   Chapter 16, Problem 35PS

Chapter
Section
Textbook Problem

The trimethylammonium ion, (CH3)3NH+, is the conjugate acid of the weak base trimethylamine, (CH3)3N. A chemical handbook gives 9.80 as the pKa value for (CH3)3NH+. What is the value of for(CH3)3N?

Interpretation Introduction

Interpretation:

The dissociation constant (Kb) for weak base (CH3)3N has to be calculated

Concept introduction:

A weak base (B) undergoes dissociation in an aqueous medium. The base dissocition gives ions BH+(aq.) and OH(aq.). The conjugate acid BH+(aq.) may interact with H2O and forms BOH(aq.) and it can be represented as following equilibrium.

B(aq.)+ H2O(l)BH+(aq.)+ OH(aq.)                                                (1)

The dissociation constant for the base is Kb,

Kb=[BH+][OH][B]

Interaction of conjugate acid BH+(aq.) with water is represented as,

BH+(aq.)+ H2O(l)B(aq.)+ H3O+(aq.)                                                (2)

The dissociation constant for the conjugate acid is Ka,

Ka=[B][H3O+][BH+]

From equation (1) and (2) the net equilibrium will be

2H2O(l)H3O+(aq.)+ OH1(aq.)                                                           (3)

The equation (3) is known as the auto-ionization of water molecule. The dissociation constant for water is written as follows,

Kw=[H3O+][OH]

Thus from equation (1) and (2), there is an important relation established between ionization constant Kb for base and ionization constant Ka of conjugate acid i.e.

Kw=Ka×Kb                                                                                               (4)

The ionisation constant of water has a constant value i.e. Kw=1×1014. On taking negative logarithm of equation (4), a new relation established in terms of pKa,pKb and pKw as follows,

log(Kw)=log(Ka×Kb)

pKw= pKa+pKb (5)

Explanation

The value of Kb for the weak base (CH3)3N is calculated below.

Given:

Value of pKa for conjugate acid of given weak base (CH3)3N is 9.80.

Value of Kw for water is 1.0×1014.

By using the relation, pKa=log(Ka). The Ka value for conjugate acid is calculated.

Ka=10(pKa)=10(9.80)=1

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