Add strong acid
H⁺ + A⁻ → HA. Subtract the added acid amount from A⁻ and add the same amount to HA, provided A⁻ is not exhausted.
SKILLS · CHEMISTRY & PHYSICS
Buffers: react first, then calculate
Adding strong acid or base changes the amounts of both partners. Complete that reaction before calculating the new pH.
01 · THE IDEA
First neutralize the added strong acid or base. Then use the remaining conjugate pair.
pH ≈ pKa + log₁₀(base / acid)
Use HA for weak acid and A⁻ for its conjugate base. When both occupy the same final volume, their mole ratio equals their concentration ratio. Assume ideal dilute behavior and appreciable amounts of both partners unless stated otherwise.
Add strong base
OH⁻ + HA → A⁻ + H₂O. Subtract the added base amount from HA and add it to A⁻, provided HA is not exhausted.
Dilution and exhaustion
Dilution preserves the ratio approximately while lowering buffer capacity per unit volume. If one partner is consumed completely, the usual buffer approximation no longer applies; account for excess strong reagent and the relevant equilibrium.
CHANGE ONE PART OF THE MODEL
Watch both buffer partners change
Start with 6 mmol A⁻ and 2 mmol HA; pKa = 7.2. Add strong acid in negligible volume. Both partners remain present over this slider’s range.
A⁻ remaining6 mmol
HA after reaction2 mmol
Approximate pH7.68
pH ≈ 7.2 + log₁₀(6 / 2)
02 · WORKED EXAMPLES
Follow the setup, then the numbers
1. Strong acid changes both sides
A buffer has pKa = 7.2 and initially contains 6 mmol A⁻ and 2 mmol HA. Add 2 mmol H⁺ in negligible volume.
- A⁻ after reaction = 6 − 2 = 4 mmol.
- HA after reaction = 2 + 2 = 4 mmol.
- Base/acid = 4/4 = 1; log₁₀1 = 0.
- pH ≈ 7.2.
Carry this forward: Subtracting acid from base alone would miss the newly formed HA.
2. Strong base reverses the changes
A separate buffer has pKa = 6.5, 2 mmol A⁻, and 6 mmol HA. Add 1 mmol OH⁻ in negligible volume. Use log₁₀0.60 ≈ −0.22.
- A⁻ becomes 3 mmol and HA becomes 5 mmol.
- pH ≈ 6.5 + log₁₀(3/5) = 6.28.
- The pH rose from about 6.02, although it remains below pKa because HA still exceeds A⁻.
Carry this forward: “pH rises” does not necessarily mean “pH exceeds pKa.”
03 · TRY IT IN DIFFERENT FORMS
Practice in different forms
For each question, identify the requested quantity and write the setup before calculating. Use a hint if needed, then choose an answer to see the explanation.
Untimed learning practice. These responses stay on this page and reset when you leave or reload; they do not change your test results.
QUESTION 1 / 4 · C/P
Acid added to a buffer
A buffer with pKa = 6.5 has 8 mmol A⁻ and 4 mmol HA. Add 2 mmol strong acid in negligible volume. Under the buffer approximation, what is the final pH?
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TAKE INTO A FRESH PASSAGE
Your setup checklist
- Convert added strong acid or base to moles.
- Use the reaction to update both HA and A⁻.
- Check for exhaustion or excess strong reagent.
- Use the new base/acid ratio and check the direction of the pH change.
Before calculating, explain what each term represents and what the requested answer would mean.
Concept references
All scenarios and questions here are original learning exercises. These references support the underlying concepts; the simplified assumptions are stated within each example.