pH and pKa are important solution parameters in many areas of chemistry, including calculations involving acid-base equilibria. pH is the universal measure of acidity, defined as the negative logarithm, to the base 10, of the "hydrogen ion concentration" of a solution, and is expressed as: pH = -log [H3O+]. The brackets denote concentration and the "+" sign denotes the charge of the hydrogen ion. pKa is the negative logarithm, to the base 10, of the "dissociation constant" of a weak acid. For example, the dissociation of a weak acid "HA" is written: Ka = [H3O+][A-] / [HA], where A- is the "conjugate base" of the acid. Therefore, pKa = -log Ka. Every weak acid has a unique pKa value. Use the Henderson-Hasselbalch equation to calculate the pH of a buffer solution, which is a solution of a weak acid and its conjugate base, when the pKa of the acid is known. This equation is expressed: pH = pKa + log ([base]/[acid]).

Assume you have a buffer solution that was prepared by adding 25.0 ml of a 0.1 M sodium hydroxide (NaOH) solution to 75.0 ml of an 0.1 M solution of acetic acid (CH3COOH), where "M" denotes molar concentration. Note that acetic acid reacts with NaOH to form the conjugate base, CH3C00H-, as follows: CH3COOH + NaOH = CH3C00- + Na + H20. In order to calculate the pH, it is necessary to calculate the amounts of acid and conjugate base in the buffer solution following the reaction.

Calculate the initial moles of base and acid in the buffer solution. For example, moles of NaOH = 25.0 ml x 0.1 mole/liter x 1 liter/1000 ml = 0.0025 moles; moles of CH3COOH = 75.0 ml x 0.10 mole/liter x 1 liter/1000 ml = 0.0075 moles.

Note that, upon mixing the solutions, the CH3COOH consumes the OH- (hydroxyl) ions associated with the NaOH, so that what remains is 0.0050 moles of CH3COOH (acid), 0.0025 moles of CH3COO- (base) and 0 moles of OH-.

Substitute the pKa of the acid (4.74 for acetic acid) and the acid and base concentrations into the Henderson-Hasselbalch equation to calculate the pH of the buffer solution. For example, pH = 4.74 + log (0.0025/0.005) = 4.74 + log 0.5 = 4.44.