MCAT content library / Chemistry & physics / Titration curves and equivalence
Separate stoichiometric equivalence from indicator endpoint and use the curve region to decide whether stoichiometry, buffering, or conjugate-base chemistry controls pH.
About 17 minutes, including weak-acid and concentration calculations.
Useful foundations
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Equivalence is a stoichiometric condition
The equivalence point occurs when titrant and analyte have reacted in the stoichiometric ratio specified by the balanced reaction. For a monoprotic acid HA titrated with OH⁻, equivalence occurs when moles OH⁻ added equal the initial moles HA.
That definition is independent of indicator color. It is also independent of whether the equivalence-point pH happens to be 7.
Concept: endpoint is an observed signal
An endpoint is the experimentally observed signal used to stop or interpret the titration—for example, an indicator color change or an instrumental threshold. A good indicator changes color within the steep pH region close to the equivalence point, but endpoint and equivalence are not conceptually identical.
If an indicator changes too early or too late, it introduces systematic titration error. Choosing an indicator therefore requires the expected curve, not just the words “acid” and “base.”
Worked application: a weak acid with strong base
An original titration contains 25.00 mL of 0.100 M monoprotic weak acid HA with pKa = 4.76. It is titrated with 0.100 M NaOH.
Initial moles HA:
(0.02500 L)(0.100 mol/L) = 2.50 × 10⁻³ mol.
At equivalence, the same amount of OH⁻ is required. With 0.100 M NaOH:
V = n/C = (2.50 × 10⁻³ mol)/(0.100 mol/L) = 0.0250 L = 25.0 mL.
| NaOH added | Dominant reasoning |
|---|---|
| 0 mL | Weak-acid equilibrium |
| 12.5 mL | Half-equivalence buffer; [HA] = [A⁻] |
| 25.0 mL | Equivalence; conjugate base A⁻ dominates acid–base behavior |
| >25.0 mL | Excess strong base controls pH |
Pattern: half-equivalence reveals pKa
At half-equivalence, exactly half of the initial HA has been converted to A⁻, so [HA] = [A⁻] after accounting for the common volume. The Henderson–Hasselbalch relation becomes:
pH = pKa + log([A⁻]/[HA]) = pKa.
In this example, the pH at 12.5 mL added base is therefore approximately 4.76 under the ideal buffer assumptions.
This is useful in reverse: if a titration curve clearly identifies half-equivalence, the pH at that point estimates the weak acid’s pKa.
Rule: equivalence does not always mean pH 7
At equivalence for a strong acid–strong base titration at 25°C under ideal dilute conditions, the resulting solution is approximately neutral. But a weak acid titrated with strong base leaves its conjugate base at equivalence. That conjugate base reacts with water and makes the equivalence solution basic.
Likewise, a weak base titrated with strong acid leaves a conjugate acid at equivalence and produces an acidic solution. Therefore, “equivalence means pH 7” is not a general rule.
Redox titrations use the same stoichiometric idea with electron-transfer reactions rather than proton transfer. The equivalence condition comes from the balanced redox stoichiometry; an observed endpoint is still a measurement signal that should occur near that condition.
ACTIVE PRACTICE · ACID–BASE CHEMISTRY
Find Equivalence Before Interpreting pH
Start with the lesson’s original monoprotic weak acid: 25.00 mL of 0.100 M HA, pKa = 4.76. The base is initially 0.100 M NaOH. Distinguish the stoichiometric equivalence point from an indicator’s observed endpoint.
Initial HA: 2.50 mmol. OH⁻ added: mmol. Equivalence volume: mL.
50.0% of initial HA neutralized
Current region:.
Approximate pH = 4.76 (pH = pKa at half-equivalence).
Henderson–Hasselbalch values are shown only for the intermediate buffer region (approximately 10–90% neutralization) under ideal assumptions. Near the start and equivalence, use weak-acid or conjugate-base equilibrium instead; past equivalence, excess strong base governs. The model does not calculate those pH values.
Check your reasoning
Choose an answer before opening its feedback.
Check 1
20.0 mL of a monoprotic acid requires 16.0 mL of 0.250 M NaOH to reach equivalence. What is the acid concentration?
A. 0.200 M
Correct. Base moles = 0.0160×0.250 = 0.00400 mol; divide by 0.0200 L acid.
B. 0.125 M
Review this choice. This does not use the 1:1 mole stoichiometry correctly.
C. 0.3125 M
Review this choice. This corresponds to an incorrect volume ratio.
D. 4.00 M
Review this choice. 0.00400 is moles, not molarity.
Check 2
At half-equivalence of a weak monoprotic acid titrated with strong base, what relationship holds under the ideal buffer model?
A. pH = pKa
Correct. [A⁻] = [HA].
B. pH must equal 7
Review this choice. Half-equivalence is governed by the acid’s pKa.
C. All acid has been consumed
Review this choice. Half remains as HA.
D. No conjugate base is present
Review this choice. Half the initial acid has been converted to A⁻.
Check 3
A weak acid is titrated with a strong base. Why is the equivalence-point pH typically above 7?
A. The conjugate base of the weak acid reacts with water.
Correct. It generates OH⁻ through base hydrolysis.
B. Equivalence always contains excess strong base.
Review this choice. At equivalence the stoichiometric strong base has been consumed.
C. The indicator forces the pH above 7.
Review this choice. Indicator is a measurement tool, not the cause of the equilibrium pH.
D. Weak acids cannot donate protons.
Review this choice. Weak acids donate protons incompletely, not never.
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