# Acid-Base Titrations

Titrations can be used to find the concentration of an acid or alkali from the relative volumes used and the concentration of one of the two reactants.

You should be able to carry out calculations involving neutralisation reactions in aqueous solution given the balanced equation or from your own practical results.

NB: 1dm³ = 1 litre = 1000ml = 1000 cm³, so dividing cm³/1000 gives dm³.
and other useful formulae or relationships are:
moles = molarity (mol/dm³) x volume (dm³=cm³/1000),
molarity (mol/dm³) = mol / volume (dm³=cm³/1000),
1 mole = RMM (formula mass) in grams.
In most volumetric calculations of this type, you first calculate the known moles of one reactant from a volume and molarity. Then, from the equation, you relate this to the number of moles of the other reactant, and then with the volume of the unknown concentration, you work out its molarity.
Example: Given the equation NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) – you need to do this first if it isn’t given to you so you can calculate molar ratios of reactants.

25 cm³ of a sodium hydroxide solution was pipetted into a conical flask and titrated with 0.2M hydrochloric acid. Using a suitable indicator – phenolphthalein is good – weakly basic is fuchsia pink and acidic is colorless – it was found that 15 cm³ of acid was required to neutralise the alkali.

Calculate the molarity of the sodium hydroxide and concentration in g/dm³.
moles HCl = (15/1000) x 0.2 = 0.003 mol
moles HCl = moles NaOH (1 : 1 in equation)
so there is 0.003 mol NaOH in 25 cm³
scaling up to 1000 cm³ (1 dm³), there are …
0.003 x (1000/25) = 0.12 mol NaOH in 1 dm³
molarity of NaOH is 0.12M or mol/dm³
since mass = moles x RMM, and RMM of NaOH = 23 + 16 + 1 = 40
concentration in g/dm³ is 0.12 x 40 = 4.41g/dm³