Unit 8 · Topic 8.9 Beta

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), is the Ka expression in log form.

Practice 5: Mathematical Routines

Question set for this topic

Part 1 · Hook

Why this matters

A biology lab needs a phosphate buffer at exactly pH 7.40 for an enzyme that stops working outside a narrow range. Nobody finds that pH by trial and error. One line of algebra, the Henderson-Hasselbalch equation, tells you which acid to use and how much of each component to weigh out.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is log(2.0)?

  1. 0.30
  2. 0.69
  3. 2.0
  4. −0.30
Show the answer

log(2.0) = 0.301. (0.69 is ln 2.)

  • Correct: 0.30:
  • 0.69:
  • 2.0:
  • −0.30:

2. In a buffer, the pH depends on what?

  1. The ratio [A⁻]/[HA]
  2. The total concentration of acid only
  3. The volume of the solution
  4. The amount of water
Show the answer

[H₃O⁺] = Ka × [HA]/[A⁻]: only the ratio matters.

  • Correct: The ratio [A⁻]/[HA]:
  • The total concentration of acid only:
  • The volume of the solution:
  • The amount of water:

3. Which form predominates at a pH one unit below pKa?

  1. HA
  2. A⁻
  3. Equal amounts
  4. Neither
Show the answer

Below pKa, the protonated form wins, by 10 : 1 at one unit below.

  • Correct: HA:
  • A⁻:
  • Equal amounts:
  • Neither:

Part 4 · See it

See it first

pH against log([A⁻]/[HA]) for pKa 4.74: a straight line of slope 1 through (0, 4.74), (1, 5.74) and (−1, 3.74). The shaded band is ratios from 0.1 to 10. A note: use the base-10 log; ln would give pKa + 2.30 for a 10:1 ratio.
pH against log([A⁻]/[HA]): a line of slope 1 that crosses pKa where the ratio is 1. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

  1. Ka = [H₃O⁺][A⁻]/[HA] holds in a buffertaking −log of both sides gives pH = pKa + log([A⁻]/[HA])
  2. The ratio is 1 when the two components are equallog 1 = 0, so pH = pKa
  3. Every tenfold change in the ratiomoves the pH by exactly one unit (base-10 log)
  4. A useful buffer needs both components in real amountspick an acid whose pKa is within about 1 of the target pH, then set the ratio

Part 6 · Key ideas

Key ideas

  • pH = pKa + log([A⁻]/[HA]), base on top, base-10 log (never ln).
  • Moles can replace concentrations: both share one volume, which cancels.
  • Weak base buffer (NH₃/NH₄⁺): use pKa of the conjugate acid, 14.00 − pKb.
  • To make a buffer: choose pKa ≈ target pH, find the ratio 10^(pH − pKa), then the amounts (or the volume of NaOH for partial neutralization).

Part 7 · Misconception

A common mistake

The wrong idea: The Henderson-Hasselbalch equation uses the natural log, like the integrated rate law for a first-order reaction.

What actually happens: It is the base-10 log, because pH and pKa are base-10 logs. Using ln moves the answer by a factor of 2.303 in the log term, a frequent point lost on recent exams.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Data table

Weak acids for making buffers

A lab stocks each weak acid below and a sodium salt of its conjugate base (for NH₄⁺, the stock is NH₃ and NH₄Cl).

Ka and pKa at 25 °C
Weak acidFormulaKapKa
Formic acidHCOOH1.8 × 10⁻⁴3.74
Acetic acidCH₃COOH1.8 × 10⁻⁵4.74
Dihydrogen phosphateH₂PO₄⁻6.2 × 10⁻⁸7.21
Ammonium ionNH₄⁺5.6 × 10⁻¹⁰9.25

1. A student needs a buffer at pH 7.40. Which acid and conjugate base pair is the best choice?

  1. H₂PO₄⁻ and HPO₄²⁻
  2. CH₃COOH and CH₃COO⁻
  3. NH₄⁺ and NH₃
  4. HCOOH and HCOO⁻
Show the answer

Choose the acid whose pKa is closest to the target pH: H₂PO₄⁻ has pKa 7.21, within 0.2 of 7.40, so the ratio needed is near 1.

  • Correct: H₂PO₄⁻ and HPO₄²⁻: Right: pKa 7.21, closest to 7.40.
  • CH₃COOH and CH₃COO⁻: pKa 4.74 is 2.7 units away; the ratio would need to be about 450 : 1, a poor buffer.
  • NH₄⁺ and NH₃: pKa 9.25 is 1.85 units above the target; the ratio would be about 1 : 70.
  • HCOOH and HCOO⁻: pKa 3.74 is far too low.

2. What is the pH of a buffer that is 0.25 M HCOOH and 0.15 M HCOONa?

Type a number.

Show the answer

pH = pKa + log([HCOO⁻]/[HCOOH]) = 3.74 + log(0.15/0.25) = 3.74 − 0.22 = 3.52. More acid than base, so the pH is below pKa, as it should be.

  • Answer: 3.52

3. What is the pH of a buffer that is 0.30 M NH₃ and 0.20 M NH₄Cl?

Type a number.

Show the answer

The acid of the pair is NH₄⁺ (pKa 9.25) and the base is NH₃. pH = 9.25 + log(0.30/0.20) = 9.25 + 0.18 = 9.43.

  • Answer: 9.43

4. For an acetic acid / sodium acetate buffer at pH 5.00, what ratio [CH₃COO⁻]/[CH₃COOH] is needed?

Type a number.

Show the answer

log([A⁻]/[HA]) = pH − pKa = 5.00 − 4.74 = 0.26; [A⁻]/[HA] = 10^0.26 = 1.8.

  • Answer: 1.8

5. How many grams of NH₄Cl (53.49 g/mol) must be dissolved in 0.500 L of 0.200 M NH₃ to make a buffer at pH 9.00? Include the unit.

Type a number and its unit.

Show the answer

pH = pKa + log([NH₃]/[NH₄⁺]): 9.00 = 9.25 + log(ratio), so [NH₃]/[NH₄⁺] = 10^−0.25 = 0.562. mol NH₃ = 0.500 L × 0.200 M = 0.100 mol, so mol NH₄⁺ = 0.100/0.562 = 0.178 mol. Mass = 0.178 mol × 53.49 g/mol = 9.51 g.

  • Answer: 9.51 g

Experimental setup

Making a buffer from one acid and NaOH

A student must make a buffer of pH 5.04 from 50.0 mL of 0.200 M acetic acid (pKa = 4.74) and 0.100 M NaOH. The plan is to add NaOH from a buret to the acid in a beaker, then transfer to a 250.0 mL volumetric flask and dilute to the mark with distilled water.

6. How many mL of 0.100 M NaOH should be added? Include the unit.

Type a number and its unit.

Show the answer

Ratio needed: [A⁻]/[HA] = 10^(5.04 − 4.74) = 10^0.30 = 2. With x mol OH⁻: x/(0.0100 − x) = 2, so x = 0.00666 mol. Volume = 0.00666 mol ÷ 0.100 mol/L = 0.0666 L = 66.6 mL.

  • Answer: 66.6 mL

Part 9 · Summary

Summary

The Henderson-Hasselbalch equation, pH = pKa + log([A⁻]/[HA]), is the Ka expression in log form. It gives a buffer's pH from its ratio, and works backwards to design a buffer: pick an acid with pKa near the target, find the ratio 10^(pH − pKa), and turn it into amounts. Use base-10 log, and pKa of the conjugate acid for a weak base buffer.

Part 10 · Up next

What comes next

Part 11 · Connections

Connections