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How does a titration use a reaction of known stoichiometry to find an unknown concentration?

Topic 4.6 Introduction to Titration: use titration data and reaction stoichiometry to determine the concentration of an unknown solution, distinguishing the equivalence point from the endpoint.

A focused answer to AP Chemistry Topic 4.6, covering the titration method, the equivalence point versus the endpoint, and how to use moles, the reaction mole ratio and volume to calculate an unknown concentration, with full worked examples.

Generated by Claude Opus 4.89 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this topic is asking
How a titration works
Equivalence point versus endpoint
The calculation
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What this topic is asking

The College Board (Topic 4.6) wants you to understand a titration as a controlled reaction of known stoichiometry used to find an unknown concentration, and to distinguish the equivalence point (where the reactants have combined in their stoichiometric ratio) from the endpoint (the observed signal used to estimate it). The calculation combines molarity (Topic 3.7) with the mole ratio from a balanced equation (Topic 4.5).

How a titration works

The power of the method is that the reaction's balanced equation provides an exact mole ratio between titrant and analyte. Measuring the titrant's volume tells you its moles, the ratio converts that to the analyte's moles, and the analyte's known volume then gives its concentration. A common example is a strong acid titrated with a strong base of known molarity.

Equivalence point versus endpoint

This distinction is a favorite exam point. The equivalence point is a theoretical, stoichiometric quantity; the endpoint is an experimental estimate of it. The pH at the equivalence point is exactly 7 only for a strong acid with a strong base; for a weak acid with a strong base it is above 7. Choosing an indicator whose color change spans the equivalence pH minimizes the error between endpoint and equivalence point.

The calculation

The calculation always follows the same chain: moles of titrant ( $n = MV$ ), then moles of analyte (apply the mole ratio from the equation), then concentration of analyte ( $M = n / V$ ). Keep volumes in liters and watch the mole ratio: it is $1:1$ for HCl with NaOH but not for a diprotic acid such as $\text{H}_2\text{SO}_4$ with NaOH, which reacts $1:2$ .

Finding a concentration from titration data

A $20.00$ mL sample of NaOH is titrated to the equivalence point with $0.200$ M HCl, requiring $25.00$ mL. The reaction is $\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}$ . Find the NaOH concentration.

step 1 Find the moles of titrant

$n(\text{HCl}) = MV = (0.200)(0.02500) = 5.00 \times 10^{-3}$ mol.

step 2 Apply the mole ratio

HCl and NaOH react $1:1$ , so $n(\text{NaOH}) = 5.00 \times 10^{-3}$ mol.

step 3 Find the analyte concentration

$M(\text{NaOH}) = \dfrac{n}{V} = \dfrac{5.00 \times 10^{-3}}{0.02000} = 0.250$ M.

step 4 Check the reasoning

A smaller volume of NaOH ( $20.00$ mL) than HCl ( $25.00$ mL) was used at the equivalence point, so the NaOH must be more concentrated than the HCl, which it is ( $0.250 > 0.200$ M).

Try this

Q1. Calculate the moles of base needed to reach equivalence with $0.0100$ mol of a monoprotic acid. [1 point]

Cue. $0.0100$ mol (the $1:1$ ratio means equal moles of base).

Q2. State the difference between the equivalence point and the endpoint of a titration. [2 points]

Cue. The equivalence point is where reactants have combined in their stoichiometric ratio; the endpoint is the observed signal (such as an indicator change) used to estimate it.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). A

25.00

mL sample of hydrochloric acid (HCl) of unknown concentration is titrated with

0.150

M sodium hydroxide (NaOH). The equivalence point is reached after

30.00

mL of NaOH is added. The reaction is

\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}

. (a) Calculate the moles of NaOH used. (b) Determine the moles of HCl that reacted. (c) Calculate the concentration of the HCl. (d) Distinguish the equivalence point from the endpoint.

Show worked answer →

A 4-point quantitative FRQ on titration.

(a) Moles NaOH (1 point): $n = MV = (0.150)(0.03000) = 4.50 \times 10^{-3}$ mol.
(b) Moles HCl (1 point): the mole ratio is $1:1$ , so $n(\text{HCl}) = 4.50 \times 10^{-3}$ mol.
(c) Concentration (1 point): $M = \dfrac{n}{V} = \dfrac{4.50 \times 10^{-3}}{0.02500} = 0.180$ M.
(d) Equivalence vs endpoint (1 point): the equivalence point is when stoichiometrically equal amounts of acid and base have reacted; the endpoint is the observed signal (such as an indicator color change) used to estimate it. They coincide for a well-chosen indicator but are not identical concepts.

Markers reward correct moles of titrant, applying the mole ratio, the unknown concentration, and a correct distinction between equivalence point and endpoint.

AP 2021 (style)1 marksSection I (multiple choice). In an acid-base titration, the equivalence point is the point at which (A) the indicator changes color (B) the moles of acid and base have reacted in their stoichiometric ratio (C) the pH equals 7 in every case (D) the burette is empty. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

The equivalence point is when the acid and base have reacted in exactly their stoichiometric ratio (the moles match the equation). The indicator color change is the endpoint (an estimate of it), and the pH at equivalence is 7 only for a strong acid with a strong base, not in every case.

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Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)