Chapter 15, Problem 15.61QA

Interpretation Introduction

To:

a) Determine aminoethanol is a stronger or weaker base than ethylamine.

b) Calculate $pH$ of $1.67\times {10}^{-2}M$ aminoethanol.

c) Calculate the $\left[O{H}^{-}\right]$ concentration of $4.25\times {10}^{-4}M$ aminoethanol.

Answer to Problem 15.61QA

Solution:

a) Aminoethanol is a weaker base than ethylamine.

b) $pH=10.86$

c) $\left[O{H}^{-}\right]=1.000\times {10}^{-4}M$

Explanation of Solution

1) Concept:

We are calculating the  $\mathrm{p}{\mathrm{K}}_{\mathrm{b}}$ using given ${\mathrm{K}}_{\mathrm{b}}$ for aminoethanol. Then we can compare the $\mathrm{p}{\mathrm{K}}_{\mathrm{b}}$ of both the bases. The greater the ${\mathrm{K}}_{\mathrm{b}}$ value, stronger is the base. Therefore, the lower the value of $\mathrm{p}{\mathrm{K}}_{\mathrm{b}}$, stronger is the base.

$K_b=\frac{{\left[Product\right]}^{coefficents}}{{\left[Reactant\right]}^{coefficents}}$

Here, ${\mathrm{K}}_{\mathrm{b}}$ is equilibrium constant of base, $\left[\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{d}\mathrm{u}\mathrm{c}\mathrm{t}\mathrm{s}\right]$ is the concentration of products, $\left[\mathrm{R}\mathrm{e}\mathrm{a}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{t}\right]$ is the concentration of reactants, and coefficient are the coefficient of reactants and product from the balanced reaction.

2) Formulae:

i) $K_b=\frac{{\left[Product\right]}^{coefficents}}{{\left[Reactant\right]}^{coefficents}}$

ii) $p{K}_b= -\mathrm{l}\mathrm{o}\mathrm{g}\left[K_b\right]$

iii) $pH= -\mathrm{l}\mathrm{o}\mathrm{g}\left[H3O^{+}\right]$

iv) $pOH= -\mathrm{l}\mathrm{o}\mathrm{g}\left[O{H}^{-}\right]$

v) $pH+pOH=14$

3) Given:

i) $K_b of aminoethanol=3.1\times {10}^{-5}$

ii) $p{K}_{b}of ethylamine=3.36$

iii) $\left[HOC{H}_{2}C{H}_{2}N{H}_2\right]=1.67\times {10}^{-2}M$

iv) $\left[HOC{H}_{2}C{H}_{2}N{H}_2\right]=4.25\times {10}^{-4}M$

4) Calculations:

a) To compare the Kb values of aminoethanol and ethylamine.

$p{K}_b= -\mathrm{l}\mathrm{o}\mathrm{g}\left[K_b\right]$

$p{K}_b= -\log \left[3.1\times {10}^{-5}\right]=4.51$

Here we see that $\mathrm{p}{\mathrm{K}}_{\mathrm{b}}$ of aminoethanol is greater than the $\mathrm{p}{\mathrm{K}}_{\mathrm{b}}$ of ethylamine. Therefore, aminoethanol is a weaker base than ethylamine.

b) To calculate the pH of aminoethanol from the given concentration and RICE table.

RICE table for the given reaction:

Reaction	$HOC{H}_{2}C{H}_{2}N{H_2}_{aq)} + H_2{O}_{\left(l\right)} \rightleftharpoons H_{2}OC{H}_{2}C{H}_{2}N{H_2^{+}}_{aq)}+ O{H}_{\left(aq\right)}^{-}$
	$\left[HOC{H}_{2}C{H}_{2}N{H}_2\right]\left(M\right)$		$\left[H_{2}OC{H}_{2}C{H}_{2}N{H}_2^{+}\right] \left(M\right)$	$\left[O{H}^{-}\right] \left(M\right)$
Initial(I)	$1.67\times {10}^{-2}$		$0$	$0$
Change(C)	$-x$		$+x$	$+x$
Equilibrium(E)	$(1.67\times {10}^{-2}-x)$		$x$	$x$

Therefore, ${\mathrm{K}}_{\mathrm{b}}$ expression for this reaction is

$K_b=\frac{\left[H_{2}OC{H}_{2}C{H}_{2}N{H}_2^{+}\right]\left[O{H}^{-}\right]}{\left[HOC{H}_{2}C{H}_{2}N{H}_2\right]}=\frac{\left(x\right)\left(x\right)}{(1.67\times {10}^{-2}-x)}$

$3.1\times {10}^{-5}=\frac{x^2}{(1.67\times {10}^{-2}-x)}$

To solve for $x$, we make the simplifying assumption that $x$ will be small compared to $1.67\times {10}^{-2}$ M because ${ \left[HOC{H}_{2}C{H}_{2}N{H}_2\right]}_{initial}$ is $1.67\times {10}^{-2}/(3.1\times {10}^{-5}) =538.71$ times the $K_b$ value, which is much greater than our $500\times$ guideline.

$3.1\times {10}^{-5}=\frac{x^2}{1.67\times {10}^{-2}}$

$3.1\times {10}^{-5}\times 1.67\times {10}^{-2}=x^2$

$x^2=5.18\times {10}^{-7}$

$x=7.20\times {10}^{-4} M$

So check the reliability of this $>500$ test. Calculation is given below:

The ratio of the smaller initial partial pressure to the value of $x is \frac{1.67\times {10}^{-2}}{7.20\times {10}^{-4}}=23$

This value is less than $500$. Therefore, our assumption is valid. So, we can use $x=7.19\times {10}^{-4}M$ for further calculations.

$\left[O{H}^{-}\right]=x=7.20\times {10}^{-4} M$

$pOH= -\log \left[7.20\times {10}^{-4}\right]=3.14$

$pH+3.14=14$

$pH=14-3.14=10.86$

c) To calculate the [OH-] from the given concentration.

RICE table for the given reaction:

Reaction	$HOC{H}_{2}C{H}_{2}N{H_2}_{aq)} + H_2{O}_{\left(l\right)} \rightleftharpoons H_{2}OC{H}_{2}C{H}_{2}N{H_2^{+}}_{aq)}+ O{H}_{\left(aq\right)}^{-}$
	$\left[HOC{H}_{2}C{H}_{2}N{H}_2\right]\left(M\right)$		$\left[H_{2}OC{H}_{2}C{H}_{2}N{H}_2^{+}\right] \left(M\right)$	$\left[O{H}^{-}\right] \left(M\right)$
Initial(I)	$4.25\times {10}^{-4}$		$0$	$0$
Change(C)	$-x$		$+x$	$+x$
Equilibrium(E)	$(4.25\times {10}^{-4}-x)$		$x$	$x$

Therefore, $K_b$ expression for this reaction is

$K_b=\frac{\left[H_{2}OC{H}_{2}C{H}_{2}N{H}_2^{+}\right]\left[O{H}^{-}\right]}{\left[HOC{H}_{2}C{H}_{2}N{H}_2\right]}=\frac{\left(x\right)\left(x\right)}{(4.25\times {10}^{-4}-x)}$

$3.1\times {10}^{-5}=\frac{x^2}{(4.25\times {10}^{-4}-x)}$

$3.1\times {10}^{-5}\times \left(4.25\times {10}^{-4}-x\right)=x^2$

$1.32\times {10}^{-8}-\left(3.1\times {10}^{-5}\right)x=x^2$

$x^2+\left(3.1\times {10}^{-5}\right)x-1.32\times {10}^{-8}=0$

Solving this quadratic equation for x yields a positive value and a negative value.

We can get two values of $x\mathrm{ }$ as  $\mathrm{x}=1.000\times {10}^{-4}\mathrm{ }\mathrm{a}\mathrm{n}\mathrm{d}\mathrm{ }\mathrm{x}=\mathrm{ }-1.31\times {10}^{-4}$

We will use the positive value of $x$ for further calculation because concentration is never negative.

$\left[O{H}^{-}\right]=x=1.000\times {10}^{-4}M$

Conclusion:

We have compared the $p{K}_b$ values to determine the strength of given base. Using RICE table and $K_b$ expression, we have calculated the $\left[O{H}^{-}\right], pOH and pH$.