pH

pH AND THE HENDERSON-HASSELBACH EQUATION

pH is defined as minus the log of the hydrogen ion concentration in a solution i.e.

pH = - log [H⁺]

[H⁺] is the concentration of H⁺ ions in moles/litre.

Water dissociates to produce hydronium and hydroxyl ions spontaneously (to a certain extent) i.e.

2H₂O = H₃O⁺ + OH^-

The dissociation coefficient Kw is equal to the product of the concentrations of OH^- and H₃O⁺ i.e.

Kw = [H₃O⁺] x [OH^-]

For pure water each of these concentrations is equal to 10^-7 moles/litre, so that Kw equals 10^-14.

Pure water is neutral where the concentrations of [H⁺] and [OH^-] are equal to each other and both are 10^-7. Because pH equals –log [H⁺], the pH of water is 7 and a neutral pH is 7.

The pH scale therefore starts at 1 (equivalent to 10^-1 moles/L of H⁺) and goes to 14 (equivalent to 10^-14 moles/L of H⁺).

With this exponential scale, a change in pH of one unit represents a tenfold change in H⁺ ion concentration. For example, a solution with a pH of 4 has 100 times the amount of H⁺ ions as a solution with a pH of 6.

Note also that the scale is such that the lower the pH, the more acidic the solution. If you add acid to a solution, the pH will decrease. If you add alkali to a solution, the pH will increase.

HENDERSON-HASSELBACH EQUATION

This equation is used to calculate the pH of buffer solutions, and is a very important concept in clinical chemistry.

A buffer consists of a solution of an acid and its conjugate base e.g acetic acid and sodium acetate. The pH of a mixture of these two substances can be calculated from the concentrations of each and the pKa of the acid thus:

If you have solutions of an acid and its conjugate base with equal concentrations and you mix equal volumes of these, then log 1 = 0 and the pH of the solution is the same as the pKa. If you do not know the pKa of an acid, then carry out the procedure above, i.e. mixing equal volumes of equal concentration solutions, and measure the pH, which will give you a pKa value. Most pKa values can be obtained from tables in chemistry textbooks.

For most buffer solutions however, different concentrations of acid and base solution are used e.g.

1.    Mix 50 ml of 0.2 mol/l acetate with 50 ml of 0.1 mol/l acetic acid. pKa of acetic acid is 4.76. pH of the resultant solution is:

pH = 4.76 + log (0.2/0.1)
= 4.76 + log 2
= 4.76 + 0.3010
= 5.061

In this case the volumes did not need to be taken into account because they were the same. If they were different however, they must be included e.g.

Mix 50 mL of 0.2 mol/l acetate with 100 mL of 0.1 mol/l acetic acid. pKa of acetic acid is 4.76. pH of the resultant solution is:

pH = 4.76 + log 1
pH = 4.76 + 0
pH = 4.76

This must be correct because in this example you have added more acid (100 ml compared to 50 ml) and the resultant pH is lower.