A-level Chemistry/WJEC/Module 3/Acid-Base Equilibria

Calculating the pH of a strong alkali

For a strong alkali we have the concentration of hydroxide ions, [OH^–], but to find the pH we need the concentration of hydrogen ions, [H⁺].

The two concentrations are related by the equation

K_w = [H⁺][OH^–]

K_w = 1 x 10^-14 mol² dm^-6 at 25 °C, a value in your Data Booklet.

To find [H⁺] from [OH^–] we can use:

[H⁺] = K_w/[OH^–]

We can also define pOH along the lines of pH:

pOH = -log₁₀[OH^–]

pH = pK_w - pOH

Example: To find the pH of 0.2 mol dm^-3 NaOH solution:

[OH^–] = 0.2 mol dm^-3

[H⁺] = K_w/[OH^–]

[H⁺] = 1 x 10^-14/0.2 = 5 x 10^-14 mol dm^-3

pH = -log₁₀[H⁺] = -log₁₀(5 x 10^-14) = 13.30

Alternatively:

pOH = -log₁₀[OH^–] = -log₁₀(0.2) = 0.70

pH = pK_w - pOH = 14 - 0.70 = 13.30

Calculating the pH of a weak acid solution

The pH of a weak acid solution can be calculated approximately using the following formulae:

pH = -log₁₀ √(K_a x [HA]_o) = 0.5 x (pK_a - log₁₀[HA]_o)

[HA]_o is the original concentration of acid, which we assume is the same as the concentration of acid after it has partly dissociated.

Example; If pK_a = 4.7 and c = 0.010 mol dm^-3 then either:

K_a = 10^-pKa = 10^-4.7 = 2.00 x 10^-5 mol dm^-3

√(K_a x [HA]_o) = √(2.00 x 10^-5 x 0.010)

= √(2.00 x 10^-7)

= 4.47 x 10^-4

pH = -log₁₀ (4.47 x 10^-4) = 3.35

or:

- log₁₀[HA]_o = - log₁₀(0.010) = 2.00

pH = 0.5 x (pK_a - log₁₀[HA]_o)

= 0.5 x (4.7 + 2) = 0.5 x 6.7 = 3.35

Calculating the pH of a buffer solution

The pH of a buffer solution can be calculated approximately using the following formula:

pH = pK_a + log₁₀([A^–]/[HA])

The ratio [A^–]/[HA] is equal to the number of moles of each species: [A^–]/[HA] = n(A^–)/n(HA). This is because the two solutes are in the same volume of solution. This means that buffers are not affected by dilution.

In general, we can assume that if we mix a weak acid and its salt, [HA] is the acid concentration and [A^–] is the salt concentration. Both assumptions are slightly incorrect, but the errors tend to cancel each other out.

If we part-neutralise a weak acid with a strong base, [A^–] is the concentration of neutralised acid and [HA] is concentration of the excess acid. If you half-neutralise a weak acid then [A^–] = [HA]. This makes log₁₀([A^–]/[HA]) = 0 and so pH = pK_a.

Simulated titration of an acidified solution of a weak acid (pK_a = 4.7, c = 0.010 mol dm^-3) with alkali. The end point is when 1.0 ml of alkali are added. When half-neutralised with 0.50 ml of alkali, the pH = pK_a = 4.7 (green lines). The pure acid has a pH of 3.35.

Indicators

Indicators are weak acids or bases which have strong colours and/or the conjugate acid/base has a different, strong, colour.

Examples:

• Methyl red is red in acidic solution (the HA molecule) but above pH 5.1 it forms the yellow A^– ion.

• Methyl orange is red in acidic solution (the HA molecule) but above pH 3.5 it forms the yellow A^– ion.

• Phenolphthalein is colourless in acidic/neutral solution (the H₂A molecule) but above pH 9.5 it forms the pink A^2– ion.

In an exam, you will be given details about any indicators so there is no need to learn the details.

The pK_a value of the indicator is sometimes referred to as pK_in. You need to choose an indicator that rapidly changes colour at the endpoint of a titration i.e. the pK_in is at the steepest part of the pH/titre curve.

The most rapid pH change in the graph above is around pH 7, so we need to choose an indicator with a similar pK_in value.