Step I : Get the average titre of the acid
Step II : Calculate moles of HCl in this volume
We know that, concentration of the acid is 0.0987M
These are the moles of HCl that reacted with barium hydroxide solution.
Step III : Write balanced equation for the reaction and calculate moles of barium hydroxide that reacted.
Ba(OH)2(aq) + 2HCl(aq) BaCl2(aq) + 2H2O(l)
From the reaction mole ratio (Moles of base:moles of acid) is 1:2
Meaning 1 mole of Ba(OH)2 reacts with 2 moles of HCl
These moles of Ba(OH)2 were in 25cm3
How many moles were there in the 250cm3 of the stock solution?
By cross multiplication we get;
These were the number of moles in the 3.632g of solid sample that was dissolved.
Step IV : Calculate RFM of the solid sample of Ba(OH)2 . xH2O
From the formula; we get,
Step V : Now Determine the value of x in Ba(OH)2. xH2O
(Atomic mass, Ba=137.3, H=1.0, O=16.0)
Ba(OH)2.xH2O = 315.911083
173.3 + 2(16+1) + x(2+16) = 315.911083
137.3 + 34 + 18x = 315.911083
18x = 315.911083 - 171.3
18x = 144.61
Divide both sides by 18
The value of x is approximately 8
The formula of hydrated barium hydroxide is therefore Ba(OH)2 . 8H2O