Buffer solutions resist pH changes

Buffer solutions are aqueous solutions that maintain a nearly constant pH when small amounts of acids or bases are added. They contain a mixture of:

• A weak acid and its conjugate base, or
• A weak base and its conjugate acid.

When a strong acid like HCl is added to a buffer, it is neutralised by the weak conjugate base:

H⁺_(aq) + A^-_(aq) ➔ HA_(aq)

Similarly, adding a strong base like NaOH is neutralised by the weak conjugate acid:

OH^-_(aq) + HA_(aq) ➔ A^-_(aq) + H₂O_(l)

Since the products are a weak acid and weak base, which only dissociate slightly, the [H⁺] and hence the pH stay nearly constant. This neutralising action is called the buffer action.

Components of a buffer solution

A buffer solution must contain both the weak acid and weak base of a conjugate acid-base pair in sufficient concentrations.

Some common types of buffer solutions include:

The conjugate acid-base pair work together to resist changes in pH. For example, consider the ethanoate buffer system:

CH₃COOH ⇌ H⁺_(aq) + CH₃COO^-_(aq)

• If H⁺ is added, most of these additional H⁺ ions react with CH₃COO^- to form CH₃COOH. The equilibrium position shifts to the left, and [H⁺] decreases back towards its initial value.
• If OH^- is added, most of these additional OH^- ions react with H⁺ to form water: OH^-_(aq) + H⁺_(aq) ➔ H₂O_(l). This lowers [H⁺], causing more CH₃COOH to dissociate and replenish H⁺. The equilibrium position shifts to the right, and [H⁺] increases back towards its initial value.

This minimises the change in [H⁺] and pH.

Preparing buffers in the lab

There are several ways to prepare a buffer solution in the lab.

For example, an ethanoate buffer can be made by:

1. Dissolving solid CH_3_COONa and liquid CH_3_COOH in water - This method directly combines the conjugate acid-base pair in their pure forms to create the buffer solution.
2. Mixing solutions of CH_3_COONa and CH_3_COOH - This method combines pre-made solutions of the conjugate acid-base pair to create the buffer solution.
3. Reacting excess CH_3_COOH with NaOH - This method generates the conjugate base (CH_3_COO^−^) in situ via the equation: CH₃COOH_(aq) + OH⁻_(aq) ➔ CH₃COO⁻_(aq) + H₂O_(l), while the excess CH_3_COOH acts as the conjugate acid to form the buffer pair.
4. Reacting excess CH_3_COONa solution with HCl This method generates the conjugate acid (CH_3_COOH) in situ via the equation: CH₃COO⁻_(aq) + H⁺_(aq) ➔ CH₃COOH_(aq), while the excess CH_3_COO^−^ acts as the conjugate base to form the buffer pair.

Regardless of the method, the same two essential buffer components are produced - CH_3_COOH and CH_3_COO^-.

The Henderson-Hasselbalch equation

The pH of a buffer depends on:

1. The ratio of the conjugate acid and base concentrations - A higher proportion of base to acid gives a higher pH.
2. The acid dissociation constant (K_a) - A larger K_a (smaller pK_a) means the acid dissociates more, giving a lower pH.

The Henderson-Hasselbalch equation quantifies this relationship:

$pH = pK_a + log\frac{[A^{-}]}{[HA]}$

Where:

• pK_a = $-logK_a$ of the weak acid.
• [A^-] = concentration of conjugate base.
• [HA] = concentration of conjugate acid.

The Henderson-Hasselbalch equation allows us to calculate the pH of a buffer if the composition is known.

Some important points about buffer pH:

• Buffer pH is independent of dilution since [A⁻] and [HA] change by the same factor, leaving their ratio constant. However, infinite dilution will eventually produce pH 7 at 25°C.
• Buffer capacity (ability to resist pH changes) is limited. If too much strong acid/base is added, the weak conjugate is used up and buffer action is lost.

Worked example 1 - Calculating buffer pH

Calculate the pH of an ethanoate buffer containing 0.150 mol dm^-3^ ethanoic acid (CH₃COOH) and 0.250 mol dm^-3^ sodium ethanoate (CH₃COONa) at 25°C. The pKa of ethanoic acid at 25°C is 4.76. Give your answer to 2 decimal places.

Step 1: Equation

$pH = pK_a + log\frac{[A^{-}]}{[HA]}$

Step 2: Substitution and correct evaluation

$pH = 4.76 + log\frac{0.250}{0.150}=4.98$

The higher concentration of conjugate base CH_3_COO^-^ compared to CH_3_COOH gives a pH slightly above the pK_a_.

Worked example 2 - Calculating [base] : [acid] ratio

Calculate the ratio of concentrations of conjugate base and acid for a buffer with a pH of 10.90 prepared by reacting ethylamine (C_2_H_5_NH_2_) with HCl at 25°C. The pK_b of ethylamine at 25°C is 3.28.

Step 1: Write buffer equation

C_2_H_5_NH_2(aq)_ + HCl_(aq)_ ➔ C_2_H_5_NH_3_Cl_(aq)_

C_2_H_5_NH_2(aq)_ + H^+^_(aq)_ ➔ C_2_H_5_NH_3_^+^_(aq)_

Step 2: Identify the conjugate acid-base pair

Conjugate base (A^-) = C_2_H_5_NH_2_

Conjugate acid (HA) = C_2_H_5_NH_3_^+^

Step 3: Calculate pK_a of C_2_H_5_NH_3_^+^

$pK_a = 14.0 - pK_b = 14.0 - 3.28=10.72$

Step 4: Equation

$pH = pK_a + log\frac{[A^{-}]}{[HA]}$

Step 5: Rearrange equation

$\frac{[A^{-}]}{[HA]}=10^{(pH-pK_a)}$

Step 6: Substitution and correct evaluation

$\frac{[A^{-}]}{[HA]}=\frac{[C_2H_5NH_2]}{[C_2H_5NH_3^{+}]}=10^{(10.90-10.72)}=10^{0.18}=1.51$

Therefore the ratio of [C_2_H_5_NH_2_] : [C_2_H_5_NH_3_^+^] = 1.51 : 1

The buffer pH is slightly above the pK_a_ of the conjugate acid (10.72), so the ratio of conjugate base to acid is greater than 1.