General Chemistry · Aqueous Equilibria · BUFFER ACTION

Buffer action — same NaOH, different pH change

Add 5.0 mL of 4.0 M NaOH to 1 L of pure water and to 1 L of an acetic acid buffer solution, side by side. The pH of both beakers is calculated by solving the charge-balance equation exactly. Dilute the buffer components or add more base, and find for yourself the moment when the buffer capacity breaks down.

Learning objectives After this activity you can …

explain, from the reactions of the components, the difference in pH change when the same OH⁻ is added to a buffer and to water.

calculate the pH of a buffer after adding a strong base in two steps: stoichiometry → equilibrium.

show by experiment how the amount of buffer components relates to the buffer capacity.

Key summary 1 and 3 — shown after you submit your prediction.
Key summary 1 — A strong base (OH⁻) reacts completely with the weak acid HA of the buffer and turns it into A⁻.
Key summary 2 — shown after you submit your prediction.
Key summary 2 — The pH depends only on the log of the ratio [A⁻]/[HA], so the pH change is small (4.74 → 4.80).
Key summary 3 — When HA runs out, the leftover OH⁻ sets the pH directly, just as in water (the buffer breaks down).

Estimated time 15–20 min · Prerequisites: pH and pOH, Ka of a weak acid, amounts in neutralization reactions (limiting reactant)

Progress 0% · Predict → Observe → Check → Complete

Level Basic Standard Advanced

Two beakers compared

Buffer pH —
ΔpH —

Pure water pH —
ΔpH —

Amounts of components (bars)

Stoichiometry step (mol after reaction)—

pH vs volume added (solid: buffer · dashed: water)

Point where a buffer component is used up (vertical dotted line)—

Entry check Before you start — prerequisites and key terms

What happens when OH⁻ from a strong base meets the weak acid CH₃COOH?

They react almost completely to form CH₃COO⁻ and H₂O It is a weak acid, so they do not react Only half reacts, then it stops at equilibrium

Key terms

Buffer solution: a solution that contains a weak acid and its conjugate base (or a weak base and its conjugate acid) in similar amounts, so its pH changes little when a small amount of strong acid or strong base is added. · Conjugate acid–base pair: HA/A⁻, which differ by one H⁺ (here CH₃COOH/CH₃COO⁻). · Henderson–Hasselbalch equation: pH = pKa + log([A⁻]/[HA]). · Buffer capacity: the amount of strong acid or strong base a buffer can take up before its pH changes greatly. · Charge balance: total positive charge = total negative charge in the solution.

Predict Think first, before you observe

You add 5.0 mL of 4.0 M NaOH to 1 L of pure water (pH 7.00) and to 1 L of a 0.300 M CH₃COOH/0.300 M CH₃COO⁻ buffer (pH 4.74). How large is the rise in pH?

Both rise by a similar, large amount The buffer rises much less The buffer rises more The pH of the buffer does not change at all

ConfidenceMedium

The explanation opens after you lock in a prediction

Volume of 4.0 M reagent added 0.0 mL

0 mL100 mL

Reagent added NaOH

strong base NaOHstrong acid HCl

Concentration of each buffer component 0.300 M

0.005 M (dilute)0.500 M (concentrated)

Actions

Explanation notes (open after prediction)

The two-step calculation (stoichiometry → equilibrium) and the explanation of the pH changes in water and in the buffer open after you submit your prediction. For now, think about what the added OH⁻ meets in each beaker.

Two-step calculation

① Stoichiometry: OH⁻ reacts completely with HA (K = 1/K_b ≈ 10⁹, practically complete). 0.020 mol OH⁻ → HA 0.300→0.280 mol, A⁻ 0.300→0.320 mol. ② Equilibrium: use the remaining HA/A⁻ to find the pH. The volume (1.005 L) cancels in the ratio.

pH = 4.74 + log(0.320 / 0.280) = 4.80

Why water changes so much · buffer breakdown

Water has no HA to take up the OH⁻, so [OH⁻] = 0.020/1.005 = 0.0199 M → pOH 1.70 → pH 12.30. A buffer also jumps in pH once its HA is completely used up: the leftover OH⁻ then sets the pH (buffer capacity exceeded). Soils resist pH change in a similar way: a soil with a larger buffer capacity needs more lime to change its pH.

[Na⁺] + [H⁺] = [A⁻] + [OH⁻] (+ [Cl⁻])

Limits of this model — activity coefficients ignored (ideal solution), 25 °C · K_w = 1.0×10⁻¹⁴, K_a(CH₃COOH) = 1.8×10⁻⁵, volumes are additive, CO₂ uptake from air ignored. The starting pH 7.00 of "pure water" is an ideal value; real distilled water is about 5.6 because of CO₂. The pH on screen is a numerical solution of the charge balance, without approximation (x is not neglected).

Check Formative check — pass all items to complete

Q1What mainly happens to the OH⁻ added to the buffer?

It reacts with CH₃COO⁻ and disappears It reacts with CH₃COOH to form CH₃COO⁻ and H₂O It does not react and stays in the solution It combines with Na⁺ and precipitates

Hint

OH⁻ is a base. A base reacts with something that can give it an H⁺.

Q2 · TaskKeep NaOH at 5.0 mL (±0.5) and lower only the buffer concentration until the buffer ΔpH is 1.0 or more. Then press [Check]. (buffer capacity breakdown)

Current buffer ΔpH: — · concentration 0.300 M

Q3 · Calculate—

Hint

① mol of OH⁻ added = 4.0 × V(L). ② HA decreases by that amount and A⁻ increases. ③ pH = 4.74 + log(n_A⁻/n_HA).

Q4 · Fill in the blankFill in the blank: the largest amount of strong acid or strong base that a buffer can take up without a large pH change is called the buffer [＿＿].

Hint

The word means "how much a container can hold".

Q5 · Apply (advanced, not required for completion)For 1.000 L of the 0.300 M/0.300 M acetic acid buffer, what is the smallest volume (mL) of 4.0 M NaOH needed to use up all the CH₃COOH?

Reflect Reflect — metacognition · completion

My prediction result: Not predicted yet

In one sentence: why does the buffer change its pH much less than water, with the same NaOH?

I can explain which component reacts with the added OH⁻ I can calculate the pH in two steps: stoichiometry → equilibrium I know that buffer capacity depends on the concentration of the components

Apply Use the idea in a new context (not required for completion)

Transfer task: blood keeps its pH at about 7.4 with the H₂CO₃/HCO₃⁻ buffer system. During hard exercise, lactic acid (H⁺) floods in. Which component takes up the H⁺, and what happens when HCO₃⁻ runs out (acidosis)? Explain, linking it to the strong acid HCl mode of this simulation.

Next up: acid–base titrations — the buffer region and pH = pKa at the half-equivalence point, then solubility equilibria (Ksp).

COMPLETE · Completed — all checks passed.

Worksheet — Buffer action: same NaOH, different pH change

Name ________ Class ________ Date ________

1. Predict — Which beaker has the smaller rise in pH? Why?

2. Stoichiometry — Reaction: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. After the reaction, n(HA) = mol, n(A⁻) = mol

3. Equilibrium — buffer pH = 4.74 + log( / ) = · water pH =

4. Fill in the blank — When a strong base is added, the [] component of the buffer uses up the OH⁻. The limit of what it can take up is called the buffer [].

5. Conclusion — If the buffer concentration is lowered to 0.020 M, what happens with the same NaOH?

Answer key (end of teacher printout) — 1. the buffer (HA uses up the OH⁻) · 2. 0.280 / 0.320 · 3. 0.320/0.280 → 4.80, water 12.30 · 4. weak acid (HA) / capacity · 5. all 0.020 mol of HA is used up, and the pH jumps to about 8.7 (buffer breakdown).

Teacher notes — open the lesson plan

Lesson plan · 5E phase and time

Explore and Explain, 15–20 min. Whole-class demonstration (?mode=teacher) or individual work.

Demonstration in 3 steps

① Standard condition (0.300 M, NaOH 5.0 mL): buffer ΔpH 0.06 vs water ΔpH 5.30.
② In the bar chart, HA↓ and A⁻↑ — "Where did the OH⁻ go?"
③ Lower the concentration to 0.020 M, or add 75 mL of NaOH or more, to show the buffer breakdown.

Three key questions

"The amount of OH⁻ is the same. Why does only one beaker reach pH 12?"
"Why did the buffer pH change at all, even by 0.06?"
"Can a buffer be used forever? What sets the limit?"

Common misconceptions

MC-buf-01: "it rises as much as water" → use the bars to show that OH⁻ is used up by HA.
MC-buf-02: "it does not change at all" → the ratio 0.320/0.280 is not 1, so the pH rises by 0.06.
MC-buf-03: "capacity is unlimited" → HA 0.300 mol = NaOH 75 mL is the limit.

Answer key (teacher only)

Basis and sources

Buffer 4.74→4.80, water 7.00→12.30.
K_a(CH₃COOH) = 1.8×10⁻⁵ (common reference value). The Henderson–Hasselbalch approximation (pKa 4.74) and the exact charge-balance solution of this simulation agree to two decimal places (4.745→4.803).
Verification: multi-point comparison with an external Python charge-balance solver (scipy brentq) — error < 0.001 pH.

Prerequisites

pH and pOH, Ka, limiting-reactant calculations, the common-ion effect.

University general chemistry · buffer solutionsEngine E2 equation-driven (exact charge balance)Verified: checked against an external solver (2026-10-01)

Teacher mode — class controls