   # Nicotine, C 10 H 14 N 2 , 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) K b1 is 7.0 × 10 −7 and K b2 is 1.1 × 10 −10 . Calculate the approximate pH of a 0.020 M solution. ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640

#### Solutions

Chapter
Section ### Chemistry & Chemical Reactivity

9th Edition
John C. Kotz + 3 others
Publisher: Cengage Learning
ISBN: 9781133949640
Chapter 16, Problem 100GQ
Textbook Problem
232 views

## 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 of Solution

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

Nic(aq) + H2O(l) NicH+(aq) + OH-(aq) Kb1=7.0×10-7NicH+(aq) + H2O(l) NicH2+2(aq) + OH- Kb2=1.1×10-10

From the ionization values Kb2 <  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.

Let’s calculate the hydroxide ion concentration from first ionization step.

Nic(aq) + H2O(l) NicH+(aq) + OH-(aq) Kb1=7.0×10-7

Equilibrium expression:

Kb1[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.

Nic(aq)         +   H2O (aq)   NiCH++(aq)+ OH-(aq)I     0.020 M                --                      --                   --C       -x                       --                    + x                + xE   (0.020-x)               --                      x                    x

Therefore,

Kb1= 7.0×10-7Kb1[NicH+][OH-][Nic]

Substitute the value in ICE table

7.0×10-7=(x)(x)(0.020-x)x is very small,(0.020-x) approximately equals to 0.020

Solve and find the x value.

7.0×107 = (x)2(0.020)          x2=  (0.020)(7.0×10-7)          x  =  (0.020)(7.0×10-7)              = 1.18×10-4

Therefore,

[OH-] = 1.18 ×10-4 M[NicH+]= 1.18 ×10-4 M

Let’s calculate the hydronium ion concentration by using hydroxide ion concentration

Kw= [H3O+][OH-][OH-] =  1

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