Buffer Solutions and pH Calculations: Acid-Base Equilibria

Acid-Base Equilibria

Buffer Solutions

A buffer solution is a solution that resists changes in pH when small amounts of acid or base are added. Buffers typically consist of a weak acid and its conjugate base, or a weak base and its conjugate acid.

• Weak Acid Example: Propionic acid (C2H5COOH)

• Conjugate Base Example: Sodium propionate (C2H5COONa)

The effectiveness of a buffer depends on the concentrations of the acid and its conjugate base.

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is used to calculate the pH of a buffer solution:

$\mathrm{pH} = \mathrm{p}K_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right)$

• $[\text{A}^-]$ = concentration of the conjugate base (e.g., propionate ion)

• $[\text{HA}]$ = concentration of the weak acid (e.g., propionic acid)

• $K_a$ = acid dissociation constant of the weak acid

• $\mathrm{p}K_a = -\log K_a$

Example: Calculating Buffer pH

Given:

• 11.52 g sodium propionate ($C_2H_5COONa$) in 1.00 L solution

• 0.090 mol propionic acid ($C_2H_5COOH$) in 1.00 L solution

• $K_a$ for propionic acid = $1.3 \times 10^{-5}$

Step 1: Calculate moles of sodium propionate

• Molar mass of $C_2H_5COONa$ = 96.06 g/mol

• Moles = $\frac{11.52\ \text{g}}{96.06\ \text{g/mol}} = 0.12$ mol

Step 2: Use Henderson-Hasselbalch equation

• $[\text{A}^-] = 0.12$ M (since 1.00 L)

• $[\text{HA}] = 0.090$ M

• $\mathrm{p}K_a = -\log(1.3 \times 10^{-5}) = 4.89$

• $\mathrm{pH} = 4.89 + \log\left(\frac{0.12}{0.090}\right) = 4.89 + \log(1.33) = 4.89 + 0.12 = 5.01$

Example: The pH of the buffer is approximately 5.01.

Buffer Response to Addition of Strong Base (KOH)

When a strong base (e.g., KOH) is added to a buffer, it reacts with the weak acid component, converting it to its conjugate base:

• $\text{KOH} + \text{HA} \rightarrow \text{A}^- + \text{H}_2O$

Steps:

1. Subtract moles of KOH added from moles of HA.

2. Add moles of KOH added to moles of A-.

3. Recalculate pH using the Henderson-Hasselbalch equation.

Example: If 0.020 mol KOH is added:

• New $[\text{HA}] = 0.090 - 0.020 = 0.070$ mol

• New $[\text{A}^-] = 0.12 + 0.020 = 0.14$ mol

• $\mathrm{pH} = 4.89 + \log\left(\frac{0.14}{0.070}\right) = 4.89 + \log(2) = 4.89 + 0.30 = 5.19$

Example: The new pH after adding KOH is approximately 5.19.

Buffer Response to Addition of Strong Acid (HBr)

When a strong acid (e.g., HBr) is added to a buffer, it reacts with the conjugate base component, converting it to the weak acid:

• $\text{HBr} + \text{A}^- \rightarrow \text{HA} + \text{Br}^-$

Steps:

1. Subtract moles of HBr added from moles of A-.

2. Add moles of HBr added to moles of HA.

3. Recalculate pH using the Henderson-Hasselbalch equation.

Example: If 0.010 mol HBr is added:

• New $[\text{A}^-] = 0.12 - 0.010 = 0.11$ mol

• New $[\text{HA}] = 0.090 + 0.010 = 0.10$ mol

• $\mathrm{pH} = 4.89 + \log\left(\frac{0.11}{0.10}\right) = 4.89 + \log(1.1) = 4.89 + 0.04 = 4.93$

Example: The new pH after adding HBr is approximately 4.93.

Summary Table: Buffer Calculations

Situation	[A-] (mol)	[HA] (mol)	pH
Original Buffer	0.12	0.090	5.01
After 0.020 mol KOH	0.14	0.070	5.19
After 0.010 mol HBr	0.11	0.10	4.93

Key Points:

• Buffers resist changes in pH upon addition of small amounts of strong acid or base.

• The Henderson-Hasselbalch equation is essential for buffer pH calculations.

• Always adjust the moles of acid and base after addition of strong acid/base before recalculating pH.