Buffer pH Calculations

Learning Objectives (OCR Spec 5.1.3)

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

• Derive the buffer pH equation from Ka
• Calculate the pH of a buffer given [HA] and [A-]
• Calculate the pH after adding acid or base to a buffer
• Use the approximation [HA]/[A-] = ratio of moles, not just concentrations
• Choose a suitable weak acid for a target buffer pH
• Apply the Henderson-Hasselbalch form (equivalent)

Derivation

Start from the acid dissociation constant:

Ka = [H+][A-] / [HA]

Rearranging:

[H+] = Ka x [HA] / [A-]

Taking -log10 of both sides:

pH = -log10(Ka) - log10([HA]/[A-]) = pKa + log10([A-]/[HA])

This is the Henderson-Hasselbalch equation. At OCR A-Level you can use whichever form you prefer - the [H+] form is often easier for numerical work, while the pH form is useful when you want to see the pH = pKa structure.

Approximation: Initial Concentrations ~ Equilibrium Concentrations

When you mix HA and A- in comparable amounts, very little additional dissociation happens because Le Chatelier pushes the equilibrium back to the left (the A- side is already well-populated). So we can use the initial concentrations of HA and A- as the equilibrium concentrations. This is the standard approximation for buffer calculations and is accurate to within a percent or so for normal buffers.

Worked Example 1: Simple Buffer

Calculate the pH of a buffer made by dissolving 0.100 mol of CH3COOH and 0.100 mol of CH3COONa in 1.00 dm3 of water. Ka(CH3COOH) = 1.74 x 10^-5 mol dm-3.

[CH3COOH] = 0.100 mol dm-3 [CH3COO-] = 0.100 mol dm-3

[H+] = Ka x [HA]/[A-] = 1.74 x 10^-5 x (0.100 / 0.100) = 1.74 x 10^-5 mol dm-3 pH = -log10(1.74 x 10^-5) = 4.76 (2 dp)

Key observation: when [HA] = [A-] the ratio is 1 and [H+] = Ka. Therefore pH = pKa at equal concentrations. This is the half-equivalence point of a titration (lesson 9).

Worked Example 2: Unequal Concentrations

Calculate the pH of a buffer containing 0.200 mol dm-3 CH3COOH and 0.100 mol dm-3 CH3COONa. Ka = 1.74 x 10^-5 mol dm-3.

[H+] = Ka x [HA]/[A-] = 1.74 x 10^-5 x (0.200 / 0.100) = 3.48 x 10^-5 mol dm-3 pH = -log10(3.48 x 10^-5) = 4.46 (2 dp)

With more acid than salt, [H+] is higher (pH lower) than when they were equal. Sensible: more HA means more dissociation in absolute terms.

Worked Example 3: More Salt than Acid

Calculate the pH of a buffer containing 0.100 mol dm-3 CH3COOH and 0.300 mol dm-3 CH3COONa.

[H+] = 1.74 x 10^-5 x (0.100 / 0.300) = 5.80 x 10^-6 mol dm-3 pH = -log10(5.80 x 10^-6) = 5.24 (2 dp)

With more salt (A-), [H+] falls and pH rises. The ratio [HA]:[A-] = 1:3 pushes the equilibrium to the left, reducing H+.

The Ratio of Moles Works Too

Because both HA and A- are in the same volume, the ratio [HA]/[A-] is equal to the ratio of moles of HA to moles of A-. You do not need to calculate separate concentrations if you know the moles:

Worked Example 4: A buffer contains 0.150 mol of methanoic acid (pKa = 3.75) and 0.050 mol of sodium methanoate in 500 cm3 of solution. Calculate the pH.

Ka = 10^-3.75 = 1.78 x 10^-4 mol dm-3 Ratio of moles HA:A- = 0.150 : 0.050 = 3 : 1

[H+] = Ka x (mol HA / mol A-) = 1.78 x 10^-4 x 3 = 5.34 x 10^-4 mol dm-3 pH = -log10(5.34 x 10^-4) = 3.27 (2 dp)

Notice that we did not need to work out concentrations - the volume cancels.

Worked Example 5: Buffer Made by Partial Neutralisation

25.0 cm3 of 0.200 mol dm-3 NaOH is added to 50.0 cm3 of 0.200 mol dm-3 ethanoic acid. Calculate the pH. Ka = 1.74 x 10^-5 mol dm-3.

Step 1: Calculate moles. Moles CH3COOH = 0.200 x 0.0500 = 0.0100 mol Moles NaOH = 0.200 x 0.0250 = 0.00500 mol

Step 2: React. NaOH consumes half the ethanoic acid, producing ethanoate. Moles CH3COOH left = 0.0100 - 0.00500 = 0.00500 mol Moles CH3COO- formed = 0.00500 mol

Step 3: Substitute into buffer equation. Ratio = 0.00500 / 0.00500 = 1 [H+] = Ka x 1 = 1.74 x 10^-5 mol dm-3 pH = 4.76

This is the half-neutralisation point. The pH equals the pKa of ethanoic acid.

Worked Example 6: Adding Acid to a Buffer

A 1.00 dm3 buffer contains 0.200 mol of CH3COOH and 0.150 mol of CH3COO-. Calculate the initial pH, and then the pH after adding 0.020 mol of HCl (assume no volume change).

Initial pH: [H+] = 1.74 x 10^-5 x (0.200 / 0.150) = 2.32 x 10^-5 mol dm-3 pH = -log10(2.32 x 10^-5) = 4.63

After adding 0.020 mol HCl: The added H+ reacts with CH3COO- to form CH3COOH: CH3COO- + H+ -> CH3COOH

Moles CH3COO- = 0.150 - 0.020 = 0.130 mol Moles CH3COOH = 0.200 + 0.020 = 0.220 mol