Chapter 16.7, Problem 16.15WE

Caffeine, the stimulant in coffee and tea, is a weak base that ionizes in water according to the equation

${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}\text{(}aq\text{)}+{\text{H}}_{\text{2}}\text{O(}l\text{)}\rightleftarrows {\text{HC}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_2^{+}\text{(}aq{\text{)+OH}}^{-}(aq)$

A 0.15-M solution of caffeine at 25°C has a pH of 8.45. Determine the K_b of caffeine.

Interpretation Introduction

Interpretation:

The ${\text{K}}_b$ of the given 0.15 M solution of caffeine has to be calculated

Concept Information:

Base ionization constant: ${\text{K}}_{\text{b}}$

The ionization of a weak base $\text{B}$ is given by the below equation.

${\text{B}}_{(aq)}+{\text{H}}_{\text{2}}{\text{O}}_{(l)}\stackrel{}{\to }{\text{HB}}^{\text{+}}{}_{(aq)}+{\text{OH}}^{\text{-}}{}_{(aq)}$

The equilibrium expression for the ionization of weak base $\text{B}$ will be,

${\text{K}}_b=\frac{[{\text{HB}}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[\text{B}]}$

Where,

$K_b$ is base ionization constant,

$[{\text{OH}}^{-}\text{]}$ is concentration of hydroxide ion

$[{\text{HB}}^{\text{+}}\text{]}$  is concentration of conjugate acid

$[\text{B]}$ is concentration of the base

pOH definition:

The $\text{pOH}$ of a solution is defined as the negative base-10 logarithm of the hydroxide ion ${\text{[OH}}^{\text{-}}\text{]}$ concentration. $\text{pOH}$ scale is analogous to pH scale.

$\text{pOH}\text{=}{\text{-log[OH}}^{\text{-}}\text{]}$

Relationship between pH and pOH

$\text{pOH}$ 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.

The relationship between the hydronium ion concentration and the hydroxide ion concentration is given by the equation,

$\text{pH}\text{+}\text{pOH}\text{=}\text{14,}\text{at}\text{25}{}^{\text{o}}\text{C}$

As $\text{pOH}$ and $\text{pH}$ are opposite scale, the total of both has to be equal to 14.

To Calculate: The ${\text{K}}_b$ of the given 0.15 M solution of caffeine

Answer to Problem 16.15WE

Answer

The ${\text{K}}_b$ of the given 0.15 M solution of caffeine is $5.3\times {10}^{-11}$

Explanation of Solution

Given data:

Caffeine is the stimulant in tea and coffee.

Caffeine is a weak base which ionizes in water as follows,

${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}\text{O}\stackrel{}{\rightleftarrows }{\text{HC}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}{{}^{\text{+}}}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}$

The pH of the given 0.15 M solution of caffeine is 8.45

Calculation strategy:

Using the given pH, the pOH can be found, then from pOH obtained, the hydroxide ion concentration  is calculated.

From the obtained hydroxide ion concentration, use reaction stoichiometry to determine the other equilibrium concentrations, then evaluate ${\text{K}}_b$

Calculation of pOH:

The pOH can be calculated using the following formula as follows,

$\begin{array}{l}\text{ pH+pOH}\text{= 14}\\ \text{ pOH}\text{= 14}\text{.00-8}\text{.45}\\ \text{= 5}\text{.55}\end{array}$

Therefore, the pOH of the given caffeine solution is 5.55

Calculation of hydroxide ion:

The hydroxide ion can be calculated as follows,

$\begin{array}{l}\text{pOH}{\text{= -log[OH}}^{\text{-}}]\\ {\text{[OH}}^{\text{-}}\text{]}{\text{= 10}}^{\text{-pOH}}\\ [{\text{OH}}^{\text{-}}]={\text{10}}^{\text{-5}\text{.55}}\\ =2.82\times {10}^{-6}M\end{array}$

Therefore, the concentration of $[{\text{OH}}^{\text{-}}]$= $2.82\times {10}^{-6}M$

Calculation of ${\text{K}}_b$

The concentration of hydroxide ion for the given weak base can be found out from its equilibrium reaction with water.

Based on the stoichiometry, ${\text{[HC}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}{}^{\text{+}}{\text{]=[OH}}^{\text{-}}\text{]}$

If the concentration of ${\text{OH}}^{\text{-}}$ is $2.82\times {10}^{-6}M$ at equilibrium, it must mean that $2.82\times {10}^{-6}M$ of the base is ionized.

We summarize the changes as follows,

	${\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}{}_{\text{(aq)}}{\text{ + H}}_{\text{2}}\text{O}\stackrel{}{\rightleftarrows }{\text{HC}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}{{}^{\text{+}}}_{\text{(aq)}}{\text{+ OH}}^{\text{-}}{}_{\text{(aq)}}$
Initial $(M)$	0.15 $-2.82\times {10}^{-6}$ $(0.15-2.82\times {10}^{-6})$	$0.00$	$0.00$
Change $(M)$		$+2.82\times {10}^{-6}$	$+2.82\times {10}^{-6}$
Equilibrium $(M)$		$2.82\times {10}^{-6}$	$2.82\times {10}^{-6}$

${\text{[C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}]$= (0.15- $2.82\times {10}^{-6}$) M

$\approx 0.15$ M

$\begin{array}{l}{\text{K}}_b=\frac{[{\text{HC}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}{}^{\text{+}}][{\text{OH}}^{\text{-}}]}{[{\text{C}}_{\text{8}}{\text{H}}_{\text{10}}{\text{N}}_{\text{4}}{\text{O}}_{\text{2}}]}\\ {\text{K}}_b=\frac{{(2.82\times {10}^{-6})}^2}{(0.15)}\\ =\text{ 5}\text{.3}\times {10}^{-11}\end{array}$

Therefore, the base ionization constant ${\text{K}}_b$ of the given caffeine solution is $\text{ 5}\text{.3}\times {10}^{-11}$

Conclusion

The ${\text{K}}_b$ of the given 0.15 M solution of caffeine was calculated