction

Interpretation: The $K_{\text{b}}$ value of aminoethanol, ${\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2$ is given to be $3.1\times {10}^{-5}$.

A comparison in the basic strength of aminoethanol and ethylamine is to be stated.

The $\text{pH}$ of $1.67\times {10}^{-2}\text{M}$ solution of aminoethanol is to be calculated.

The concentration of $\left[{\text{OH}}^{-}\right]$ in $4.25\times {10}^{-4}\text{M}$ solution of aminoethanol is to be calculated.

Concept introduction: The weak bases are the substances that don’t ionize completely in to an anion and its contrary part. The equilibrium constant helps to find out the association between concentration of reaction and product at equilibrium point.

To determine: If the aminoethanol is a stronger or weaker base than ethylamine.

Answer to Problem 15.57QP

Solution

Aminoethanol is a weaker base than that of ethylamine.

Explanation of Solution

Explanation

Given

The equilibrium constant, $K_{\text{b}}$ of aminoethanol is $3.1\times {10}^{-5}$.

The $\text{p}{K}_{\text{b}}$ value of ethylamine is $3.36$.

The ionization of aminoethanol is shown by the equation,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\rightleftharpoons {\text{CH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}+{\text{OH}}^{-}$

The $\text{p}{K}_{\text{b}}$ of solution of aminoethanol is calculated by the formula,

$\text{p}{K}_{\text{b}}={10}^{-K_{\text{b}}}$ (1)

Substitute the value of $K_{\text{b}}$ in equation (1).

$\begin{array}{c}\text{p}{K}_{\text{b}}={10}^{-3.1\times {10}^{-5}}\\ =0.999\end{array}$

Therefore, on comparing the $\text{p}{K}_{\text{b}}$ value of aminoethanol and ethylamine, it is clear that aminoethanol is a weaker base than that of ethylamine due to less $\text{p}{K}_{\text{b}}$ value of aminoethanol.

(b)

Interpretation Introduction

To determine: The $\text{pH}$ of $1.67\times {10}^{-2}\text{M}$ solution of aminoethanol.

Answer to Problem 15.57QP

Solution

The $\text{pH}$ of $1.67\times {10}^{-2}\text{M}$ solution of aminoethanol is $\underset{\_}{10.857}$.

Explanation of Solution

Explanation

Given

The equilibrium constant, $K_{\text{b}}$ of aminoethanol is $3.1\times {10}^{-5}$.

The total concentration, $\left[{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\right]$ of solution of aminoethanol is $1.67\times {10}^{-2}\text{M}$.

The ionization of aminoethanol is shown by the equation,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\rightleftharpoons {\text{CH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}+{\text{OH}}^{-}$

The calculation of concentration of $\left[{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\right]$ and $[{\text{OH}}^{-}]$ is shown by the $\text{ICE}$ table.

$\begin{array}{cccc}& \left[{\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2\right]& \left[{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\right]& \left[{\text{OH}}^{-}\right]\\ \text{Initial}& 1.67\times {10}^{-2}\text{M}& 0& 0\\ \text{Change}& -x& x& x\\ \text{Equilbrium}& \left(1.67\times {10}^{-2}-x\right)\text{M}& x& x\end{array}$

The equilibrium constant is calculated by the formula,

$K_{\text{b}}=\frac{\left[{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\right]\times \left[{\text{OH}}^{-}\right]}{{\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2}$ (1)

Substitute the values of equilibrium constant, concentration of ions and compound in equation (1).

$\begin{array}{c}3.1\times {10}^{-5}\text{M}=\frac{\left(x\right)\times \left(x\right)}{\left(1.67\times {10}^{-2}\text{M}\right)-x}\\ \approx \frac{x^2}{1.67\times {10}^{-2}\text{M}}\end{array}$ (2)

In equation (2), $x$ is very small as compared to $\left(1.67\times {10}^{-2}\text{M}\right)$. So, it is neglected.

$\begin{array}{c}{x}^2=\left(1.67\times {10}^{-2}\text{M}\right)\times \left(3.1\times {10}^{-5}M\right)\\ x=\sqrt{0.0000005177}\\ =7.195\times {10}^{-4}\text{M}\end{array}$

Thus, the concentration of both ions $\left[{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ in the solution is

$7.195\times {10}^{-4}\text{M}$.

The $\text{pOH}$ is calculated by the formula,

$\text{pOH}=-\text{log}\left[{\text{OH}}^{-}\right]$ (3)

Substitute the value of $\left[{\text{OH}}^{-}\right]$ in equation (3).

$\begin{array}{c}\text{pOH}=-\text{log}\left(7.195\times {10}^{-4}\text{M}\right)\\ =3.143\end{array}$

The $\text{pH}$ is calculated by using $\text{pOH}$ shown as,

$\begin{array}{c}\text{pH}=14-\text{pOH}\\ =\text{14}-\text{3}\text{.143}\\ =\underset{\_}{10.857}\end{array}$

Therefore, the $\text{pH}$ of aminoethanol is $\underset{\_}{10.857}$.

(c)

Interpretation Introduction

To determine: The concentration of $\left[{\text{OH}}^{-}\right]$ in $4.25\times {10}^{-4}\text{M}$ solution of aminoethanol.

Answer to Problem 15.57QP

Solution

The concentration of $\left[{\text{OH}}^{-}\right]$ in the $4.25\times {10}^{-4}\text{M}$ solution is $\underset{\_}{1.15\times 10^{-4}M}$.

Explanation of Solution

Explanation

Given

The equilibrium constant, $K_{\text{b}}$ of aminoethanol is $3.1\times {10}^{-5}$.

The concentration, $\left[{\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\right]$ of solution of aminoethanol is $4.25\times {10}^{-4}\text{M}$.

The ionization of aminoethanol is shown by the equation,

${\text{OHCH}}_2{\text{CH}}_2{\text{NH}}_2\rightleftharpoons {\text{CH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}+{\text{OH}}^{-}$

The calculation of concentration of $\left[{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\right]$ and $[{\text{OH}}^{-}]$ is shown by the $\text{ICE}$ table.

$\begin{array}{cccc}& \left[{\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2\right]& \left[{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\right]& \left[{\text{OH}}^{-}\right]\\ \text{Initial}& 4.25\times {10}^{-4}\text{M}& 0& 0\\ \text{Change}& -x& x& x\\ \text{Equilbrium}& \left(4.25\times {10}^{-4}-x\right)\text{M}& x& x\end{array}$

The equilibrium constant is calculated by the formula,

$K_{\text{b}}=\frac{\left[{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\right]\times \left[{\text{OH}}^{-}\right]}{{\text{HOCH}}_2{\text{CH}}_2{\text{NH}}_2}$ (1)

Substitute the values of equilibrium constant, concentration of ions and compound in equation (1).

$\begin{array}{c}3.1\times {10}^{-5}\text{M}=\frac{\left(x\right)\times \left(x\right)}{\left(4.25\times {10}^{-4}\text{M}\right)-x}\\ \approx \frac{x^2}{4.25\times {10}^{-4}\text{M}}\end{array}$ (2)

In equation (2), $x$ is very small as compared to $\left(1.67\times {10}^{-2}\text{M}\right)$. So, it is neglected.

$\begin{array}{c}{x}^2=\left(4.25\times {10}^{-4}\text{M}\right)\times \left(3.1\times {10}^{-5}M\right)\\ x=\sqrt{0.000000013175}\\ =\underset{\_}{1.15\times 10^{-4}M}\end{array}$

Thus, the concentration of both ions $\left[{\text{CH}}_2{\text{CH}}_2{\text{NH}}_3{}^{+}\right]$ and $\left[{\text{OH}}^{-}\right]$ in the solution is

$\underset{\_}{1.15\times 10^{-4}M}$.

Conclusion

1. a) Aminoethanol is a weaker base than that of ethylamine.
2. b) The $\text{pH}$ of $1.67\times {10}^{-2}\text{M}$ solution of aminoethanol is $\underset{\_}{10.857}$.
3. c) The concentration of $\left[{\text{OH}}^{-}\right]$ in the $4.25\times {10}^{-4}\text{M}$ solution is $\underset{\_}{1.15\times 10^{-4}M}$.