AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/8/8-4)
Unit 8 · Topic 8.4
8.4 Acid-Base Reactions and Buffers
When you mix an acid and a base, let the strong one react completely first, then look at what's left. Leftover strong acid or base sets the pH; leftover weak acid together with its conjugate base forms a buffer; and if the weak acid and strong base are exactly used up, the conjugate base that remains makes the solution basic.
Key terms
- neutralization
- excess reagent
- buffer
- conjugate base
- weak acid–strong base reaction
Step 1: react first, in moles
Reactions between a strong acid or base and anything else go essentially to completion. So the first step is a stoichiometry problem, not an equilibrium problem. Convert everything to moles (or millimoles: mL × M = mmol), react them, and see what's left.
Then decide which situation you're in by looking at the major species remaining.
Checklist before you calculate
- List the moles of every acid and base before mixing.
- Write the reaction between the strongest acid and the strongest base present, and run it to completion.
- List the major species left (ignore spectator ions like Na⁺, K⁺, Cl⁻ and NO₃⁻).
- Match the leftovers to one case below, and only then pick an equation.
Strong acid plus strong base
The net ionic reaction is H₃O⁺(aq) + OH⁻(aq) → 2H₂O(l). Whichever is in excess sets the pH: divide the leftover moles by the total volume to get the concentration, then find pH. If neither is left, the solution is neutral (pH 7.00 at 25 °C), because the remaining ions like Na⁺ and Cl⁻ don't affect pH.
Weak acid plus strong base
The reaction is HA(aq) + OH⁻(aq) → A⁻(aq) + H₂O(l), and it goes to completion. Three outcomes are possible.
| What's left after reacting | Type of solution | How to find pH |
|---|---|---|
| Some HA and some A⁻ | Buffer | Henderson–Hasselbalch: pH = pKa + log([A⁻]/[HA]) |
| Only A⁻ (equal moles reacted) | Weakly basic | Kb of A⁻ = Kw/Ka, then a weak-base ICE calculation |
| A⁻ plus excess OH⁻ | Basic | Excess OH⁻ ÷ total volume, then pOH and pH |
Weak base plus strong acid
The reaction is B(aq) + H₃O⁺(aq) → HB⁺(aq) + H₂O(l). It mirrors the case above. With leftover B and some HB⁺, you have a buffer. With equal moles, only HB⁺ remains, which is a weak acid, so the solution is slightly acidic: use Ka = Kw/Kb. With excess strong acid, the leftover H₃O⁺ sets the pH.
Example: equal amounts of NH₃ and HCl leave NH₄⁺. A 0.10 M NH₄⁺ solution has a pH of about 5.13.
Weak acid plus weak base
Neither reacts completely. The proton transfer HA + B ⇌ A⁻ + HB⁺ reaches an equilibrium. It favors the side with the weaker acid and weaker base. If HA is a stronger acid than HB⁺, products are favored (K > 1).
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Excess strong acid
50.0 mL of 0.100 M HCl is mixed with 30.0 mL of 0.100 M NaOH. What is the pH?
Show the solutionHide the solution
- Step 1: Moles: HCl = 50.0 mL × 0.100 M = 5.00 mmol H₃O⁺; NaOH = 30.0 × 0.100 = 3.00 mmol OH⁻.
- Step 2: React: 3.00 mmol OH⁻ uses up 3.00 mmol H₃O⁺, leaving 2.00 mmol H₃O⁺.
- Step 3: Total volume = 80.0 mL, so [H₃O⁺] = 2.00 mmol ÷ 80.0 mL = 0.0250 M.
- Step 4: pH = −log(0.0250) = 1.60.
Answer: pH = 1.60
- Example 2Calculator allowed
Weak acid partly neutralized: a buffer forms
50.0 mL of 0.200 M acetic acid (Ka = 1.8 × 10⁻⁵) is mixed with 20.0 mL of 0.200 M NaOH. What is the pH?
Show the solutionHide the solution
- Step 1: Moles: CH₃COOH = 10.0 mmol; OH⁻ = 4.0 mmol.
- Step 2: React completely: CH₃COOH + OH⁻ → CH₃COO⁻ + H₂O. After: 6.0 mmol CH₃COOH and 4.0 mmol CH₃COO⁻.
- Step 3: Both members of the conjugate pair are present, so it's a buffer. Both are in the same 70.0 mL, so you can use the mole ratio directly.
- Step 4: pKa = −log(1.8 × 10⁻⁵) = 4.745 (keep the extra digit until the end).
- Step 5: pH = 4.745 + log(4.0 / 6.0) = 4.745 − 0.176 = 4.57.
Answer: pH ≈ 4.57
- Example 3Calculator allowed
Trap: equal moles isn't neutral
25.0 mL of 0.100 M acetic acid is mixed with 25.0 mL of 0.100 M NaOH. A student says the pH is 7.00 because the acid and base exactly neutralize each other. Find the actual pH (Ka of acetic acid = 1.8 × 10⁻⁵).
Show the solutionHide the solution
- Step 1: 2.50 mmol acid reacts with 2.50 mmol OH⁻, leaving 2.50 mmol CH₃COO⁻ in 50.0 mL: [CH₃COO⁻] = 0.0500 M.
- Step 2: Acetate is a weak base: CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻, with Kb = Kw/Ka = (1.0 × 10⁻¹⁴) ÷ (1.8 × 10⁻⁵) = 5.6 × 10⁻¹⁰.
- Step 3: x² / 0.0500 = 5.6 × 10⁻¹⁰, so x = [OH⁻] = 5.3 × 10⁻⁶ M.
- Step 4: pOH = 5.28, so pH = 14.00 − 5.28 = 8.72.
- Step 5: Only strong acid plus strong base gives pH 7 at equal moles. Here the leftover conjugate base makes the solution basic.
Answer: pH ≈ 8.72 (basic, not neutral)
Common mistakes
- Starting with an equilibrium (ICE) table before letting the strong acid or base react completely.
- Calling an equal-moles mixture of a weak acid and a strong base neutral.
- Forgetting to use the total volume after mixing when finding a leftover concentration.
- Using Ka instead of Kb (or the reverse) for the conjugate species left at the end.
On the exam
- Free-response questions often ask you to identify the major species after mixing before calculating anything. List them; it shows you know which case you're in.
- Use millimoles to keep numbers clean when volumes are in mL.
Connected topics
Videos
Check yourself
4 questions on 8.4 Acid-Base Reactions and Buffers. Pick an answer to see if you got it, and why.
25.0 mL of 0.10 M HCl is mixed with 15.0 mL of 0.10 M NaOH at 25 °C. What is the pH of the resulting solution?
Equal volumes of 0.10 M NH₃ and 0.10 M HCl are mixed at 25 °C. Which of the following best describes the resulting solution?
0.010 mol of CH₃COOH and 0.015 mol of NaOH are dissolved in water to make 250 mL of solution at 25 °C. Which of the following is closest to the pH?
Which of the following is the best net ionic equation for the reaction that occurs when NaOH(aq) is added to CH₃COOH(aq)?
0 of 4 answered