Chapter 16, Problem 102GQ

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# Nicotine, C10H14N2, has two basic nitrogen atoms (Figure 16.12), and both can react with water. Nic(aq) + H 2 O( l )  ⇄  NicH + (aq) + OH − (aq) NicH + (aq) + H 2 O( l )  ⇄  NicH 2 2 + (aq) + OH − (aq) Kb1 is 7.0 × 10−7 and Kb2 is 1.1 × 10−10. Calculate the approximate pH of a 0.020 M solution.

Interpretation Introduction

Interpretation:

The pH of 0.020 M solution of Nicotine has to be determined.

Concept introduction:

Equilibrium constants:

The equilibrium constant is used to quantitative measurement of the strength of the acid and bases in the water.

Ka is an acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb is base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Explanation

Nicotine has to basic hydrogens both are involved in the reaction with water.

Nic(aq)Â +Â H2O(l)Â â‡ŒNicH+(aq)Â +Â OH-(aq)Â Kb1=7.0Ã—10-7NicH+(aq)Â +Â H2O(l)Â â‡ŒNicH2+2(aq)Â +Â OH-Â Kb2=1.1Ã—10-10

From the ionization values Kb2Â <Â Â Kb1.

Most of hydroxide ions are formed in the first step. Therefore, hydroxide ion concentration from the first ionization step is calculated and it.is also checked that hydroxide ion concentration from the second step in negligible.

Letâ€™s calculate the hydroxide ion concentration from first ionization step.

Nic(aq)Â +Â H2O(l)Â â‡ŒNicH+(aq)Â +Â OH-(aq)Â Kb1=7.0Ã—10-7

Equilibrium expression:

Kb1=Â [NicH+][OH-][Nic]

Enter the ICE table the concentrations before equilibrium is established, the change that occurs as the reaction proceeds to equilibrium and the concentrations when equilibrium has been achieved.

The initial concentration of Nicotine is 0.020.

Â Â Â Â Â Â Nic(aq)Â Â Â Â Â Â Â Â Â +Â Â Â H2OÂ (aq)Â Â Â â‡ŒNiCH++(aq)+Â OH-(aq)IÂ Â Â Â Â 0.020Â MÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â --Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â --Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â --CÂ Â Â Â Â Â Â -xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â --Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +Â xEÂ Â Â (0.020-x)Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â --Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x

Therefore,

Kb1=Â 7.0Ã—10-7Kb1=Â [NicH+][OH-][Nic]

Substitute the value in ICE table

Â Â 7.0Ã—10-7=(x)(x)(0.020-x)xÂ isÂ veryÂ small,(0.020-x)Â approximatelyÂ equalsÂ toÂ 0.020

Solve and find the x value.

7.0Ã—10âˆ’7Â =Â (x)2(0.020)Â Â Â Â Â Â Â Â Â Â x2=Â Â (0.020)(7.0Ã—10-7)Â Â Â Â Â Â Â Â Â Â xÂ Â =Â Â (0.020)(7.0Ã—10-7)Â Â Â Â Â Â Â Â Â Â Â Â Â Â =Â 1.18Ã—10-4

Therefore,

[OH-]Â =Â 1.18Â Ã—10-4Â M[NicH+]=Â 1.18Â Ã—10-4Â M

Letâ€™s calculate the hydronium ion concentration by using hydroxide ion concentration

Kw=Â [H3O+][OH-][OH-]Â =Â Â 1

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