CALCULATING pH AND pOH
Introduction
Liquid water is not a static system but an ever changing one in which some molecules are spontaneously
breaking up and re-forming according to the following equilibrium.
H20(l)  H+(aq) + OH-(aq)
The equilibrium constant for this reaction, Kw, is 1.0x10-14, well towards the left, with [H+] and [OH-] both
equal to 1.0x10-7. If we add additional H+ ions to water the equilibrium responds by shiftling left and
reducing the concentration of [OH-] and if we add [OH-], the equilibrium shifts left and lowers the
concentration of H+. Importantly [H+] multiplied by [OH-] will always equal Kw.
You will already be familiar with the pH scale as a measure of acidity/basicity but are probably less familiar
with the similar but opposite pOH scale. pH is the ‘power of hydrogen’ and is simply –log10[H+], a pH less
than seven means acid and a pH greater than 7 means alkali. pOH is the ‘power of hydroxide’ and,
similarly, is –log10[OH-] with a number less than 7 being alkali and greater than 7 acid. You certainly will not
be familiar with ‘pKw’ but this is the same kind of thing; pKw = -log10(Kw) = 14.
The last two paragraphs can then be put together to allow us to calculate the pH and pOH of any solution
as follows:
 [H+].[OH-] = Kw = 1.0x10-14
 pH + pOH = pKw = 14
For example: What are the pH and pOH of 0.005 mol dm-3 H2SO4(aq)?
[H+]
= 0.005 x 2 = 0.01 mol dm-3
[OH-] = Kw / [H+] = 1.0x10-14 / 0.01 = 1.0x10-12
pH
= -log10(0.01) = 2
pOH = -log10(1x10-12) = 12,
pH = 14 – pOH = 14-12 = 2
pOH = 14 – pH = 14 - 2 = 12
Activity
Calculate the pH and pOH for each of the following mixtures make the mixture up and test with a pH probe
to check you are correct (you can deduce pOH from pH using the above). Hint: you will need to use some
stoichiometry first.
All solutions are 1.0 M.
1.
2.
3.
4.
Add 25 cm3 HCl to 10 cm3 NaOH
Add 0.50 g Mg(OH)2 to 15 cm3 H2SO4.
Add 15 cm3 HCl and 10 cm H2SO4 to 40 cm3 NaOH
Add 1.0g Ba(OH)2 to 10 cm3 HNO3 and 30 cm3 water