Titrations questions Flashcards
(2 cards)
An impure sample of barium hydroxide of mass 1.6524g was allowed to react
with 100cm3 of 0.200mol dm-3 hydrochloric acid. When the excess acid was titrated against sodium hydroxide, 10.9cm3 of sodium hydroxide solution was required. 25.0cm3 of the sodium hydroxide required 28.5cm3 of the hydrochloric acid in a separate titration. Calculate the percentage purity of the sample of barium hydroxide.
You still have not said how much of the excess acid was titrated against the NaOH - all of it?
OK lets assume all of it..
25.0 cm3 of the sodium hydroxide required 28.5 cm3 of the hydrochloric acid
moles acid = moles base = 0.0285 x 0.2 = 0.0057
Thus molarity of NaOH = 0.0057/0.025 = 0.228 M
When this titrated the excess acid 10.9 ml was used…
moles of excess acid = 0.0109 x 0.228 = 0.002485
Initial moles of acid = 0.1 x 0.2 = 0.02
therefore moles acid reacted = 0.02 - 0.002485 = 0.0175 mol
HCl reacts with Ba(OH)2 in a 2 :1 ratio
2HCl + Ba(OH)2 –> BaCl2 + 2H2O
thus moles of Ba(OH)2 = 0.0175/2 = 0.00876
mass = moles x RMM = 0.00876 x 171 = 1.4975 g
original mass weighed 1.6524 so percentage purity = 1.4975/1.6524 = 90.6%
A solution of metal carbonate, M2CO3, was prepared by dissolving 7.46g of the anhydrous solid in water to give 1000cm3 of solution. 25.0cm3 of this solution reacted with 27.00cm3 of 0.100 mol dm-3 hydrochloric acid. Calculate the relative formula mass of M2CO3 and hence the relative atomic mass of the metal
1) Write out the equation first
2HCl + M2CO3 –> 2MCl + CO2 + H2O
2) moles of HCl = 0.027 x 0.1 = 0.0027
2 moles HCl is equivalent to 1 mole M2CO3
In titre moles M2CO3 = 0.0027/2
Moles per litre = 0.0027 x 40/2 = 0.054 moles
this is equivalent to 7.46 g
therefore 1 mole = 7.46/0.054 =138
Mass of CO3 = 60
Therefore remaining mass = 2 x M = 138 - 60 = 78
Therefore M = 39 (Potassium)
