# Acid/base Dissociation Constants and Kw

This is part of the HSC Chemistry course under the topic Quantitative Analysis.

### HSC Chemistry Syllabus

• calculate and apply the dissociation constant (Ka) and pK(pKa = –log10 (Ka)) to determine the difference between strong and weak acids (ACSCH098)

Kw (equilibrium constant of the self-ionisation of water) is the product of Ka of a weak acid and Kb of its conjugate base

$$K_w = K_a \times K_b$$

Derivation (do not memorise): For a generic acid-base system,

$$\boldsymbol{Acid}: \hspace{0.5cm} HA_{(aq)} + H_2O_{(l)} \leftrightharpoons A^-_{(aq)} + H_3O^+_{(aq)} \hspace{0.5cm} K_a = \frac{[A^-][H_3O^+]}{[HA]}$$

$$\boldsymbol{Base}: \hspace{0.5cm} A^-_{(aq)} + H_2O_{(l)} \leftrightharpoons HA_{(aq)} + OH^-_{(aq)} \hspace{0.5cm} K_b = \frac{[HA][OH^-]}{[A^-]}$$

$$K_a \times K_b = \frac{[{A^-}][H_3O^+]}{[HA]} \times \frac{[HA][OH^-]}{[A^-]}$$

$$K_a \times K_b = [H_3O^+_{(aq)}][OH^-_{(aq)}] = 1.0 \times 10^{-14}$$

This means at 25℃ (298 K):

$$K_a = \frac{10^{-14}}{K_b} \hspace{3cm} K_b = \frac{10^{-14}}{K_a}$$

The mathematical implication of this relationship between acid and base dissociation constants is that:

 Weak acid Conjugate weak base High Ka & low pKa High strength Low strength Low Ka & high pKa Low strength High strength

 Weak base Conjugate weak acid High Kb & low pKb High strength Low strength Low Kb & high pKb Low strength High strength