# Conjugate Acids of Bases - KaKb and Kw

Part I, Part II

### Skills to develop

• Explain conjugate acids of bases.
• Evaluate Ka of the conjugate acid of a base.
• Treat the conjugate acid of a base as an acid in numerical calculations.
• Reverse the role of acid and base for the previous skills.

# Conjugate Acids of Bases

The conjugation of acids and bases have been discussed earlier. After losing a proton, the acid species becomes the conjugate base. A base and its protonated partner also form a conjugated acid-base pair. These relationships have been represented by

H+ + Base = Conjugate_acid of Base+
Acid = H+ + Conjugate_base of Acid-
For example: NH3 + H2O = NH4+ + OH-
HAc = H+ + Ac-
Thus, NH4+ and NH3 are a pair of conjugate acids and bases, as are HAc and Ac-.

## Ka Values of Conjugate Acids of Bases

We have used Ka and Kb as the acidic and basic constants of acids and bases. Can an acidic constant, Ka, be assigned to the conjugate acid of a base? If so, what is the relationship between Ka of the conjugate acid and Kb of the base? We are going to derive the relationship here. Note that water always plays a role in the conjugation acid-base pair.

Let BH+ be the conjugate acid of a base, then the expression for the acidic constant Ka for the conjugate acid:

BH+ = B + H+ can be written as
```          [B] [H+]
Ka  =  ---------
[BH+]

[B] [H+]  [OH-]
=  --------- ------
[BH+]    [OH-]

[B]
=  -----------  [H+] [OH-]
[BH+] [OH-]

1
=  --- Kw
Kb

```
Thus, Ka Kb = Kw Furthermore, - log ( Ka) - log (Kb) = -log (Kw)

and at 298 K, we have

p Ka + pKb = 14.

Examples 1

The Ka for HCO3- is 4.7E-11, what is the conjugate base and its Kb?

Solution
The conjugate base is CO32-.

Kb = (1E-14)/(4.7E-11)
= 2.1E-4

DIscussion
The Kb so calculated is for the reaction,

```     CO32- + H2O = HCO3- + OH-

[HCO3-] [OH-]
Kb = -------------
[CO32-]
```
The anion CO32- is a rather strong base, and the large value calculated for Kb agrees with the fact.

Example 2

The Kb for the anion of oxalic acid, COO- | COOH is 1.8E-10. What is Ka for the oxalic acid (COOH)2?

Solution
The Ka for oxalic acid is

Ka = (1E-14) / (1.8E-10)
= 5.6E-5

DIscussion
The calculation regarding Ka and Kb conversion is simple, but understanding what problems require this type of conversion is difficult. The concept is rather useful, and it further broadens the concept of acid and base.

## Kb Values of Conjugate Bases of Acids

We can also calculate the Kb value of the conjugate base from the Ka value of its conjugate acid. The principle is the same as that used to calculate the Ka values of the conjugate acid of a base as we have just discussed.

Let A- be the conjugate base of an acid HA, then the expression for the equilibrium constant for the reaction:

A- + H2O = HA + OH-

can be written as

```         [HA] [OH-]
Kb  =  ----------
[A-]
```
Multiplying the numerator and denominator with [H+] leads to,
```         [HA] [OH-] [H+]
Kb  =  ---------- ----
[A-]     [H+]

Rearrangement gives

[HA]
Kb  =  ---------- [OH-] [H+]
[A-] [H+]

[HA]
=  --------- Kw
[A-] [H+]

Kw
= ----
Ka
```
Thus, Ka Kb = Kw

and this formula is the same as the one derived for the conjugate acid of a base. Again, at 298 K, we have

Ka Kb = 1E-14

and the value for Kw is larger than 1E-14 at higher temperatures. Kw is smaller at temperature less than 298 K.

### Applications

The concept of conjugate acid and base pairs are very useful for the consideration of acidity and basicity of salts. The applications of the relationship, Ka Kb = Kw are further illustrated on the topic of Hydrolysis. Hydrolysis reactions are reactions of cations or anions of salts with water. As a result of these reactions, a salt solution is either acidic or basic.

### Confidence Building Questions

• Calculate Kb for the acetate ion from the Ka for acetic acid of 1.8E-5.

Consider...
Kb = (1e-14)/(1.8e-5) = 5.6E-10 If Kb for the acetate ion is 5.6E-10, what is Ka for acetic acid?

• The Ka for trimethylammonium ion (CH3)3NH+ is 1.6E-10. Calculate Kb for its conjugate base.

Consider...
Ka = (1e-14)/(1.6e-10) = 6.25E-5 You know all about conjugate acid-base pairs now. Learning is a pleasure.

• At some temperature, Kw = 1e-13. Calculate the Kb value for the acetate ion. (Ka for acetic acid is 9.5E-5 at the same temperature).

Consider...
Kb = 1e-13/9.5e-5 The acidic constants are dependent on temperature.