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Chemistry 2e · Acid-Base Equilibria

Acid-Base Titrations

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On this page 8 sections
  1. In 30 seconds
  2. Why this matters
  3. The college version
  4. Eli explains
  5. Key takeaway
  6. Check yourself
  7. Study tools
  8. Sources & references

In 30 seconds

A is a controlled neutralization: you slowly add a solution of known concentration (the ) from a buret to a measured volume of solution with unknown concentration (the ), until the reaction is exactly complete. At that point — the — the moles of acid and base added are stoichiometrically equivalent, and the volume of titrant used lets you calculate the unknown concentration.

The workhorse relationship is stoichiometry, not pH:

mol titrant = concentration × volume

n = M × V

For a monoprotic acid–base reaction (HA + OH− → A− + H2O), one mole of OH− neutralizes one mole of HA, so at equivalence:

Ma Va = Mb Vb

The equivalence point is not always pH 7: strong acid + strong base → pH 7; weak acid + strong base → pH > 7 (the conjugate base hydrolyzes); weak base + strong acid → pH < 7 (the conjugate acid hydrolyzes). Choosing an whose color-change range brackets the equivalence-point pH is the practical skill — the (observed color change) should match it as closely as possible.

Why this matters

  • Analytical chemistry everywhere: Titration determines the concentration of acids in wine and vinegar, bases in cleaning products, hardness of water, and active ingredients in pharmaceuticals — a standard quality-control technique.
  • Clinical and biological chemistry: Quantifying acids in blood/urine, measuring enzyme activity, and calibrating reagents all rely on titration logic. The same Ma Va = Mb Vb reasoning underlies how "normal" saline and buffer compositions are verified.
  • The titration curve is a map of acid-base behavior: The shape of the pH-vs-volume curve reveals whether the acid is strong or weak, its Ka, and the stoichiometry of polyprotic species — one experiment, many answers.
  • Exams: Titration calculations (find unknown molarity, find Ka from half-equivalence, choose an indicator) are guaranteed exam material in general chemistry and health-science programs.

The college version

Core Concepts

The setup and the equivalence point

A buret delivers titrant in measured increments into a flask (or beaker) containing the analyte plus a few drops of indicator. The equivalence point is the theoretical point where moles of H3O+ and OH− added are stoichiometrically equal. The endpoint is where the indicator actually changes color. A good indicator changes color right at the equivalence point, making endpoint ≈ equivalence point.

Titration curves: three classic shapes

  • Strong acid–strong base: pH starts low (e.g., 1 for 0.1 M HCl), rises slowly, then jumps almost vertically through pH 7 at equivalence; the curve is nearly symmetric. Equivalence pH = 7. Any indicator changing color between ~4 and ~10 (phenolphthalein, bromothymol blue, methyl red) works.
  • Weak acid–strong base: The curve starts higher (pH ≈ 2.9 for 0.1 M acetic acid) and shows a — a flat stretch around pKa where pH changes slowly — before the steep rise. Equivalence pH > 7 (the acetate ion is a weak base). At the half-equivalence point (half the base needed), pH = pKa, which is how a titration can measure Ka. Use phenolphthalein (color change ~8.2–10); methyl red would change too early.
  • Weak base–strong acid: Mirror image. Starts basic (0.1 M NH3 has pH ≈ 11.1), buffer region around pKa of NH4+ (9.25), steep drop, equivalence pH < 7 (ammonium is a weak acid). Use methyl red (change ~4.4–6.2) or bromothymol blue; phenolphthalein would change too late.

Indicators: how they work

An indicator is itself a weak acid whose acid and base forms have different colors: HIn (color A) ⇌ H+ + In− (color B). The eye sees the acid color when pH is below roughly pKa,In - 1 and the base color above pKa,In + 1, with a transition range of about 2 pH units around its pKa. Rule: pick an indicator whose transition range contains the equivalence-point pH.

Polyprotic acids and specialized titrations

A diprotic acid like H2SO3 or H2CO3 shows two equivalence points and two buffer regions, one per proton. This is how chemists determine both the concentration and the identity of a polyprotic species. (Indicators and curves for these are more complex; general chemistry usually treats them qualitatively.)

Practical notes

General safety principles apply whenever handling acids, bases, and glassware: wear eye protection, work in a ventilated area, and add reagents slowly to avoid splashing. Never taste or touch samples; treat all solutions as potentially hazardous.

How It Works / Step-by-Step Process

Worked example 1: strong acid–strong base — find the unknown concentration

Problem. 25.00 mL of HCl of unknown concentration requires 30.00 mL of 0.100 M NaOH to reach the equivalence point. Find the molarity of the acid.

Solution.

  1. Convert volumes to liters: Vb = 30.00 mL = 0.03000 L.
  2. Compute moles of titrant (formula first, then substitution): nb = Mb Vb = (0.100 mol/L)(0.03000 L) = 3.00 × 10-3 mol NaOH Dimensional check: (mol/L)(L) = mol ✓
  3. Stoichiometry: HCl + NaOH → NaCl + H2O is 1:1, so na = nb = 3.00 × 10-3 mol HCl.
  4. Solve for the unknown: Ma = naVa = 3.00 × 10-3 mol0.02500 L = 0.120 M Dimensional check: mol/L = M ✓
  5. Equivalence pH is 7 (strong + strong); any mid-range indicator works.

Worked example 2: weak acid–strong base — pH at equivalence and indicator choice

Problem. 25.00 mL of 0.100 M CH3COOH (Ka = 1.8 × 10-5) is titrated with 0.100 M NaOH. (a) What volume of base reaches equivalence? (b) Estimate the pH at equivalence and choose an indicator.

Solution (a):

  1. Moles of acid: na = (0.100 mol/L)(0.02500 L) = 2.50 × 10-3 mol.
  2. 1:1 reaction → need 2.50 × 10-3 mol NaOH. Solve the volume formula for Vb: Vb = nbMb = 2.50 × 10-3 mol0.100 mol/L = 0.0250 L = 25.00 mL

Solution (b):

  1. At equivalence all acid is converted to acetate. Total volume = 25.00 + 25.00 = 50.00 mL = 0.0500 L, so: [CH3COO−] = 2.50 × 10-3 mol0.0500 L = 0.0500 M
  2. Acetate is a weak base: Kb = Kw/Ka = (1.0 × 10-14)/(1.8 × 10-5) = 5.6 × 10-10.
  3. For a weak base, [OH−] ≈ Kb [A−]: [OH−] = (5.6 × 10-10)(0.0500) = 2.8 × 10-11 = 5.3 × 10-6 M
  4. pOH = -log(5.3 × 10-6) = 5.28, so pH = 14.00 - 5.28 = 8.72.
  5. Equivalence pH ≈ 8.7 (> 7). Choose phenolphthalein (transition 8.2–10), which brackets 8.7. Methyl red would change color far too early.

Worked example 3: weak base–strong acid — indicator choice and half-equivalence

Problem. 0.100 M NH3 is titrated with 0.100 M HCl. (a) At half-equivalence, what is the pH? (b) What is the equivalence pH (qualitatively) and which indicator fits?

Solution (a):

  1. At half-equivalence, half the NH3 has become NH4+: [NH3] = [NH4+].
  2. Use Henderson–Hasselbalch with pKa of NH4+ = 9.25: pH = pKa + log[NH3][NH4+] = 9.25 + log(1) = 9.25
  3. Half-equivalence pH = pKa = 9.25 — the buffer region of this titration.

Solution (b):

  1. At equivalence the solution is pure NH4+, a weak acid → pH < 7 (about 5.3 for these concentrations, mirroring example 2's arithmetic with Ka = 5.6 × 10-10).
  2. Indicator: methyl red (transition ~4.4–6.2) brackets pH ~5.3. Phenolphthalein would signal the endpoint in the buffer region, long before equivalence.

Common Confusions

Do Not ConfuseWithDifference
Equivalence pointEndpointEquivalence = stoichiometric completion (calculated); endpoint = color change (observed). A poor indicator makes them differ
Equivalence pH = 7Always trueOnly for strong acid + strong base; weak acid + strong base is > 7, weak base + strong acid is < 7
pH at half-equivalencepH at equivalenceHalf-equivalence → buffer with pH = pKa; equivalence → hydrolysis product determines pH
Ma Va = Mb VbAny acid-base ratioOnly valid for 1:1 reactions; diprotic acids need mole-ratio factors (2:1)
Using any indicatorMatching indicator to curveAn indicator whose range misses the equivalence pH gives a wrong endpoint — and a wrong answer
Concentration of titrantAmount of titrantYou need both: n = M × V. Many errors come from plugging mL instead of L into molarity formulas
Eli, the EliExplains learning guide

Eli explains

The same idea, in plain words

Explain it like I’m 10

A titration is like figuring out how much salt is in a glass of water by adding one teaspoon of pure saltwater at a time until the taste is exactly balanced — the number of teaspoons tells you how much was there. In chemistry, the "taste" is a color change from an indicator. The instant the color flips, you stop and read the buret: the volume you added is the answer to the math problem.

Key takeaways

  • Equivalence point: moles of acid and base are stoichiometrically equal; calculated from Ma Va = Mb Vb (1:1 reactions).
  • Endpoint ≠ equivalence point: endpoint is the observed color change; aim for an indicator whose range brackets the equivalence pH.
  • Equivalence pH: strong+strong = 7; weak acid+strong base > 7; weak base+strong acid < 7.
  • Half-equivalence point of a weak acid: pH = pKa — the easiest way to measure Ka from a titration curve.
  • Buffer region on a weak-acid curve: flat pH stretch around pKa, from the conjugate pair formed during titration.
  • Indicator rule of thumb: phenolphthalein (8.2–10) for weak acid + strong base; methyl red (4.4–6.2) for weak base + strong acid; any mid-range indicator for strong + strong.
  • Always use stoichiometry with units (dimensional analysis) — volume in L, concentration in mol/L, product in mol.
  • General lab safety: eye protection, slow addition, no tasting — applies to all titrations.

Check yourself

5 review questions from the chapter. Try each one, then open the answer.

  1. Define equivalence point and endpoint. Why must they nearly coincide?

    Show answer

    Equivalence point is where moles of titrant are stoichiometrically equal to moles of analyte; endpoint is where the indicator changes color. They must nearly coincide so the volume read at the endpoint reflects the true equivalence volume.

  2. A 20.00 mL sample of HCl needs 25.00 mL of 0.150 M NaOH. Find MHCl.

    Show answer

    nb = (0.150 mol/L)(0.02500 L) = 3.75 × 10-3 mol; 1:1 so Ma = 3.75 × 10-3 mol/0.02000 L = 0.1875 M.

  3. Is the equivalence-point pH of a weak acid titrated with strong base above, below, or equal to 7? Why?

    Show answer

    Above 7, because the conjugate base (e.g., acetate) hydrolyzes to produce OH−.

  4. What is the pH at the half-equivalence point of a weak acid titration? What does it allow you to measure?

    Show answer

    pH = pKa of the weak acid; it lets you determine the acid's Ka directly from the titration curve.

  5. Which indicator would you choose for a weak base–strong acid titration, and why?

    Show answer

    Methyl red (transition ~4.4–6.2), because the equivalence pH is below 7 (the conjugate acid, e.g., NH4+, is acidic); phenolphthalein would change too early in the buffer region.

Keep learning

Ready to build on this? Continue to the next lesson.

Study tools & related lessonsKey vocabulary · Related

Key vocabulary

titration
Controlled addition of known-concentration solution to find an unknown concentration
titrant
Solution of known concentration delivered from the buret
analyte
The solution of unknown concentration being analyzed
equivalence point
Point where added moles of titrant are stoichiometrically equal to moles of analyte
endpoint
Point where the indicator changes color
indicator
Weak acid whose two forms have different colors
buffer region
Flat pH region near pKa on a weak-acid/base titration curve

Sources & references

  1. openstax.org — Chemistry 2e

This lesson was adapted from the open educational references above; their licenses and attributions are preserved. See Copyright & Licensing.

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