Chapter 16, Problem 95GQ

### 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

# The local anaesthetic novocaine is the hydrogen chloride salt of an organic base, procaine. C 13 H 20 N 2 O 2 (aq) + HCl(aq)  →  [HC 13 H 20 N 2 O 2 ] + Cl − (aq) The pKa for novocaine is 8.85. What is the pH of a 0.0015M solution of novocaine?

Interpretation Introduction

Interpretation:

The pH of a 0.0015M solution of novocaine 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: It is an acid constant for equilibrium reactions.

HA + H2OH3O++ A-Ka[H3O+][A-][HA]

Kb: It is a base constant for equilibrium reaction.

BOH + H2OB++ OH-Ka[B+][OH-][BOH]

Ion product constant for wter  Kw= [H3O+][OH-]       =1.00×10-14 pH = -log[H3O+]pOH= -log[OH-]

Relation between pH and pOH

pH + pOH =14

Explanation

The given PKa of novocaine is 8.85.

Letâ€™s calculate the Ka of novocaine.

Pka=Â -logKaPka=Â 8.85Ka=Â -log(8.85)=Â 1.4Ã—10-9

Therefore,

Ka=1.4Ã—10-9

The chemical equilibrium reaction is as follows.

[HC13H20N2O2]+Â (aq)Â +Â H2O(l)Â â‡ŒH3O+(aq)Â +Â C13H20N2O2

Equilibrium expression:

Ka=Â [H3O+][C13H20N2O2][HC13H20N2O2]+

The initial concentration of is 0.0015M.

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 [HC13H20N2O2]+Â (aq)Â +Â H2O(l)Â â‡ŒH3O+(aq)Â +Â C13H20N2O2IÂ Â Â Â Â Â Â Â Â 0.0015MÂ 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 Â Â Â Â Â Â Â Â Â --CÂ 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 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 xEÂ Â Â Â Â Â Â Â (0

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