### Henderson-Hasselbalch Equation

**pH = pK**

_{a}+ log ([A^{-}]/[HA])[A^{-}] = molar concentration of a conjugate base

[HA] = molar concentration of a undissociated weak acid (M)

The equation can be rewritten to solve for pOH:

**pOH = pK _{b} + log ([HB^{+}]/[ B ])**

[HB^{+}] = molar concentration of the conjugate base (M)

[ B ] = molar concentration of a weak base (M)

### Example Problem Applying the Henderson-Hasselbalch Equation

Calculate the pH of a buffer solution made from 0.20 M HC_{2}H

_{3}O

_{2}and 0.50 M C

_{2}H

_{3}O

_{2}

^{-}that has an acid dissociation constant for HC

_{2}H

_{3}O

_{2}of 1.8 x 10

^{-5}.

Solve this problem by plugging the values into the Henderson-Hasselbalch equation for a weak acid and its conjugate base.

pH = pK_{a} + log ([A^{-}]/[HA])

pH = pK_{a} + log ([C_{2}H_{3}O_{2}^{-}] / [HC_{2}H_{3}O_{2}])

pH = -log (1.8 x 10^{-5}) + log (0.50 M / 0.20 M)

pH = -log (1.8 x 10^{-5}) + log (2.5)

pH = 4.7 + 0.40

pH = 5.1