Chemistry: Atoms First
Chemistry: Atoms First
3rd Edition
ISBN: 9781259638138
Author: Julia Burdge, Jason Overby Professor
Publisher: McGraw-Hill Education
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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

Chapter 16.7, Problem 16.15WE, Caffeine, the stimulant in coffee and tea, is a weak base that ionizes in water according to the

C 8 H 10 N 4 O 2 ( a q ) + H 2 O( l ) HC 8 H 10 N 4 O 2 + ( a q )+OH ( a q )

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

Expert Solution & Answer
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Interpretation Introduction

Interpretation:

The Kb of the given 0.15 M solution of caffeine has to be calculated

Concept Information:

Base ionization constant: Kb

The ionization of a weak base B is given by the below equation.

B(aq)+H2O(l)HB+(aq)+OH-(aq)

The equilibrium expression for the ionization of weak base B will be,

Kb=[HB+][OH-][B]

Where,

Kb is base ionization constant,

[OH] is concentration of hydroxide ion

[HB+]   is concentration of conjugate acid

[B] is concentration of the base

pOH definition:

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

pOH=-log[OH-]

Relationship between pH and pOH

pOH is similar to pH .  The only difference is that in pOH the concentration of hydroxide ion is used as a scale while in 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,

pH+pOH=14,at25oC

As pOH and pH are opposite scale, the total of both has to be equal to 14.

To Calculate: The Kb of the given 0.15 M solution of caffeine

Answer to Problem 16.15WE

Answer

The Kb of the given 0.15 M solution of caffeine is 5.3×1011

Explanation of Solution

Given data:

Caffeine is the stimulant in tea and coffee.

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

C8H10N4O2(aq) + H2OHC8H10N4O2+(aq)+ OH-(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 Kb

Calculation of pOH:

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

   pH+pOH = 14   pOH = 14.00-8.45 = 5.55

Therefore, the pOH of the given caffeine solution is 5.55

Calculation of hydroxide ion:

The hydroxide ion can be calculated as follows,

pOH = -log[OH-][OH-] = 10-pOH[OH-] =10-5.55 =2.82×106M

Therefore, the concentration of [OH-] = 2.82×106M

Calculation of Kb

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

Based on the stoichiometry, [HC8H10N4O2+]=[OH-]

If the concentration of OH- is 2.82×106M at equilibrium, it must mean that 2.82×106M of the base is ionized.

We summarize the changes as follows,

  C8H10N4O2(aq) + H2OHC8H10N4O2+(aq)+ OH-(aq)
Initial (M)

0.15

2.82×106

(0.152.82×106)

0.00 0.00
Change (M) +2.82×106 +2.82×106
Equilibrium (M) 2.82×106 2.82×106

[C8H10N4O2] = (0.15- 2.82×106 ) M

0.15 M

Kb = [HC8H10N4O2+][OH-][C8H10N4O2]Kb = (2.82×106)2(0.15) = 5.3×1011

Therefore, the base ionization constant Kb of the given caffeine solution is  5.3×1011

Conclusion

The Kb of the given 0.15 M solution of caffeine was calculated

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Chapter 16 Solutions

Chemistry: Atoms First

Ch. 16.2 - Predict the relative strengths of the oxoacids in...Ch. 16.2 - Prob. 3PPACh. 16.2 - Based on the information in this section, which is...Ch. 16.2 - Prob. 3PPCCh. 16.2 - Arrange the following organic acids in order of...Ch. 16.2 - Arrange the following acids in order of increasing...Ch. 16.2 - Prob. 16.2.3SRCh. 16.3 - Prob. 16.4WECh. 16.3 - The concentration of hydroxide ions in the antacid...Ch. 16.3 - The value of Kw at normal body temperature (37C)...Ch. 16.3 - Prob. 4PPCCh. 16.3 - Calculate [OH] in a solution in which [H3O+] =...Ch. 16.3 - Prob. 16.3.2SRCh. 16.4 - Determine the pOH of a solution at 25C in which...Ch. 16.4 - Determine the pOH of a solution at 25C in which...Ch. 16.4 - Determine the pOH of a solution at 25C in which...Ch. 16.4 - Prob. 5PPCCh. 16.4 - Calculate the hydroxide ion concentration in a...Ch. 16.4 - Prob. 6PPACh. 16.4 - Prob. 6PPBCh. 16.4 - Prob. 6PPCCh. 16.4 - Determine the pH of a solution at 25C in which...Ch. 16.4 - Determine [H+] in a solution at 25C if pH = 5.75....Ch. 16.4 - 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