pH Calculations for Weak Acids

Learning Objectives (OCR Spec 5.1.3)

By the end of this lesson you should be able to:

• Calculate the pH of a weak monobasic acid from Ka and [HA]
• Apply the standard approximations ([H+] = [A-]; [HA] ~ initial)
• Justify when the approximations are valid
• Calculate Ka from pH and [HA] (reverse problem)
• Handle percentage dissociation calculations

The Key Expression

For a weak acid HA at equilibrium in water:

HA(aq) <=> H+(aq) + A-(aq) Ka = [H+][A-] / [HA]

To calculate [H+] and hence pH we need to express all three equilibrium concentrations in terms of the initial acid concentration c and a single unknown.

Setting Up the ICE Table

Let c be the initial concentration of the weak acid and let x be the concentration of H+ at equilibrium.

	HA	H+	A-
Initial	c	0	0
Change	-x	+x	+x
Equilibrium	c - x	x	x

Notice two features:

• [H+] = [A-] = x, because each dissociation produces one H+ and one A- together.
• [HA] at equilibrium = c - x, because x mol dm-3 of HA has dissociated.

Substituting into the Ka expression:

Ka = x^2 / (c - x)

This is a quadratic in x, but we normally avoid solving it by making two approximations.

The Two Standard Approximations

Approximation 1: [H+] = [A-]

When the acid is appreciably acidic (pH < 6 or so), the H+ from the acid dominates the H+ from water self-ionisation, so [H+] from water is negligible and [H+] = [A-] exactly. This approximation is excellent for almost any practical weak acid.

Approximation 2: [HA] at equilibrium ~ c

If only a small fraction of HA has dissociated, then c - x ~ c. This approximation is valid when x is less than about 5% of c, i.e. when the acid is less than 5% dissociated. For most weak acids at ordinary concentrations (Ka < 10^-3, c > 0.01 mol dm-3), the approximation works.

With both approximations:

Ka = x^2 / c x = sqrt(Ka x c) [H+] = sqrt(Ka x c) pH = -log10 sqrt(Ka x c) = -0.5 x log10(Ka x c)

Or equivalently:

pH = 0.5(pKa - log10 c)

But you should normally work from first principles rather than memorise this formula.

Worked Example 1: Ethanoic Acid

Calculate the pH of 0.100 mol dm-3 CH3COOH. Ka = 1.74 x 10^-5 mol dm-3.

Step 1: Write equilibrium and Ka expression. CH3COOH <=> H+ + CH3COO- Ka = [H+][CH3COO-] / [CH3COOH] = 1.74 x 10^-5

Step 2: Apply approximations. [H+] = [CH3COO-] = x; [CH3COOH] ~ 0.100.

Step 3: Substitute. 1.74 x 10^-5 = x^2 / 0.100 x^2 = 1.74 x 10^-6 x = 1.32 x 10^-3 mol dm-3

Step 4: Calculate pH. pH = -log10(1.32 x 10^-3) = 2.88 (2 dp)

Step 5: Check the approximation. x / c = 1.32 x 10^-3 / 0.100 = 0.0132 = 1.3%

Only 1.3% of the ethanoic acid has dissociated - much less than 5%, so the approximation is safe.

Worked Example 2: Methanoic Acid

Calculate the pH of 0.0500 mol dm-3 HCOOH. Ka = 1.60 x 10^-4 mol dm-3.

Ka = x^2 / 0.0500 x^2 = 1.60 x 10^-4 x 0.0500 = 8.00 x 10^-6 x = 2.83 x 10^-3 mol dm-3 pH = -log10(2.83 x 10^-3) = 2.55 (2 dp)

Check: x / c = 2.83 x 10^-3 / 0.0500 = 0.057 = 5.7%. This is just above the 5% threshold. The full quadratic would give a marginally more accurate pH (~2.56), but at A-Level precision 2.55 is acceptable.

Worked Example 3: Benzoic Acid

Calculate the pH of 0.250 mol dm-3 benzoic acid C6H5COOH. pKa = 4.20.

First convert: Ka = 10^-4.20 = 6.31 x 10^-5 mol dm-3.

Ka = x^2 / 0.250 x^2 = 6.31 x 10^-5 x 0.250 = 1.578 x 10^-5 x = 3.97 x 10^-3 mol dm-3 pH = -log10(3.97 x 10^-3) = 2.40 (2 dp)

Check: x / c = 3.97 x 10^-3 / 0.250 = 1.6%. Good.

Worked Example 4: Very Weak Acid

Calculate the pH of 0.100 mol dm-3 HCN. Ka = 4.9 x 10^-10 mol dm-3.

Ka = x^2 / 0.100 x^2 = 4.9 x 10^-11 x = 7.0 x 10^-6 mol dm-3 pH = -log10(7.0 x 10^-6) = 5.15 (2 dp)

A very weak acid like HCN barely moves the pH from 7. Check: x/c = 7.0 x 10^-5 = 0.007% - approximation extremely safe.

Warning: For extremely weak acids (Ka < 10^-10) or very dilute solutions, the contribution from water self-ionisation ([H+] from water is ~10^-7) can become comparable to [H+] from the acid. In that case the simple formula overestimates acidity and the water contribution must be included. This is beyond OCR A-Level.

Calculating Ka from pH

The reverse problem is common: given the pH and initial concentration of a weak acid, find Ka and pKa.

Worked Example 5

A 0.100 mol dm-3 solution of propanoic acid CH3CH2COOH has pH = 2.94. Calculate Ka and pKa.

[H+] = 10^-2.94 = 1.148 x 10^-3 mol dm-3 [A-] = [H+] = 1.148 x 10^-3 mol dm-3 [HA] ~ 0.100 mol dm-3