Unit 8 · Topic 8.4 · about 30 minutes · 7 parts · 13 scored checks, 2 of them multiple choice, plus 2 from earlier lessons
AP Chemistry 8.4: Acid-Base Reactions and Buffers
Unit 8: 11 to 15% of the multiple-choice section
What you'll be able to do
- Work out what is in a mixture of an acid and a base after they react, and use the concentrations of the major species to explain the solution's pH.
Key vocabulary
| Term | What it means | The exam trap |
|---|---|---|
| Excess reagent | The reactant with more moles than its partner can react with, so some of it is left once the other is used up. | Find which reagent runs out before you decide what the pH depends on. |
| Buffer | A solution with a weak acid and its conjugate base, or a weak base and its conjugate acid, both present in significant amounts. | Name every species that is left after the reaction, not just the ones you started with. |
| Henderson-Hasselbalch equation | The relationship pH = pKa + log([base form] / [acid form]) for a solution that holds a conjugate pair. | Say which species is the base form and which is the acid form before you divide. |
| Conjugate acid and conjugate base | A pair of species that differ by one H⁺: the acid has it and the base does not. | A conjugate partner has its own acid-base character, so ask how it behaves before you decide what a solution does. |
| Quantitative reaction | A reaction that goes essentially to completion, so the limiting reagent is used up and you can subtract moles. | The final volume belongs to the whole mixture, not to either starting solution. |
Tap to open the picture full sizePart 1 Think first Cleaning up a spill
Commit to a guess first. Being wrong now makes the right idea stick better.
Predict first
A custodian cleans up a spill of drain cleaner, a sodium hydroxide solution, with vinegar, an acetic acid solution. She adds exactly enough vinegar to use up all the hydroxide, no more and no less, and then dips pH paper in the liquid. What does the paper show?
In short. When an acid and a base are mixed, a strong reagent reacts completely, and the pH is set by whatever the reaction leaves behind.
- For a strong acid and a strong base, subtract moles and find the pH of the excess reagent in the total volume.
- A weak acid with some strong base leaves a buffer, only the conjugate base, or excess hydroxide ion, and a weak base with strong acid works the same way in reverse.
- A weak acid and a weak base do not react completely: they reach an equilibrium with all four species present.
Part 2 Learn it What is left after mixing
2a. Strong acid plus strong base
When an acid and a base are mixed, count moles first and find concentrations last.
A strong acid and a strong base react completely: H⁺(aq) + OH⁻(aq) → H₂O(l). The reagent with fewer moles is used up, and the other, the excess reagent, is what is left.
- Find the moles of each, n = M × V, with V in liters.
- Subtract. The excess is the larger amount minus the smaller.
- Divide the excess by the total volume of the mixed solution.
- Find the pH: pH = −log[H⁺] if acid is in excess, or pOH = −log[OH⁻] and pH = 14.00 − pOH if base is in excess.
If the moles are equal, nothing strong is left and the pH is 7.00.
Example (a teaching example): 20.0 mL of 0.100 M HCl is mixed with 10.0 mL of 0.100 M NaOH. There are 2.00 × 10⁻³ mol of H⁺ and 1.00 × 10⁻³ mol of OH⁻, so 1.00 × 10⁻³ mol of H⁺ is left in 30.0 mL. [H⁺] = 0.0333 M and the pH is 1.48.
Quick check · Calculate
A student mixes 50.0 mL of 0.200 M HCl with 30.0 mL of 0.100 M NaOH. What is the pH of the mixture? Give your answer with two decimal places.
2b. Weak acid plus strong base
A strong base also reacts with a weak acid essentially to completion: HA(aq) + OH⁻(aq) ⇌ A⁻(aq) + H₂O(l). So the first move is the same: count moles and subtract. What is left depends on which reagent was in excess.
- Weak acid in excess. HA and its conjugate base A⁻ are both present, which is a buffer. Its pH comes from the Henderson-Hasselbalch equation: pH = pKa + log(n(A⁻) / n(HA)), where pKa = −log Ka. Both species are in one solution, so the ratio of moles is the ratio of concentrations and the volume cancels.
- Strong base in excess. The leftover OH⁻ sets the pH, found as in the last chunk. The weak A⁻ adds almost nothing beside it.
- Equal moles. The acid is used up and only A⁻ is left. It is a weak base: A⁻(aq) + H₂O(l) ⇌ HA(aq) + OH⁻(aq). The pH is slightly above 7. Use Kb = Kw/Ka, then [OH⁻] = √(Kb × [A⁻]).
A later topic, Henderson-Hasselbalch Equation, returns to the buffer case.
Quick check · Sort it
Each card describes a weak acid mixed with a strong base. Tap a card, then the bin for what is in the solution once the reaction is complete.
Weak acid and its conjugate base
Only the conjugate base
Excess hydroxide ion
2c. Weak base plus strong acid
This is the same pattern turned around. A strong acid reacts essentially completely with a weak base B: B(aq) + H₃O⁺(aq) ⇌ HB⁺(aq) + H₂O(l). Ammonia is the standard case, NH₃ + H₃O⁺ ⇌ NH₄⁺ + H₂O. Count moles, subtract, and see what is left.
- Weak base in excess. B and its conjugate acid HB⁺ are both present, a buffer. The Henderson-Hasselbalch equation uses the pKa of the conjugate acid, with the base on top: pH = pKa + log(n(B) / n(HB⁺)). For NH₄⁺, pKa = 9.25.
- Strong acid in excess. The leftover H₃O⁺ sets the pH.
- Equal moles. Only HB⁺ is left. It is a weak acid: HB⁺(aq) + H₂O(l) ⇌ B(aq) + H₃O⁺(aq). The pH is slightly below 7.
The conjugate acid of a weak base is a weak acid, and the conjugate base of a weak acid is a weak base. That is why a mixture with nothing left over is not always neutral.
Quick check · Decide and justify
A student mixes 25.0 mL of 0.100 M ammonia, NH₃, with 25.0 mL of 0.100 M hydrochloric acid. Is the final solution acidic, neutral or basic?
Which reason supports your decision?
2d. Weak acid plus weak base
When both partners are weak, nothing forces the reaction to completion. A weak acid and a weak base transfer a proton to each other and settle into an equilibrium:
HA(aq) + B(aq) ⇌ A⁻(aq) + HB⁺(aq)
At equilibrium all four species are present in measurable amounts, so you cannot subtract moles and read off what is left. Acetic acid with ammonia is an example:
CH₃COOH(aq) + NH₃(aq) ⇌ CH₃COO⁻(aq) + NH₄⁺(aq)
A double arrow, ⇌, is the signal. With a strong acid or base in the mixture, a single complete reaction does most of the work. With two weak partners, the equilibrium itself is the whole story.
Quick check · Multiple choice
Acetic acid, CH₃COOH, a weak acid, is mixed with ammonia, NH₃, a weak base. Which equation represents the reaction?
Part 3 Watch it done A buffer from a titration flask
Predict first, then watch each step, then finish the calculation yourself.
A student mixes 40.0 mL of 0.150 M acetic acid, CH₃COOH, with 25.0 mL of 0.100 M NaOH. The pKa of acetic acid is 4.74. What is the pH of the mixture?
Predict first
Before you open the first step, predict: will the pH be above or below 4.74?
Step 1 · Find the moles that go in
n(CH₃COOH) = (0.0400 L)(0.150 M) = 6.00 × 10⁻³ mol. n(OH⁻) = (0.0250 L)(0.100 M) = 2.50 × 10⁻³ mol.
Step 2 · Let the strong base react completely
The OH⁻ is limiting, so all 2.50 × 10⁻³ mol of it is used up. It converts 2.50 × 10⁻³ mol of CH₃COOH into CH₃COO⁻, which leaves 3.50 × 10⁻³ mol of CH₃COOH. Both the weak acid and its conjugate base are present: a buffer.
Why this step?
Why can the moles of CH₃COO⁻ and CH₃COOH be used in the ratio instead of their concentrations?
Both species are dissolved in the same 65.0 mL of solution. Dividing each by that volume gives the concentrations, and the same volume appears in the top and the bottom of the ratio, so it cancels.
Your step · Calculate
Finish it. After the reaction there are 3.50 × 10⁻³ mol of CH₃COOH and 2.50 × 10⁻³ mol of CH₃COO⁻. The pKa of acetic acid is 4.74. What is the pH? Give your answer with two decimal places.
The pH came out below the pKa because there was more acid than conjugate base. Had the base been added until the two were equal, the pH would equal the pKa. Past that, the conjugate base outnumbers the acid and the pH rises above the pKa.
Part 4 Try it Back to the spill
Put the spill into numbers. These are teaching examples with acetic acid, Ka = 1.8 × 10⁻⁵ (pKa 4.74), in 25.0 mL of 0.100 M solution, mixed with 0.100 M NaOH.
Try it · Calculate
Now the strong base is in excess. A student mixes 25.0 mL of 0.100 M CH₃COOH with 40.0 mL of 0.100 M NaOH. What is the pH? Give your answer with two decimal places.
Try it · Calculate
A student mixes 25.0 mL of 0.100 M CH₃COOH with 25.0 mL of 0.100 M NaOH. For acetic acid, Ka = 1.8 × 10⁻⁵. What is the pH? Give your answer with two decimal places.
Part 5 Explain it Ammonia with hydrochloric acid
On the exam this is handwritten, so answer on paper first. Nothing you type here leaves this page. Then check your answer against the model and each scoring point.
Free response is 50% of the exam score. This one is shaped like a short-answer question, which is worth 4 points.
Free-response practice · Short-answer style
A student mixes 30.0 mL of 0.100 M NH₃(aq) with 10.0 mL of 0.100 M HCl(aq). The pKa of NH₄⁺ is 9.25.
(a) Represent the reaction that occurs when the solutions are mixed with a net ionic equation.
(b) Calculate the moles of NH₃ and of NH₄⁺ in the mixture after the reaction.
(c) Calculate the pH of the mixture.
(d) The student adds more HCl until the moles of HCl added equal the moles of NH₃ originally present. Predict whether the pH is above, equal to or below 7, and justify your prediction.
Model answer
(a) NH₃(aq) + H₃O⁺(aq) ⇌ NH₄⁺(aq) + H₂O(l). Writing H⁺ in place of H₃O⁺ is also accepted.
(b) The mixture starts with (0.0300 L)(0.100 M) = 3.00 × 10⁻³ mol of NH₃ and (0.0100 L)(0.100 M) = 1.00 × 10⁻³ mol of H₃O⁺. The acid is used up, so 2.00 × 10⁻³ mol of NH₃ is left and 1.00 × 10⁻³ mol of NH₄⁺ forms.
(c) pH = pKa + log(n(NH₃) / n(NH₄⁺)) = 9.25 + log(2.00 × 10⁻³ / 1.00 × 10⁻³) = 9.25 + 0.30 = 9.55.
(d) Below 7. All of the NH₃ has been converted to NH₄⁺, which is a weak acid. It reacts with water, NH₄⁺ + H₂O ⇌ NH₃ + H₃O⁺, and makes the solution slightly acidic.
Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.
Part 6 Check yourself Four mixtures
All three case questions use the table.
A student mixes 25.0 mL of a 0.100 M acid with a volume of 0.100 M NaOH. HCl is a strong acid. HF is a weak acid with pKa 3.17 (Ka = 6.8 × 10⁻⁴). These numbers are a teaching example.
| Mixture | Acid | Volume of NaOH added (mL) |
|---|---|---|
| A | HCl | 15.0 |
| B | HF | 15.0 |
| C | HF | 25.0 |
| D | HF | 35.0 |
Case question 1 · Calculate
Using the table: what is the pH of mixture B? Give your answer with two decimal places.
Case question 2 · Multiple choice
Using the table: after mixture C has reacted completely, which equilibrium sets its pH?
Case question 3 · Calculate
Using the table: what is the pH of mixture A? Give your answer with two decimal places.
The rest stand on their own and do not use the case.
Check yourself · Put it in order
A student wants the pH after a weak acid is mixed with a strong base. Put the steps in order.
Check yourself · Spot the error
A student finds the pH when 30.0 mL of 0.100 M NH₃ is mixed with 10.0 mL of 0.100 M HCl. The pKa of NH₄⁺ is 9.25. Tap the line where the reasoning first goes wrong.
Check yourself · Match
Match each mixture to what you do next.
Part 7 Keep it Bring back earlier ideas
Mixing problems start with moles from a molarity and a volume, and later steps use the same skill again. Pulling it back up is what makes it last.
From 3.7 Solutions and Mixtures · Calculate
A student dissolves 29.2 g of sodium chloride, NaCl (molar mass 58.44 g/mol), in water and adds water to make 250.0 mL of solution. What is the molarity of the solution? Give your answer to three significant figures.
Two lab situations.
From 3.7 Solutions and Mixtures · Calculate
A student uses a pipet to put 25.00 mL of 2.00 M glucose into a 100.0 mL volumetric flask, then adds water up to the mark and mixes. What is the molarity of the new solution? Give your answer to three significant figures.
In three lines
- A strong acid and a strong base react completely; subtract moles and find the pH from the excess reagent in the total volume.
- A weak acid with a strong base (or a weak base with a strong acid) also reacts completely, and what is left is a buffer, only the conjugate, or excess strong reagent.
- A weak acid and a weak base do not react completely; they reach an equilibrium with all four species present.
Exam strategy for this topic
- Section I, Multiple Choice: 60 questions in 90 minutes.
- Section II, Free Response: 7 questions in 105 minutes.
- A scientific or graphing calculator is recommended for use on both sections of the exam.
Start every mixing problem with moles from n = M × V, let the strong reagent react completely, and write down what is left before choosing a method. Give pH to the decimal places the question asks for, and keep extra digits until the last step. Use the right arrow: a single arrow for a complete reaction, a double arrow for an equilibrium.
Questions students ask
How do you find the pH when a strong acid and a strong base are mixed?
Find the moles of each, subtract, and divide the excess by the total volume. The excess acid gives pH = −log[H⁺]; the excess base gives pOH = −log[OH⁻] and pH = 14.00 − pOH. Equal moles give pH 7.00.
What is left when a weak acid reacts with a strong base?
If the acid is in excess, a mixture of the weak acid and its conjugate base (a buffer). If the base is in excess, leftover hydroxide ion. If the moles are equal, only the conjugate base.
When does a mixture form a buffer?
When a weak acid is mixed with less strong base than acid, or a weak base with less strong acid than base, so that both members of a conjugate pair remain.
Why is the pH above 7 when equal moles of a weak acid and a strong base are mixed?
All the acid is converted to its conjugate base, which is a weak base and reacts with water to make a little hydroxide ion.
Which equation do you use for a weak acid and a weak base?
HA + B ⇌ A⁻ + HB⁺, with a double arrow, because the reaction reaches an equilibrium instead of going to completion.
Course alignment, for teachers
AP Chemistry topic 8.4, Unit 8: Acids and Bases.
- Skill 5.F: Calculate, estimate, or predict an unknown quantity from known quantities by selecting and following a logical computational pathway and attending to precision (e.g., performing dimensional analysis and attending to significant figures).
This lesson maps all 4 essential knowledge statements of the topic to the part that teaches each one: 4 are checked by a scored question. The map is in the lesson file.