Chapter 16, Problem 58PS

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

Calculate the pH of a 0.12 M aqueous solution of the base aniline, C 6 H 5 NH 2 ( K b = 4 .0×10 − 10 ) . C 6 H 5 NH 2 (aq) + H 2 O( l )  ⇌ C 6 H 5 NH 3 + (aq) + OH − (aq)

Interpretation Introduction

Interpretation:

The pH of a 0.12 M aqueous solution of the base aniline has to be calculated.

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 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

The given chemical reaction is as follows,

C6H5NH2(aq)Â +Â H2O(l)Â â‡ŒC6H5NH3+(aq)Â +Â OH-(aq).

The equilibrium expression:

kb=Â [OH-][C6H5NH3+][C6H5NH2]

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.

Â Â Â Â Â Â Â C6H5NH2(aq)Â +Â H2O(l)Â â‡ŒC6H5NH3+(aq)Â +Â OHâˆ’(aq)IÂ Â Â Â Â Â Â Â Â 0.12Â 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 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 xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â +xEÂ Â Â Â Â (0.12Â -Â 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 xÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â x

Kb of C6H5NH2 is 4.0Â Ã—Â 10-10

At equilibrium, xÂ =Â [OH-]Â [C6H5NH3+]

kb=Â [OH-][C6H5NH3+][C6H5NH2]

4

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