# Conjugate Acids of Bases - Ka Kb 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 has 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

$$\mathrm{H^+ + {\color{Blue} Base} = {\color{Red} Conjugate\: acid\: of\: Base}^+}$$

$$\mathrm{{\color{Red} Acid} = H^+ + {\color{Blue} Conjugate\: base\: of\: Acid}^-}$$

For example:

$$\mathrm{{\color{Blue} NH_3} + H_2O \rightleftharpoons {\color{Red} NH_4^+} + OH^-}$$

$${\color{Red} \mathrm{HAc}} \rightleftharpoons \mathrm{H^+} + {\color{Blue} \mathrm{Ac^-}}$$

Thus, $${\color{Red} \mathrm{NH_4^+}}$$ and $${\color{Blue} \mathrm{NH_3}}$$ are a pair of conjugate acids and bases, as are $${\color{Red} \mathrm{HAc}}$$ and $${\color{Blue} \mathrm{Ac^-}}$$.

### 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 $$\ce{BH+}$$ be the conjugate acid of a base. The expression for the acidic constant Ka for the conjugate acid

$$\mathrm{{\color{Red} BH^+} = {\color{Blue} B} + H^+}$$

can be written as

\begin{align}K_{\mathrm a} &= \mathrm{\dfrac{[{\color{Blue} B}] [H^+]}{[{\color{Red} BH^+}]}} \\ &= \mathrm{\dfrac{[{\color{Blue} B}] [H^+]}{[{\color{Red} BH^+}]}\dfrac{[{\color{Green} OH^-}]}{[{\color{Green} OH^-}]}}\\ &= \mathrm{\dfrac{[{\color{Blue} B}]}{[{\color{Red} BH^+}] [OH^-]}[H^+][OH^-]}\\ &= \dfrac{1}{K_{\mathrm b}}K_{\mathrm w} \end{align}

Thus,

$$\mathrm{{\color{Red} \mathit K_a} {\color{Blue} \mathit K_b} = \mathit K_w}$$

Furthermore,

$$\mathrm{- \log ({\color{Red} \mathit K_a}) - \log ({\color{Blue} \mathit K_b}) = -\log (\mathit K_w)}$$

and at 298 K, we have

$$\mathrm{p {\color{Red} \mathit K_a} + p {\color{Blue} \mathit K_b} = 14}$$

Example 1

The Ka for $$\ce{HCO3-}$$ is 4.7E-11; what is the conjugate base and its Kb?

Solution
The conjugate base is $$\ce{CO3^2-}$$.

$$\mathrm{\mathit K_b = \dfrac{1E{-}14}{4.7E{-}11}= 2.1E{-}4}$$

DIscussion
The Kb so calculated is for the reaction,

$$\mathrm{CO_3^{2-} + H_2O \rightleftharpoons HCO_3^- + OH^-}$$

$$\mathrm{\mathit K_b = \dfrac{[HCO_3^-] [OH^-]}{[CO_3^{2-}]}}$$

The anion $$\ce{CO3^2-}$$ 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, $$\mathrm{COO^- | COOH}$$ is 1.8E-10. What is the Ka for the oxalic acid $$\mathrm{(COOH)_2}$$?

Solution
The Ka for oxalic acid is

$$\mathrm{\mathit K_a = \dfrac{1E{-}14}{1.8E{-}10}= 5.6E{-}5}$$

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 $$\mathrm{A^-}$$ be the conjugate base of an acid $$\mathrm{HA}$$. Then the expression for the equilibrium constant for the reaction

$$\mathrm{A^- + H_2O \rightleftharpoons HA + OH^-}$$

can be written as

$$\mathrm{\mathit K_b =\dfrac{[HA] [OH^-]}{[A^-]}}$$

Multiplying the numerator and denominator with $$\mathrm{[H^+]}$$ leads to

$$\mathrm{\mathit K_b =\dfrac{[HA] [OH^-]}{[A^-]}\dfrac{[H^+]}{[H^+]}}$$

Rearrangement gives

\begin{align}\mathit K_{\mathrm b} &=\mathrm{\dfrac{[HA]}{[A^-] [H^+]}[OH^-] [H^+]}\\ &=\mathrm{\dfrac{[HA]}{[A^-] [H^+]}\mathit K_w}\\ &=\mathrm{\dfrac{\mathit K_w}{\mathit K_a}}\end{align}

Thus,

$$\mathrm{\mathit K_a \mathit K_b = \mathit K_w}$$

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

$$\mathrm{\mathit K_a \mathit K_b = 1E{-}14}$$

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 is very useful for the consideration of acidity and basicity of salts. The applications of the relationship

$$\mathrm{\mathit K_a \mathit K_b = \mathit K_w}$$

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

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

Consider...
$$\mathrm{\mathit K_b = \dfrac{1e{-}14}{1.8e{-}5} = 5.6E{-}10}$$

If Kb for the acetate ion is 5.6E-10, what is Ka for acetic acid?

1. The Ka for trimethylammonium ion $$\mathrm{(CH_3)_3NH^+}$$ is 1.6E-10. Calculate Kb for its conjugate base.

Consider...
$$\mathrm{\mathit K_a = \dfrac{1e{-}14}{1.6e{-}10} = 6.25E{-}5}$$

You know all about conjugate acid-base pairs now. Learning is a pleasure.

2. At some temperature, $$\mathrm{\mathit K_w = 1e{-}13}$$. Calculate the Kb value for the acetate ion. (Ka for acetic acid is 9.5E-5 at the same temperature).

$$\mathrm{\mathit K_b = \dfrac{1e{-}13}{9.5e{-}5}}$$