Acid-Base Titrations

Acid-base titration answers a simple question: how much acid or base is in a solution whose concentration you do not know? A measured volume of titrant, whose concentration is known precisely, is added. This continues until the reaction with the analyte is exactly complete. That instant is the equivalence point. It is detected experimentally by an indicator color change or a pH meter reading, called the endpoint. The entire calculation rests on tools already familiar from Solution Stoichiometry and Neutralization Reactions: moles react in fixed ratios given by a balanced equation, and molarity converts between moles and volume. Titration simply runs that logic in reverse. Instead of predicting how much product forms, you measure the volume of titrant consumed. Then you work backward to find the unknown concentration. This article develops the master equation for titration calculations. It walks through two fully numeric examples: a simple 1:1 acid-base titration and a diprotic acid case. It also highlights the traps that catch beginners — mismatched mole ratios, unit slips between mL and L, and confusing the equivalence point with the endpoint.

Key takeaways

Titration finds an unknown concentration by measuring the volume of a titrant of known concentration needed to reach the equivalence point.
The equivalence point is where moles of acid and base have reacted in their exact stoichiometric ratio; the endpoint is what you actually observe (indicator color change).
The master relation is $M_aV_a \times (\text{mole ratio}) = M_bV_b$ , built entirely from Solution Stoichiometry.
Polyprotic acids and bases need explicit attention to the stoichiometric coefficients — a 1:1 assumption will silently give a wrong answer.
Indicators must change color near the actual pH of the equivalence point, not necessarily at pH 7.
Key idea: a titration converts a measured volume into a mole count, and stoichiometry converts that mole count into the unknown concentration.

The core insight

At the equivalence point, moles of H⁺ supplied by the acid exactly equal moles of OH⁻ (or equivalent base) supplied by the base, scaled by the reaction's stoichiometric coefficients.

The physical picture

Picture a flask holding a fixed volume of an acid solution of unknown concentration, sitting on a white tile or paper for contrast, with a few drops of indicator added. Above it, a burette — a long graduated glass tube with a stopcock — delivers a base of precisely known concentration one drop at a time. As base is added, it neutralizes acid according to the balanced equation. The reaction proceeds smoothly until the moles added become stoichiometrically equal to the moles originally present; at that instant, the solution's pH swings sharply, often by several units within a single drop. The indicator is chosen so that its own color change happens right at that pH swing. Because the color change is visible slightly before or after the true equivalence point, the volume read off the burette at color change is the endpoint — an experimental stand-in for the equivalence point, the exact stoichiometric moment. With a well-chosen indicator the two differ by a fraction of a drop, negligible for most purposes. The whole exercise only works because volume and concentration are linked by $n = MV$ . Reading the burette before and after titration gives the volume dispensed; multiplying by the known molarity gives moles of titrant used; the balanced equation converts that into moles of analyte; dividing by the analyte's original volume returns its concentration.

Figure

Titration setup: burette delivering base into a flask of acid with indicator

The core relations

For a simple monoprotic acid reacting with a monoprotic base in 1:1 ratio,

M_{\text{acid}} V_{\text{acid}} = M_{\text{base}} V_{\text{base}}

where:

$M_{\text{acid}}$ , $M_{\text{base}}$ are molarities (mol/L) of acid and base,
$V_{\text{acid}}$ , $V_{\text{base}}$ are the volumes (L) reacted at the equivalence point.

For a general acid–base reaction with stoichiometric coefficients $a$ (acid) and $b$ (base) in the balanced equation,

\frac{M_{\text{acid}} V_{\text{acid}}}{a} = \frac{M_{\text{base}} V_{\text{base}}}{b}

where:

$a$ is the number of acid "units" (e.g., moles of H⁺ per mole of acid) consumed,
$b$ is the number of base units (e.g., moles of OH⁻ per mole of base) consumed,
dividing by the coefficient converts moles of the substance into moles of reaction extent, matching the two sides.

The moles of titrant delivered at the endpoint are always

n_{\text{titrant}} = M_{\text{titrant}} \times V_{\text{titrant}}

where:

$V_{\text{titrant}}$ is read directly from the burette (initial reading subtracted from final reading),
this $n_{\text{titrant}}$ is then converted to moles of analyte using the balanced equation's mole ratio.

Derivation AFrom burette readings to unknown concentration

Goal: derive the unknown molarity of an analyte solution from the volume of titrant consumed, for a general acid-base reaction $aA + bB \rightarrow \text{products}$ .

Step 1. Write the balanced neutralization equation and identify the mole ratio between analyte (A) and titrant (B). This ratio, $b:a$ (moles of B per mole of A), comes directly from the coefficients — this is exactly the logic used in Solution Stoichiometry for any reaction.

Step 2. Record burette readings before and after titration; subtract to get the volume of titrant delivered,

V_B = V_{\text{final}} - V_{\text{initial}}.

Step 3. Convert that volume into moles of titrant using its known molarity,

n_B = M_B \, V_B.

Step 4. Use the mole ratio from Step 1 to convert moles of titrant into moles of analyte that must have been present,

n_A = n_B \times \frac{a}{b}.

Step 5. Divide by the known volume of analyte solution used (measured by pipette before titration began) to get the unknown molarity,

M_A = \frac{n_A}{V_A}.

Step 6. Combining all four steps into one line recovers the master relation,

M_A = \frac{M_B V_B}{V_A}\times\frac{a}{b},

which reduces to $M_aV_a = M_bV_b$ exactly when $a = b = 1$ .

Worked examples

Example 1 — monoprotic acid, monoprotic base (1:1 ratio). A 20.00 mL sample of hydrochloric acid of unknown concentration is titrated with 0.1024 M NaOH. The burette reads 0.55 mL initially and 31.87 mL at the phenolphthalein endpoint. Reaction:

\text{HCl}(aq) + \text{NaOH}(aq) \rightarrow \text{NaCl}(aq) + \text{H}_2\text{O}(l)

Step 1 — volume of titrant used:

V_{\text{NaOH}} = 31.87 - 0.55 = 31.32\ \text{mL} = 0.03132\ \text{L}

Step 2 — moles of NaOH delivered:

n_{\text{NaOH}} = 0.1024\ \text{mol/L} \times 0.03132\ \text{L} = 3.207\times10^{-3}\ \text{mol}

Step 3 — moles of HCl (1:1 ratio):

n_{\text{HCl}} = 3.207\times10^{-3}\ \text{mol}

Step 4 — molarity of HCl:

M_{\text{HCl}} = \frac{3.207\times10^{-3}\ \text{mol}}{0.02000\ \text{L}} = 0.1604\ \text{M}

Limiting-case check: if instead exactly 20.00 mL of NaOH were needed at the same concentration ( $a=b=1$ , equal volumes), the acid and base concentrations would come out equal — a sanity check confirming the formula behaves correctly when $V_A = V_B$ . Here $V_B$ (31.32 mL) exceeds $V_A$ (20.00 mL), correctly signaling the acid is more concentrated than the base, consistent with $M_{\text{HCl}} = 0.1604\ \text{M} > 0.1024\ \text{M} = M_{\text{NaOH}}$ .

Example 2 — diprotic acid titrated with monoprotic base (2:1 ratio). A 15.00 mL sample of sulfuric acid of unknown concentration requires 42.68 mL of 0.0890 M NaOH to reach the second equivalence point (full neutralization). Reaction:

\text{H}_2\text{SO}_4(aq) + 2\,\text{NaOH}(aq) \rightarrow \text{Na}_2\text{SO}_4(aq) + 2\,\text{H}_2\text{O}(l)

Here $a = 1$ (mole of acid) reacts with $b = 2$ (moles of base), so each mole of H₂SO₄ consumes two moles of NaOH.

Step 1 — moles of NaOH delivered:

n_{\text{NaOH}} = 0.0890\ \text{mol/L} \times 0.04268\ \text{L} = 3.799\times10^{-3}\ \text{mol}

Step 2 — moles of H₂SO₄, using the 1:2 ratio:

n_{\text{H}_2\text{SO}_4} = n_{\text{NaOH}} \times \frac{1}{2} = 1.900\times10^{-3}\ \text{mol}

Step 3 — molarity of H₂SO₄:

M_{\text{H}_2\text{SO}_4} = \frac{1.900\times10^{-3}\ \text{mol}}{0.01500\ \text{L}} = 0.1266\ \text{M}

Limiting-case check: had the ratio wrongly been treated as 1:1 (a common beginner error), the calculated concentration would double to about 0.2533 M — always check whether the acid or base is polyprotic, because that factor of 2 is easy to miss and changes the answer by exactly the ratio $b/a$ .

Limiting cases & common traps

Situation	What happens	Why
$a = b = 1$ , $V_A = V_B$	$M_A = M_B$	Master formula collapses to equal concentrations when volumes match and ratio is 1:1
Polyprotic acid/base ignored	Concentration off by factor $b/a$	Mole ratio from the balanced equation was not applied
Volume in mL not converted to L	Answer off by factor of 1000	Molarity is defined per liter; mixing units breaks the calculation
Endpoint ≠ equivalence point	Small systematic error	Indicator changes color slightly before/after the true stoichiometric point
Titrant not standardized	Systematic error in every result	$M_B$ itself must be known accurately (see Dilution of Solutions for how standard solutions are prepared)

The chemist behind it

Titration as a quantitative technique grew out of 18th- and 19th-century efforts to standardize industrial and analytical chemistry, particularly in France, where precise volumetric methods were needed to assay vinegar, soda ash, and bleaching agents. Joseph Louis Gay-Lussac formalized volumetric analysis in the 1820s, introducing burettes and systematic endpoint detection for titrating silver and chlorine solutions — work that set the template for acid-base titration as practiced today. The technique's power comes from combining a simple physical measurement (volume) with the rigor of Neutralization Reactions to extract information invisible to the eye: an exact concentration, obtained without ever weighing the unknown solution directly.

Origins · 1 minJoseph Louis Gay-Lussac and how this was discovered

Joseph Louis Gay-Lussac

1778–1850 · Volumetric analysis and quantitative titration methods, early 19th century France

Ask the tutor: How do I choose the right indicator so the endpoint actually matches the equivalence point for a weak acid–strong base titration?

What is the difference between the equivalence point and the endpoint?

The equivalence point is the exact moment moles of acid and base have reacted in their stoichiometric ratio — a theoretical point defined by the chemistry. The endpoint is the moment an indicator visibly changes color or a pH meter reading crosses a threshold, which is what the experimenter actually observes and records. A well-chosen indicator makes the two nearly identical.

Why does the acid volume have to be measured with a pipette rather than a graduated cylinder?

Titration calculations amplify any volume error directly into the final concentration, so the analyte volume $V_A$ must be as precise as the titrant volume. A volumetric pipette delivers a fixed volume far more precisely than a graduated cylinder, keeping the uncertainty in $M_A$ small.

Can titration be used for a base of unknown concentration, not just an acid?

Yes. The master relation is symmetric: you can titrate an unknown base with a standardized acid just as easily as an unknown acid with a standardized base. The same mole-ratio logic from the balanced equation applies in either direction.

Why is the titrant's own concentration trusted so precisely?

Titrants are usually prepared from primary standards — highly pure, stable solids weighed out directly — or standardized by titrating them against such a standard beforehand, sometimes after careful Dilution of Solutions from a stock. This gives $M_B$ an accuracy that then transfers into the calculated unknown concentration.

What if the acid and base react in a ratio other than 1:1?

Insert the mole ratio from the balanced equation directly into the master formula, as shown for sulfuric acid reacting 1:2 with NaOH. Skipping this step is the most common source of error in titration calculations, since it silently scales the final answer by the wrong factor.