Titration

When an acid and a base react, the ${\text{H}}^{+}$ from the acid combines with the ${\text{OH}}^{-}$ from the base to form water, while the leftover ions combine to form a salt. This is a neutralization reaction. The net ionic equation for a strong acid with a strong base is always ${\text{H}}^{+}+{\text{OH}}^{-}\to {\text{H}}_2\text{O}$, which is what makes titration work: the reaction is fast, goes to completion, and has a precise mole ratio. The salt is named from the base's cation and the acid's anion (KOH + HNO3 gives potassium nitrate). Pay attention to how many ${\text{H}}^{+}$ the acid supplies and how many ${\text{OH}}^{-}$ the base supplies per formula unit: ${\text{H}}_2{\text{SO}}_4$ provides two ${\text{H}}^{+}$, so it needs two NaOH, and ${\text{Ca(OH)}}_2$ provides two ${\text{OH}}^{-}$, so it needs two HCl. Those coefficients become the mole ratio in every titration calculation.

• A precisely known volume of one solution (often the unknown-concentration one) is measured into a flask using a pipette. The pipette is first rinsed with that solution, but the flask may be wet with distilled water — extra water does not change the MOLES of solute delivered.
• A few drops of an indicator are added — a substance that changes colour depending on pH.
• The other solution (of known, standard concentration) is loaded into a burette, which allows very precise, controlled addition (drop by drop near the end). Rinse the burette with the titrant first so leftover water cannot dilute it, and record the initial reading at the bottom of the meniscus, at eye level, to two decimal places.
• The known solution is added slowly to the flask, swirling constantly, until the indicator just changes colour — this is the endpoint of the titration. A white tile under the flask makes the first permanent faint colour easy to see.
• The volume of solution used (final reading minus initial reading) is recorded and used in calculations. The first run is usually a quick rough trial; further careful trials are repeated until at least two agree closely (typically within 0.10–0.20 mL), and those are averaged.

The core idea is exactly solution stoichiometry: use $n=cV$ on the known solution to find moles reacted, apply the mole ratio from the balanced equation, and use $n=cV$ again (rearranged) to find the unknown concentration. At the equivalence point there is no limiting reagent to find — by definition the two reactants are present in exactly the mole ratio of the equation. Three checks catch most errors. First, units: convert both volumes to litres before using $n=cV$, so the moles come out in mol rather than millimoles. Second, the ratio: write the balanced equation before calculating; a diprotic acid such as ${\text{H}}_2{\text{SO}}_4$ or a base such as ${\text{Ca(OH)}}_2$ changes the ratio from 1 : 1 to 1 : 2. Third, sense: if it took MORE titrant than the volume of sample, and the ratio is 1 : 1, the sample must be MORE concentrated than the titrant. Recognise a titration question by the words 'endpoint', 'burette', 'indicator changed colour' or 'required to neutralize'.