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II-11: Buffers and titration mathematics

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Module II-11 · Recommended before: Before buffer and titration curves

Buffers and titration mathematics

Henderson–Hasselbalch turns a concentration ratio into a pH shift. During titrations, perform reaction stoichiometry before equilibrium calculations.

Learn the method

pH = pKa + log₁₀([A⁻]/[HA])

  1. Identify acid and conjugate base.
  2. Calculate amounts after any neutralization first.
  3. Use the Henderson–Hasselbalch equation in the buffer region.
  4. At equivalence, use the appropriate conjugate-acid/base equilibrium.
  5. Check whether the answer is chemically plausible.

Common mistake: Do not apply Henderson–Hasselbalch unchanged beyond the buffer region of a titration.

Worked example

Problem: For a buffer with pKa = 4.76 and [A⁻]/[HA] = 10, find pH.

Step-by-step solution: pH = 4.76 + log₁₀(10) = 5.76.

Check the units and whether the answer is reasonable before continuing.

More applications of this mathematics

Try each extension before expanding its solution.

Buffer after neutralization

Start with 0.100 mol HA and 0.100 mol A⁻. Add 0.020 mol strong acid: A⁻→0.080 mol and HA→0.120 mol. Then pH=pKa+log(0.080/0.120)=pKa−0.176.

Titration stages

Before equivalence, account for stoichiometric neutralization. At equivalence of weak acid and strong base, calculate pH from conjugate-base hydrolysis, not from Henderson–Hasselbalch with zero HA.

Buffer capacity is not just pH

Two buffers can have the same pH and different capacities. Capacity depends on the total buffer components as well as their ratio; stoichiometry of added strong acid/base is the first step.

Independent practice — five questions

Suggested mastery threshold: four out of five. Incorrect answers display a worked explanation after submission.

Choose one response for each question, then select “Check my answers.” No timer. Hints are optional. You can retry freely. Questions are formative, not proctored.

Question 1. Use Henderson–Hasselbalch: pKa=5.30 and [A⁻]/[HA]=1. What is buffer pH?
Hint for question 1

Use the conjugate-base-to-acid ratio.

Question 2. Use Henderson–Hasselbalch: pKa=6.00 and [A⁻]/[HA]=0.1. What is buffer pH?
Hint for question 2

Use the conjugate-base-to-acid ratio.

Question 3. Use Henderson–Hasselbalch: pKa=6.50 and [A⁻]/[HA]=10. What is buffer pH?
Hint for question 3

Use the conjugate-base-to-acid ratio.

Question 4. Use Henderson–Hasselbalch: pKa=7.00 and [A⁻]/[HA]=1. What is buffer pH?
Hint for question 4

Use the conjugate-base-to-acid ratio.

Question 5. Use Henderson–Hasselbalch: pKa=7.40 and [A⁻]/[HA]=0.1. What is buffer pH?
Hint for question 5

Use the conjugate-base-to-acid ratio.

Supporting reference

OpenStax Chemistry 2e / NIST — reference for this subject. This lesson’s wording, examples and questions are original.

W3C WCAG 2.2 accessibility guidance.

Applied practice: Buffer additions and titration calculations

These additional, ungraded problems cover skills beyond the scored five-question check above. Try each before expanding its solution. Your existing score and progress records are unchanged.

  1. Challenge 1. A buffer contains 0.200 mol A⁻ and 0.300 mol HA, with pKₐ = 4.76. Add 0.050 mol HCl (assume the buffer remains valid). Find the new pH.

    Show worked solution for challenge 1

    HCl consumes 0.050 mol A⁻ and forms 0.050 mol HA. New amounts: A⁻ = 0.150 mol; HA = 0.350 mol. pH = 4.76 + log₁₀(0.150/0.350) = 4.39.

  2. Challenge 2. Mix 25.0 mL of 0.100 M HCl with 20.0 mL of 0.100 M NaOH at 25 °C. Find the pH before equivalence.

    Show worked solution for challenge 2

    Initial acid = 0.00250 mol; base = 0.00200 mol; excess H⁺ = 0.00050 mol. Total volume = 0.0450 L; [H⁺] = 0.0111 M. pH = −log₁₀(0.0111) = 1.95.

IntoChemistry original instructional content, 2026. Formative feedback is not an official course grade. Sources are linked for independent reading. Accessibility testing with your institution and assistive technology is still recommended.