   Chapter 16.10, Problem 1.3ACP

Chapter
Section
Textbook Problem

The pKa, of the conjugate acid of atropine is 4.35. How does this compare with the pKa values for the conjugate acids of ammonia, methylamine, and aniline?

Interpretation Introduction

Interpretation: To compare the pKa value of conjugate acid of atropine with conjugate acids of ammonia, methylamine, and aniline.

Concept introduction: An equilibrium constant (K) is the ratio of the concentration of products and reactants raised to appropriate stoichiometric coefficient at equilibrium.

The reaction of an acid HA with water is written as,

HA(aq)+H2O(l)H3O+(aq)+A(aq)

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

Ka=[H3O+][A][HA]

An equilibrium constant (K) with subscript a indicates that it is an equilibrium constant of an acid in water.

The reaction of any base B with water is written as,

B(aq)+H2O(l)BH(aq)+OH(aq)

The relative strength of an acid and base in water can be also expressed quantitatively with an equilibrium constant as follows:

Kb=[BH][OH][B]

An equilibrium constant (K) with subscript b indicates that it is an equilibrium constant of the base in water.

The relation between Ka and Kb is expressed as,

Ka×Kb=1014 (1)

The relation between pKa and Ka is expressed as,

pKa=logKa (2)

A conjugate acid-base pair contains two compounds that differ only by a hydrogen ion and a charge of +1.

Explanation

The calculation of pKa values of given conjugate acids; ammonium ion, methylammonium ion, and anilinium ion are calculated below.

Given:

The pKa value of protonated atropine is 4.35.

Refer to Appendix I for Kb values of given bases.

The reaction of ammonia with water is written as,

NH3+H2ONH4++OH

The Kb value of ammonia is 1.8×105.

The conjugate acid of ammonia is NH4+ ion. Substitute the value of Kb in equation (1) to calculate Ka of ammonium ion,

Ka(1.8×105)=1014Ka=0.55×109

Substitute the value of Ka in equation (2) to calculate pKa of ammonium ion,

pKa=log(0.55×109)=(9.26)=9.26

Hence, the pKa value of ammonium ion is 9.26. The value of pKa of ammonium ion is more than that of atropine.

The reaction of methylamine with water is written as,

CH3NH2+H2OCH3NH3++OH

The Kb value of methylamine is 5×104.

The conjugate acid of methylamine is CH3NH3+ ion. Substitute the value of Kb in equation (1) to calculate Ka of methylammonium ion,

Ka(5×104)=1014Ka=0

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