AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/8/8-5)
Unit 8 · Topic 8.5
8.5 Acid-Base Titrations
A titration curve tracks how the pH changes as titrant is added. You can read it to find the equivalence point (where moles of titrant equal the moles of acid or base you started with), the analyte's concentration, the pKa of a weak acid from the half-equivalence point, and the number of acidic protons in a polyprotic acid.
Key terms
- titrant
- analyte
- titration curve
- equivalence point
- half-equivalence point
- polyprotic acid
The equivalence point
In an acid–base titration, a solution of known concentration (the titrant, usually in a buret) is added to the solution being studied (the analyte). The equivalence point is where just enough titrant has been added to react with all of the analyte.
For a monoprotic acid titrated with NaOH, moles of OH⁻ added at equivalence = moles of acid originally present. This works the same whether the acid is strong or weak. On the curve, the equivalence point is the middle of the steepest, nearly vertical section.
Shapes of the curves
Strong acid titrated with strong base: the pH starts low (around 1 for 0.1 M acid), rises slowly, jumps steeply near equivalence, and the equivalence point is at pH 7.00 at 25 °C. Past equivalence, excess OH⁻ makes it level off at a high pH.
Weak acid titrated with strong base: the pH starts higher, because a weak acid makes less H₃O⁺. Soon after the start, the curve flattens into a buffer region. The steep jump is shorter, and the equivalence point is above pH 7, because the conjugate base A⁻ is the main species there and it makes the solution basic.
Weak base titrated with strong acid: the pH starts high and falls. The equivalence point is below pH 7, because the conjugate acid HB⁺ is the main species there.
The half-equivalence point gives pKa
Halfway to the equivalence point of a weak acid titration, exactly half of the HA has been converted to A⁻. So [HA] = [A⁻], and by Henderson–Hasselbalch, pH = pKa + log(1) = pKa.
To use it: find the equivalence volume, halve it, and read the pH at that volume. That pH is the pKa. For a weak base titrated with strong acid, the pH at half-equivalence equals the pKa of the conjugate acid, HB⁺.
Polyprotic acids
A polyprotic acid has more than one acidic proton. A diprotic acid, H₂A, gives up its protons one at a time, so its curve can show two equivalence points (two steep jumps) when the two pKa values are far enough apart. Counting the jumps tells you how many acidic protons there are. The second equivalence volume is twice the first.
Halfway to the first equivalence point, pH = pKa₁ ([H₂A] = [HA⁻]). Halfway between the first and second equivalence points, pH = pKa₂ ([HA⁻] = [A²⁻]).
You should be able to name the major species at any point, but you won't be asked to calculate exact concentrations of every species along a polyprotic titration curve.
| Point on an H₂A curve | Major species |
|---|---|
| Start | H₂A |
| First half-equivalence (pH = pKa₁) | H₂A and HA⁻ in equal amounts |
| First equivalence | HA⁻ |
| Second half-equivalence (pH = pKa₂) | HA⁻ and A²⁻ in equal amounts |
| Second equivalence | A²⁻ |
| Past second equivalence | A²⁻ and excess OH⁻ |
Worked examples
Try each one yourself first, then open the solution.
- Example 1Calculator allowed
Reading a weak acid titration
25.00 mL of a weak monoprotic acid HA is titrated with 0.1000 M NaOH. The equivalence point is at 18.40 mL, and the pH at 9.20 mL of NaOH added is 4.20. Find (a) the concentration of HA, (b) its Ka, and (c) whether the pH at equivalence is above, below or equal to 7.
Show the solutionHide the solution
- Step 1: (a) Moles of NaOH at equivalence = 0.01840 L × 0.1000 M = 1.840 × 10⁻³ mol, which equals the moles of HA. [HA] = 1.840 × 10⁻³ mol ÷ 0.02500 L = 0.07360 M.
- Step 2: (b) 9.20 mL is exactly half of 18.40 mL, so it's the half-equivalence point, where pH = pKa. pKa = 4.20, so Ka = 10^(−4.20) = 6.3 × 10⁻⁵.
- Step 3: (c) At equivalence, all HA has become A⁻, a weak base. A⁻ reacts with water to make OH⁻, so the pH is above 7. (A calculation gives about 8.4.)
Answer: (a) 0.07360 M; (b) Ka ≈ 6.3 × 10⁻⁵ (pKa 4.20); (c) above 7
- Example 2
Trap: species on a diprotic curve
A diprotic acid H₂A is titrated with NaOH. The curve has steep jumps at 20.0 mL and 40.0 mL. The pH at 10.0 mL is 3.1 and at 30.0 mL is 7.0. Identify pKa₁, pKa₂, and the major A-containing species at 20.0 mL.
Show the solutionHide the solution
- Step 1: Two jumps mean two acidic protons. 10.0 mL is halfway to the first equivalence point, so pKa₁ = 3.1.
- Step 2: 30.0 mL is halfway between the first (20.0 mL) and second (40.0 mL) equivalence points, so pKa₂ = 7.0. A common mistake is to use the half of the second equivalence volume (20.0 mL) instead; that's actually the first equivalence point.
- Step 3: At 20.0 mL, the first proton has been removed from every H₂A, so the major species is HA⁻.
Answer: pKa₁ = 3.1; pKa₂ = 7.0; HA⁻ is the major species at 20.0 mL
Common mistakes
- Assuming the equivalence point is always at pH 7. That's only true for strong acid with strong base.
- Confusing the equivalence point (moles equal) with the endpoint (when the indicator changes color). A good indicator makes them very close.
- Reading pKa at the equivalence point instead of the half-equivalence point.
- For a diprotic acid, using half of the total volume to find pKa₂.
On the exam
- Titration curves are a favorite. Practice identifying the equivalence and half-equivalence points, reading pKa, and naming the major species in each region.
- You may be asked to calculate the analyte's concentration from the equivalence volume. Show moles of titrant = moles of analyte for a monoprotic acid.
Connected topics
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Check yourself
5 questions on 8.5 Acid-Base Titrations. Pick an answer to see if you got it, and why.
A solution of a diprotic acid, H₂A, is titrated with NaOH. The titration curve shows a first equivalence point when 15.0 mL of NaOH has been added. At what total volume of NaOH should the second equivalence point appear?
A 25.00 mL sample of HCl is titrated with 0.1000 M NaOH. The equivalence point is reached after 18.40 mL of NaOH has been added. What is the concentration of the HCl?
Equal volumes of 0.10 M HCl and 0.10 M CH₃COOH are each titrated with 0.10 M NaOH. Which of the following is the same for both titrations?
A 25.0 mL sample of 0.100 M NH₃ is titrated with 0.100 M HCl at 25 °C. Kb for NH₃ is 1.8 × 10⁻⁵.
The pH starts at 11.13, and the equivalence point is reached at 25.0 mL of HCl, where the pH is 5.28.
Hypothetical titration
What is the pH after 12.5 mL of HCl has been added?
Which of the following best explains why the pH at the equivalence point is below 7?
0 of 5 answered