Physical Exam Qs Flashcards
A pure solid is thought to be calcium hydroxide. The solid can be identified from its relative formula mass.
The relative formula mass can be determined experimentally by reacting a measured mass of the pure solid with an excess of hydrochloric acid. The equation for this reaction is
Ca(OH)2 + 2HCl CaCl2 + 2H2O
The unreacted acid can then be determined by titration with a standard sodium
hydroxide solution.
You are provided with 50.0 cm3 of 0.200 mol dm–3 hydrochloric acid.
Outline, giving brief practical details, how you would conduct an accurate experiment
to calculate the relative formula mass of the solid using this method. (8 marks)
Stage 1: appreciation that the acid must be in excess and calculation of amount of solid that permits this
M1 Statement that there must be an excess of acid
M2 Moles of acid = 50.0 × 0.200/1000 = 1.00 × 10^–2 mol
M3 2 mol of acid react with 1 mol of calcium hydroxide therefore moles of solid weighed out must be less than half the moles of acid = 0.5 ×1.00×10^–2 = 5.00×10^–3 mol
M4 Mass of solid must be < 5.00 × 10^–3 × 74.1 =< 0.371g
Stage 2: Experimental method
M5 Measure out 50 cm3 of acid using a pipette and add the weighed amount of solid in a conical flask
M6 Titrate against 0.100 (or 0.200) mol dm–3 NaOH added from a burette and record the volume (v) when an added indicator changes colour
Stage 3: How to calculate Mr from the experimental data
M7 Moles of hydroxide = 5.00 × 10^–3 – (v × conc NaOH)/1000 = z mol
M8 Mr =mass of solid / z
