In Titration you added base to acid until an indicator changed color. Now the numbers: the volume used at that point, together with Solution Stoichiometry, gives a concentration, a molar mass or a purity.
Key idea
At the equivalence point the moles of titrant added are exactly what the balanced equation needs for the analyte: $$n_\text{titrant} = M_\text{titrant}V_\text{titrant}, \qquad n_\text{analyte} = n_\text{titrant} \times (\text{mole ratio}), \qquad M_\text{analyte} = \frac{n_\text{analyte}}{V_\text{analyte sample}}.$$ For a diprotic acid such as $\mathrm{H_2SO_4}$, each mole of acid needs 2 moles of NaOH.
Why it works
A titration is a way of counting moles without weighing them: you add a solution of known concentration drop by drop until the reaction is exactly complete, and the buret tells you the volume. The analyte's moles then follow from the ratio. Its concentration is those moles divided by the volume of the analyte you started with, because that is where those moles were.
Try a titration yourself in Mixing Lab, the Science Lab's acid–base bench.
Worked example: Concentration of an acid
$25.00\ \text{mL}$ of HCl needs $31.25\ \text{mL}$ of $0.1000\ \text{M}$ NaOH to reach the equivalence point. Find $[\mathrm{HCl}]$.
- $$n_\text{NaOH} = 0.03125\ \text{L} \times 0.1000\ \text{M} = 3.125 \times 10^{-3}\ \text{mol}$$
- $\mathrm{HCl} + \mathrm{NaOH} \to \mathrm{NaCl} + \mathrm{H_2O}$: 1 : 1, so $3.125 \times 10^{-3}\ \text{mol}$ HCl.
- $M = \dfrac{3.125 \times 10^{-3}}{0.02500\ \text{L}} = 0.1250\ \text{M}$.
Answer: $0.1250\ \text{M}$
Worked example: A diprotic acid
$20.00\ \text{mL}$ of $\mathrm{H_2SO_4}$ needs $28.40\ \text{mL}$ of $0.1500\ \text{M}$ NaOH. Find $[\mathrm{H_2SO_4}]$.
- $$n_\text{NaOH} = 0.02840 \times 0.1500 = 4.260 \times 10^{-3}\ \text{mol}$$
- $\mathrm{H_2SO_4} + 2\,\mathrm{NaOH} \to \mathrm{Na_2SO_4} + 2\,\mathrm{H_2O}$: $n_{\mathrm{H_2SO_4}} = 4.260 \times 10^{-3}/2 = 2.130 \times 10^{-3}\ \text{mol}$.
- $M = 2.130 \times 10^{-3}/0.02000 = 0.1065\ \text{M}$.
Answer: $0.1065\ \text{M}$
Worked example: Standardizing a base
$0.5106\ \text{g}$ of potassium hydrogen phthalate (KHP, $204.23\ \text{g/mol}$, one acidic H) is neutralized by $24.65\ \text{mL}$ of NaOH. Find the NaOH concentration.
- $n_\text{KHP} = 0.5106/204.23 = 2.500 \times 10^{-3}\ \text{mol}$.
- 1 : 1, so $2.500 \times 10^{-3}\ \text{mol}$ NaOH in $0.02465\ \text{L}$.
- $M = 0.1014\ \text{M}$.
Answer: $0.1014\ \text{M}$
The endpoint, where the indicator changes, is what you observe; the equivalence point, where the moles match the equation exactly, is what you want. A well-chosen indicator changes color on the steep part of the curve, so the two agree to within a drop.
Common mistake
Dividing the acid's moles by the volume of base used instead of the volume of the acid sample. In the first example, $3.125 \times 10^{-3}/0.03125 = 0.1000\ \text{M}$ just gives back the base's concentration. Check: the analyte's moles live in the analyte's volume; if your answer equals the titrant's molarity, you divided by the wrong volume.