Chapter 16, Problem 36PS

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 chromium(III) ion in water, [Cr(H2O)6]3+. Is a weak acid with pKa = 3.95. What is the value of Kb for its conjugate base, [Cr(H2O)5OH]2+?

Interpretation Introduction

Interpretation:

The dissociation constant (Kb) for weak acid [Cr(H2O)5OH]2+ has to be calculated

Concept introduction:

An acid undergoes dissociation in an aqueous medium. The acid dissocition gives  ions H3O+(aq.) and A(aq.). The conjugate base A(aq.) may interact with H2O and forms HA(aq.) and it can be represented as following equilibrium.

HA(aq.)+ H2O(l)H3O+(aq.)+ A(aq.)                                               (1)

The dissociation constant for the acid is Ka,

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

Interaction of anion A(aq.) is represented as,

A(aq.) +  H2O(l)HA(aq.)+ OH(aq.)                                               (2)

The dissociation constant for the conjugate anion is Kb,

Kb=[HA][OH][A]

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 Ka and ionization constant Kb of conjugate base, 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 conjugate base [Cr(H2O)5OH]2+ is calculated below.

Given:

Value of pKa for the given weak acid [Cr(H2O)6]3+ is 3.95.

Value of Kw for water is 1.0Ã—10âˆ’14.

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

Ka=10âˆ’(pKa)=10âˆ’(3.95)=1.12Ã—10âˆ’4

Using equation (4),Kb for conjugate base [Cr(H2O)5OH]2+ is calculated as follows,

Kw=KaÃ—Kb

Substitute the value of Ka and Kw,

1

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