Titration Calculations Revision

titration - accurate neutralisation of an acid with an alkali data obtained can be used to do calculations equation used c x v m (acid) c x v m (alkali) = concentration of acid concentration of alkali volume of acid volume of alkali moles of alkali – balanced equation moles of acid – balanced equation

A pupil found that 20 cm 3 of hydrochloric acid was neutralised by 10 cm 3 of 2 mol/l sodium hydroxide. What is the concentration of the acid? HCl + NaOH NaCl + H 2 O 1 mol c x 20 1 (acid) 2 x 10 1 (alkali) = c = 2 x c = 1 mol/l

c x v m (acid) = c x v m (alkali) c = concentration in mol/lv = volume in cm 3 or litres m = number of moles (from balanced chemical equation) – usually provided in exams

1 mol A pupil found that 20 cm 3 of hydrochloric acid was neutralised by 10 cm 3 of 2 mol/l sodium hydroxide. What is the concentration of the acid? HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) c x v m (acid) c x v m (alkali) = cx 20 1 = 2 x 10 1 = 1 mol/l

50 cm 3 of 0.4 mol/l potassium hydroxide solution exactly neutralises 20 cm 3 of sulphuric acid. What is the concentration of the acid? 2 KOH + H 2 SO 4 K 2 SO H 2 O0.5 mol/l 20 cm 3 of 0.2 mol/l sulphuric acid exactly neutralises 0.5 mol/l sodium hydroxide solution. What volume of sodium hydroxide was required? 2 NaOH + H 2 SO 4 Na 2 SO H 2 O16 cm 3