# Conjugate Acids of Bases - Ka Kb and Kw

Skills to Develop

- Explain conjugate acids of bases.
- Evaluate
*K*_{a}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 \({\color{Blue} \mathrm{Ac^-}}\).

*K*_{a} Values of Conjugate Acids of Bases

We have used *K*_{a} and *K*_{b} as the acidic and basic constants of acids and bases. Can an acidic constant, *K*_{a}, be assigned to the conjugate acid of a base? If so, what is the relationship between *K*_{a} of the conjugate acid and *K*_{b} 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 *K*_{a} 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 *K*_{a} for \(\ce{HCO3-}\) is 4.7E-11; what is the conjugate base and its *K*_{b}?

*Solution*

The conjugate base is \(\ce{CO3^2-}\).

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

*DIscussion*

The *K*_{b} 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 *K*_{b} agrees with the fact.

Example 2

The *K*_{b} for the anion of oxalic acid, \(\mathrm{COO^- | COOH}\) is 1.8E-10. What is the *K*_{a} for the oxalic acid \(\mathrm{(COOH)_2}\)?

*Solution*

The *K*_{a} for oxalic acid is

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

*DIscussion*

The calculation regarding *K*_{a} and *K*_{b} 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.

*K*_{b} Values of Conjugate Bases of Acids

We can also calculate the *K*_{b} value of the conjugate base from the *K*_{a} value of its conjugate acid. The principle is the same as that used to calculate the *K*_{a} 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 *K*_{w} is larger than 1E-14 at higher temperatures. *K*_{w} 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

**Calculate***K*_{b}for the acetate ion from the*K*_{a}for acetic acid of 1.8E-5.

**Answer ***5.6E-10*

**Consider...**

\(\mathrm{\mathit K_b = \dfrac{1e{-}14}{1.8e{-}5} = 5.6E{-}10}\)

If *K*_{b} for the acetate ion is 5.6E-10, what is *K*_{a} for acetic acid?

**The**\(\mathrm{(CH_3)_3NH^+}\)*K*_{a}for trimethylammonium ion**is 1.6E-10. Calculate***K*_{b}for its conjugate base.**Answer***6.25E-5*

**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.

**At some temperature,**\(\mathrm{\mathit K_w = 1e{-}13}\)**. Calculate the***K*_{b}value for the acetate ion. (*K*_{a}for acetic acid is 9.5E-5 at the same temperature).**Answer***1.05e-9*

**Consider...**

\(\mathrm{\mathit K_b = \dfrac{1e{-}13}{9.5e{-}5}}\)The acidic constants are dependent on temperature.

## Contributors

Chung (Peter) Chieh (Chemistry, University of Waterloo)