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General Chemistry · Acid-base equilibria

Acid-base titration practice problems, with worked answers

Every titration pH problem starts the same way: count moles of acid and base, let them react completely, and see what is left. What is left decides the method. Leftover strong acid or base sets the pH directly. A weak acid with its conjugate base is a buffer. At the equivalence point of a weak acid, only the conjugate base remains, and it makes the solution basic.

Already know the basics? Jump to the 8 practice questions or the worked examples.

One method for every point on the curve

All pH values on this page assume 25 °C, Kw = 1.0 × 10−14, and volumes that simply add when you mix. The titrant is a strong base (NaOH) added to an acid.

  1. Count millimoles. mmol = M × mL. Do this for the acid in the flask and the base added so far.
  2. React them completely. OH− reacts with the acid one to one (for a monoprotic acid) until one of them runs out. Treat this as going to completion; strong base reacts essentially completely with both strong and weak acids.
  3. Name what is left. This picks the method in the table below.
  4. Divide by the total volume (acid volume + base volume) whenever you need a concentration. Forgetting the added base volume is the most common error.
What is in the flask, and how to get the pH
PointStrong acid + strong baseWeak acid + strong base
Before any base[H3O+] = acid concentrationWeak acid alone: solve Ka = x²/(C − x), often with x small
Before equivalenceLeftover H3O+ ÷ total volumeBuffer: Henderson–Hasselbalch
Half-equivalenceNothing specialpH = pKa
EquivalencepH = 7.00 (neutral salt)Conjugate base alone: Kb = Kw/Ka, pH > 7
After equivalenceExcess OH− ÷ total volumeExcess OH− ÷ total volume (the weak base adds almost nothing)

pH = pKa + log([A−] ÷ [HA])

In the buffer region you can put millimoles straight into the Henderson–Hasselbalch ratio, because A− and HA share the same total volume. At the half-equivalence point the two amounts are equal, the log term is log 1 = 0, and pH = pKa.

Reading a titration curve and choosing an indicator

A titration curve plots pH against volume of titrant. Four features tell you most of the story:

  • Starting pH. A strong acid starts lower than a weak acid at the same concentration.
  • The flat buffer region. Only weak-acid curves have it. Its middle, the half-equivalence point, sits at pH = pKa, so you can read Ka off the graph.
  • The steep jump. The equivalence point is the middle of the near-vertical rise. Its volume tells you the acid's moles: mmol acid = Mbase × Vbase at equivalence.
  • Equivalence pH. 7.00 for strong acid + strong base at 25 °C. Above 7 for weak acid + strong base. Below 7 for weak base + strong acid.

An indicator works when its color change falls inside the steep jump. Approximate change ranges commonly listed in college texts:

Common indicators (approximate pH ranges)
IndicatorChanges color near pHGood fit
Methyl orangeabout 3 to 4.5Strong acid + strong base, or weak base + strong acid
Bromothymol blueabout 6 to 7.6Strong acid + strong base
Phenolphthaleinabout 8.2 to 10Weak acid + strong base; also works for strong acid + strong base

For a strong acid with a strong base, the jump runs from about pH 4 to 10 for 0.1 M solutions, so all three work. For a weak acid, the jump is shorter and sits above 7, so phenolphthalein is the usual choice. Methyl orange would change color long before the weak acid is used up.

Worked examples

1. Strong acid + strong base: four points

20.00 mL of 0.150 M HCl is titrated with 0.100 M NaOH. Find the pH after 0.00, 15.00, 30.00, and 35.00 mL of base.

  1. Acid: 20.00 mL × 0.150 M = 3.000 mmol HCl. Equivalence needs 3.000 mmol ÷ 0.100 M = 30.00 mL of NaOH.
  2. 0.00 mL: HCl is a strong acid, so [H3O+] = 0.150 M. pH = −log(0.150) = 0.824.
  3. 15.00 mL: base added = 1.500 mmol. Leftover H3O+ = 3.000 − 1.500 = 1.500 mmol in 35.00 mL. [H3O+] = 0.04286 M, pH = 1.368.
  4. 30.00 mL: 3.000 mmol base meets 3.000 mmol acid. Only NaCl and water remain, so pH = 7.00.
  5. 35.00 mL: base added = 3.500 mmol. Excess OH− = 0.500 mmol in 55.00 mL. [OH−] = 0.009091 M, pOH = 2.041, pH = 14.00 − 2.041 = 11.96.

Answer: pH 0.824, 1.368, 7.00, and 11.96

2. Weak acid + strong base: every region

50.00 mL of 0.100 M acetic acid (Ka = 1.8 × 10−5, pKa = 4.74) is titrated with 0.100 M NaOH. Find the pH at 0.00, 10.00, 25.00, 50.00, and 60.00 mL.

  1. Acid: 5.000 mmol. Equivalence at 50.00 mL of base, half-equivalence at 25.00 mL.
  2. 0.00 mL, weak acid alone: x²/0.100 = 1.8 × 10−5, so x = [H3O+] = 1.34 × 10−3 M and pH = 2.87. (x is about 1.3% of 0.100, so the small-x shortcut is fine.)
  3. 10.00 mL, buffer: 1.000 mmol OH− turns 1.000 mmol HA into A−. HA left = 4.000 mmol, A− = 1.000 mmol. pH = 4.745 + log(1.000/4.000) = 4.745 − 0.602 = 4.14.
  4. 25.00 mL, half-equivalence: HA = A− = 2.500 mmol, so pH = pKa = 4.74.
  5. 50.00 mL, equivalence: all 5.000 mmol is now acetate, in 100.00 mL: [A−] = 0.0500 M. Kb = 1.0 × 10−14 ÷ 1.8 × 10−5 = 5.6 × 10−10. [OH−] = √(5.56 × 10−10 × 0.0500) = 5.27 × 10−6 M. pOH = 5.28, so pH = 8.72.
  6. 60.00 mL, after equivalence: excess OH− = 6.000 − 5.000 = 1.000 mmol in 110.00 mL = 0.009091 M. pOH = 2.04, pH = 11.96.

Answer: pH 2.87, 4.14, 4.74, 8.72, and 11.96

3. Reading a curve: find Ka, the acid's concentration, and an indicator

A 25.00 mL sample of an unknown weak monoprotic acid is titrated with 0.100 M NaOH. The steep jump is centered at 18.00 mL. At 9.00 mL the pH is 3.75. Find the acid's concentration and Ka, and pick an indicator.

  1. Concentration: at equivalence, mmol acid = mmol base = 0.100 M × 18.00 mL = 1.800 mmol. 1.800 mmol ÷ 25.00 mL = 0.0720 M.
  2. 9.00 mL is half of 18.00 mL, so it is the half-equivalence point. There, pH = pKa = 3.75.
  3. Ka = 10−3.75 = 1.8 × 10−4.
  4. Indicator: at equivalence, 1.800 mmol of A− sits in 43.00 mL (0.04186 M). With Kb = 1.0 × 10−14 ÷ 1.78 × 10−4 = 5.6 × 10−11, the pH works out to about 8.2. That is basic, so choose phenolphthalein.

Answer: 0.0720 M acid, Ka = 1.8 × 10−4, phenolphthalein

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Strong acid + strong base pH calculator

For a strong monoprotic acid (such as HCl or HNO3) titrated with a strong base that gives one OH− per formula unit (such as NaOH or KOH). Assumes 25 °C, Kw = 1.0 × 10−14, and additive volumes. Not for weak acids.

Practice questions

Pick an answer, then check it. A wrong answer tells you what went wrong, so you can try again. Every question also has a full solution.

Question 130.00 mL of 0.200 M HNO3 is mixed with 20.00 mL of 0.200 M NaOH. What is the pH?

Show solution
  1. HNO3: 30.00 × 0.200 = 6.000 mmol. NaOH: 20.00 × 0.200 = 4.000 mmol.
  2. Leftover H3O+: 2.000 mmol. Total volume 50.00 mL, so [H3O+] = 0.0400 M.
  3. pH = −log(0.0400) = 1.40.
Question 225.00 mL of 0.100 M HCl is titrated with 0.100 M NaOH. What is the pH after 30.00 mL of NaOH has been added?

Show solution
  1. HCl: 2.500 mmol. NaOH: 3.000 mmol. Past equivalence by 0.500 mmol OH−.
  2. [OH−] = 0.500 mmol ÷ 55.00 mL = 0.009091 M. pOH = 2.041.
  3. pH = 14.00 − 2.041 = 11.96.
Question 3What is the pH of a 0.200 M acetic acid solution (Ka = 1.8 × 10−5) before any base is added?

Show solution

Ka = x² ÷ (0.200 − x). With x small, x² = 3.6 × 10−6 and x = 1.90 × 10−3 M. That is under 1% of 0.200 M, so the shortcut holds. pH = −log(1.90 × 10−3) = 2.72.

Question 440.00 mL of 0.100 M acetic acid (pKa = 4.74) is titrated with 0.100 M NaOH. What is the pH after 30.00 mL of base?

Show solution
  1. Acid: 4.000 mmol. Base: 3.000 mmol, which converts 3.000 mmol HA to A−.
  2. HA left: 1.000 mmol. A−: 3.000 mmol.
  3. pH = 4.74 + log(3.000 ÷ 1.000) = 4.74 + 0.48 = 5.22.
Question 5Benzoic acid has Ka = 6.3 × 10−5. What is the pH at the half-equivalence point when it is titrated with NaOH?

Show solution

At half-equivalence, half the HA has become A−, so the ratio in Henderson–Hasselbalch is 1 and log 1 = 0. pH = pKa = −log(6.3 × 10−5) = 4.20. You don't need any concentrations or volumes.

Question 620.00 mL of 0.200 M acetic acid (Ka = 1.8 × 10−5) is titrated with 0.200 M NaOH. What is the pH at the equivalence point?

Show solution
  1. 4.000 mmol acid needs 20.00 mL of base. At equivalence: 4.000 mmol CH3COO− in 40.00 mL, so 0.100 M.
  2. Kb = 1.0 × 10−14 ÷ 1.8 × 10−5 = 5.56 × 10−10.
  3. [OH−] = √(5.56 × 10−10 × 0.100) = 7.45 × 10−6 M. pOH = 5.13.
  4. pH = 14.00 − 5.13 = 8.87.
Question 7Which indicator is the best choice for titrating acetic acid with NaOH?

Show solution

A weak acid titrated with a strong base has a basic equivalence point, because the conjugate base hydrolyzes. Pick an indicator whose color change sits in the steep part of the curve above pH 7. Phenolphthalein, which turns pink near pH 8.2, fits.

Question 8Two curves are drawn for titrating 0.1 M acids with NaOH. Curve X starts at pH 1.0 and has its jump centered at pH 7. Curve Y starts at pH 2.9, has a flat region near pH 4.7, and jumps around pH 8.7. Which statement is true?

Show solution

Curve X: pH 1.0 at 0.1 M means [H3O+] = 0.1 M, so the acid is fully ionized (strong), and its equivalence point is neutral. Curve Y: higher start, a buffer plateau, and a basic equivalence point are all signs of a weak acid. The plateau's midpoint is the half-equivalence point, so pKa ≈ 4.7 (acetic acid fits). If the acid amounts are the same, the equivalence volumes match.

Answer key

  1. A, pH 1.40: 2.000 mmol H3O+ in 50.00 mL.
  2. C, pH 11.96: 0.500 mmol excess OH− in 55.00 mL.
  3. D, pH 2.72: x = √(KaC).
  4. B, pH 5.22: buffer with A−/HA = 3.
  5. A, pH 4.20: pH = pKa at half-equivalence.
  6. C, pH 8.87: acetate at 0.100 M hydrolyzes.
  7. C, phenolphthalein: equivalence pH is above 7.
  8. B: Y is a weak acid; pKa ≈ 4.7 from the half-equivalence point.

Frequently asked questions

Why is the equivalence point of a weak acid titration above pH 7?

At equivalence the weak acid has all become its conjugate base, such as acetate. That base takes protons from water and makes some OH-, so the solution is basic. Find the pH with Kb = Kw/Ka and the base's concentration in the total volume.

Is the equivalence point the same as the endpoint?

No. The equivalence point is where moles of base equal moles of acid. The endpoint is where the indicator changes color. A well-chosen indicator makes the two nearly the same.

When can I use Henderson-Hasselbalch during a titration?

Only in the buffer region, when both the weak acid and its conjugate base are present in reasonable amounts. Do not use it at the start (no conjugate base yet) or at or after equivalence (no weak acid left).

Are titrations on the MCAT?

Yes. The AAMC content outline lists acid-base titrations, including titration curves, indicators, and buffers, among the general chemistry topics. MCAT questions usually test curve reading and half-equivalence reasoning more than long calculations.

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