One drop makes the difference
In lesson 1.8.5 you added dilute hydrochloric acid drop by drop to sodium hydroxide containing phenolphthalein, and the pink disappeared right when the two had reacted exactly. Lesson 2.9.4 used burette readings to calculate a concentration. But why does the color change at exactly that drop? And does the same method work for acetic acid? This lesson follows the pH through a titration.
Neutralization
Strong acids and strong bases ionize completely, so their neutralization is really
Lesson 2.10 found that this releases about 55.8 kJ per mole of water formed. A weak acid in solution is mostly molecules, so by the rules of lesson 2.7.4 its formula stays together in the net ionic equation:
A weak acid ionizes only a little, but as OH⁻ uses up the H⁺, its equilibrium keeps shifting right until all of the acid is neutralized. Equal moles of hydrochloric and acetic acid therefore need the same amount of NaOH.
Equivalence point and end point
Lesson 2.9.4 covered the basics of an acid–base titration: the standard solution, the end point, the equivalence point, and finding a concentration from the data (for a monoprotic acid and a base with one OH⁻, MaVa = MbVb). This lesson follows how the pH changes along the way, so keep the two “points” apart: the equivalence point化学计量点 is where the acid and base have reacted exactly, and the end point is where the indicator changes color and you stop. A well-chosen indicator makes the two nearly coincide.
Titrating a strong acid
Titrate 25.00 mL of 0.100 M HCl with 0.100 M NaOH and plot pH against the volume of NaOH added: the result is a titration curve滴定曲线. The table gives points on the curve:
| NaOH added (mL) | pH, HCl | pH, acetic acid |
|---|---|---|
| 0.00 | 1.00 | 2.88 |
| 20.00 | 1.95 | 5.35 |
| 24.975 (99.9% neutralized) | 4.30 | 7.74 |
| 25.00 | 7.00 | 8.72 |
| 25.025 (100.1% neutralized) | 9.70 | 9.70 |
| 30.00 | 11.96 | 11.96 |
| 40.00 | 12.36 | 12.36 |
At first the acid is in large excess, and after 20.00 mL the pH has risen by less than 1. But near the equivalence point, going from 99.9% to 100.1% neutralized takes only 0.05 mL of NaOH, about one drop, and the pH leaps from 4.30 to 9.70. This is the steep risepH 突跃 of the curve. For a strong acid and a strong base, the pH at the equivalence point is 7.00.
Titrating a weak acid
Now titrate 25.00 mL of 0.100 M acetic acid (Ka = 1.8 × 10⁻⁵) with the same NaOH. Recall Ka from lesson 2.12.2:
Compared with HCl, the curve differs in three ways:
- It starts higher. Acetic acid ionizes only partly, so the starting pH is about 2.88.
- The middle is flat. The solution contains both CH₃COOH and CH₃COO⁻, so the pH changes slowly. After half the NaOH (12.50 mL), exactly half the acid has been neutralized and [CH₃COOH] = [CH₃COO⁻]. Putting that into Ka gives [H⁺] = Ka, so pH = pKa = −log Ka = 4.74. This is the half-equivalence point.
- The equivalence point is basic. At the equivalence point the flask holds sodium acetate solution, and from lesson 2.12.2, CH₃COO⁻ takes H⁺ from water (salt hydrolysis). So the pH is 8.72, above 7, and the steep rise is much shorter, from 7.74 to 9.70.
(An exact calculation gives pH 4.75 at the half-equivalence point, slightly above pKa, because a tiny bit of the acetic acid ionizes on its own. Treating it as pH = pKa is standard.)
Acid–base indicators
An acid–base indicator is itself a weak organic acid or base whose molecule and ion have different colors. As the pH changes, the ratio of the two forms changes, and so does the color. The color shifts gradually over a pH interval called the transition range变色范围, usually 1 to 2 pH units wide.
| Indicator | Transition range (pH) | Color change |
|---|---|---|
| methyl orange | 3.1–4.4 | red → yellow |
| methyl red | 4.4–6.2 | red → yellow |
| bromothymol blue | 6.0–7.6 | yellow → blue |
| phenolphthalein | 8.3–10.0 | colorless → pink |
(Chinese textbooks list phenolphthalein as 8.2–10.0; sources differ slightly.)
The rule for choosing an indicator: its transition range must overlap the steep rise. Then it changes color within about half a drop of the equivalence point.
- Strong acid with strong base: the steep rise is wide (4.30 to 9.70), so methyl orange, bromothymol blue, and phenolphthalein all work.
- Weak acid titrated with strong base: the equivalence point is basic and the steep rise runs from 7.74 to 9.70, so use phenolphthalein.
- Litmus changes color gradually over a wide range, so it isn’t used for titrations.
Below, titrate hydrochloric or acetic acid with NaOH. Drag the slider or press ±0.10 mL, watch the curve and the color in the flask, and switch indicators to see which ones work:
Lab technique
The key steps in a titration:
- Check. Make sure the burette doesn’t leak.
- Rinse. After washing, rinse the burette two or three times with the solution it will hold. Rinse the flask with distilled water only, not with the unknown.
- Fill and read. Fill the burette, clear the air bubble from the tip, set the level at or below 0.00 mL, and record the initial reading.
- Measure the unknown. Deliver an exact volume of the unknown into the flask with a pipet or burette and add two or three drops of indicator.
- Titrate. Control the stopcock with one hand and swirl the flask with the other, watching the color.
- End point and calculation. Stop when the last half drop changes the color for at least 30 seconds. Record the final reading, repeat two or three times, and average.
How Chinese textbooks differ
- Chinese textbooks call the equivalence point 化学计量点 and the end point 滴定终点, and use c(待测) = c(标准)·V(标准) ÷ V(待测).
- The Chinese textbook example titrates 20.00 mL of 0.1000 mol/L HCl with 0.1000 mol/L NaOH, and the pH jumps from 4.30 to 9.70 between 19.98 and 20.02 mL. The 25.00 mL here gives the same range: the steep rise depends only on going from 99.9% to 100.1% neutralized, not on the total volume.
- US textbooks and AP put a lot of weight on pH = pKa at the half-equivalence point, which gives Ka from a titration curve; Chinese high school textbooks usually don’t require it.
Example 1: concentration from titration data
25.00 mL of HCl of unknown concentration, with 2 drops of phenolphthalein, is titrated with 0.1000 M NaOH. The average of three trials is 22.40 mL of NaOH. What is the concentration of the HCl?
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The plan. At the equivalence point, use the mole ratio to get moles of HCl from the NaOH of known concentration, then divide by the volume of acid.
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Moles of NaOH. Volume in liters:
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Moles of HCl. HCl + NaOH → NaCl + H₂O is 1 : 1, so moles of HCl = 2.240 × 10⁻³ mol.
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Concentration.
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Sig figs. Every measurement has 4 significant figures, so the answer does too.
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Check. Less NaOH solution was needed than the volume of acid, and they react 1 : 1, so the acid is a bit more dilute than 0.1000 M. In one step: 0.1000 × 22.40 ÷ 25.00 = 0.08960 M.
Example 2: why not methyl orange for acetic acid?
25.00 mL of 0.100 M acetic acid is titrated with 0.100 M NaOH, once with methyl orange and once with phenolphthalein. Which indicator is suitable? What goes wrong with the other?
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The solution at the equivalence point. After 25.00 mL of NaOH, the flask holds 50.00 mL of sodium acetate solution. The volume has doubled, so [CH₃COO⁻] = 0.0500 M. CH₃COO⁻ takes H⁺ from water, so the solution is basic.
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Where pH 8.72 comes from. The base constant of acetate is Kb = Kw ÷ Ka = 1.0 × 10⁻¹⁴ ÷ 1.8 × 10⁻⁵ = 5.6 × 10⁻¹⁰. Using the same shortcut as for [H⁺] in acetic acid (2.12.2):
pOH = 5.28, so pH = 14.00 − 5.28 = 8.72.
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Where 7.74 to 9.70 comes from. Half a drop either side: at 24.975 mL (99.9% neutralized), leftover acid to acetate is about 1 : 999, so Ka gives [H⁺] = 1.8 × 10⁻⁵ ÷ 999 = 1.8 × 10⁻⁸ M and pH = 7.74. At 25.025 mL (0.1% extra), the excess OH⁻ is 0.025 mL × 0.100 M ÷ 50 mL = 5.0 × 10⁻⁵ M, so pH = 9.70. AP Unit 8 (8.5) covers these calculations fully; for now the results are enough.
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Phenolphthalein. Its range, 8.3 to 10.0, overlaps the steep rise, so it turns pink within half a drop of the equivalence point. Suitable.
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Methyl orange. Its range, 3.1 to 4.4, is far below the steep rise. The curve reaches pH 4.4 after only about 7.8 mL of NaOH, when methyl orange has already turned fully yellow and less than a third of the acid (7.8 ÷ 25.00 ≈ 31%) has been neutralized.
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Conclusion. With methyl orange you stop far too early, and the calculated concentration of acetic acid comes out much too low. A weak acid needs an indicator that changes in the basic range, like phenolphthalein.
- 20.00 mL HCl needs 16.00 mL NaOH
0.1000 × 16.00 ÷ 20.00 = 0.08000.
- 25.00 mL HCl needs 25.00 mL NaOH
Equal volumes, equal concentrations: 0.1000.
- 10.00 mL acetic acid needs 12.00 mL NaOH
Acetic acid is weak, but it needs as much base as the same moles of a strong acid: 0.1000 × 12.00 ÷ 10.00 = 0.1200.
- 20.00 mL NaOH needs 10.00 mL HCl
0.1000 × 10.00 ÷ 20.00 = 0.05000.
- 25.00 mL H₂SO₄ needs 40.00 mL NaOH
Moles of NaOH = 4.000 × 10⁻³. H₂SO₄ is diprotic, so moles of H₂SO₄ = 2.000 × 10⁻³, and 2.000 × 10⁻³ ÷ 0.02500 = 0.08000.
- 10.00 mL Ba(OH)₂ needs 32.00 mL HCl
Moles of HCl = 3.200 × 10⁻³. Ba(OH)₂ has two OH⁻, so moles of Ba(OH)₂ = 1.600 × 10⁻³, and 1.600 × 10⁻³ ÷ 0.01000 = 0.1600.
- 20.00 mL HCl needs 32.00 mL NaOH
0.1000 × 32.00 ÷ 20.00 = 0.1600.
- 20.00 mL acetic acid needs 10.00 mL NaOH
0.1000 × 10.00 ÷ 20.00 = 0.05000.
Common mistakes
Mistake 1: “At the equivalence point the pH is always 7.” Only for a strong acid with a strong base. When a weak acid is neutralized by a strong base, the salt formed (such as sodium acetate) is basic, so the equivalence point is above pH 7.
Mistake 2: “The end point and the equivalence point are the same thing.” The equivalence point is where acid and base have reacted exactly; the end point is where the indicator changes. They coincide only if the indicator is well chosen; with the wrong one, the end point can be far off.
Mistake 3: “Acetic acid is weak, so it takes less base to neutralize than the same amount of HCl.” As base is added, acetic acid’s equilibrium keeps shifting right until all of it is neutralized. Equal moles of HCl and acetic acid need the same NaOH, and both curves have their equivalence point at 25.00 mL.
Quick check
Which indicator is best for titrating acetic acid with NaOH?
At the equivalence point the flask holds basic sodium acetate solution, and the steep rise runs from 7.74 to 9.70. Only phenolphthalein, which changes in the basic range, fits.
25.00 mL of 0.100 M acetic acid is titrated with 0.100 M NaOH. What is the pH after 12.50 mL of NaOH?
12.50 mL is the half-equivalence point: [CH₃COOH] = [CH₃COO⁻], so [H⁺] = Ka and pH = pKa = 4.74. 8.72 is the equivalence-point pH and 2.88 the starting pH.
Summary
- Strong acid–strong base neutralization is H⁺ + OH⁻ → H₂O. A titration finds an unknown concentration from a standard solution: M_a V_a = M_b V_b for a monoprotic acid and a base with one OH⁻.
- The pH rises steeply near the equivalence point: 4.30 to 9.70 for a strong acid with a strong base, with pH 7.00 at the equivalence point. A weak acid (acetic acid) starts higher, has pH = pKa at half-equivalence, reaches pH 8.72 at equivalence, and has a shorter steep rise.
- The indicator’s transition range must overlap the steep rise: methyl orange or phenolphthalein for strong acid with strong base, phenolphthalein for a weak acid with a strong base.
On to AP (for later)
AP topic 4.9 uses titration data for stoichiometry. Topic 8.5 asks you to read and sketch titration curves for strong, weak, and polyprotic acids, mark the equivalence and half-equivalence points, find Ka from pH = pKa at half-equivalence, and justify a choice of indicator. Free-response questions also ask which species are present at the equivalence point and why the pH there is above or below 7.
Data sources: points on the titration curves are calculated exactly from charge balance, and the values at 20.00, 25.00, 30.00, and 40.00 mL match OpenStax Chemistry 2e, Table 14.2 (OpenStax gives 2.87 for the starting pH of acetic acid, from the x ≪ c approximation); indicator ranges from the Wikipedia article “pH indicator.”