Answer to Question #132046 in Inorganic Chemistry for Bella

Question #132046
Barium hydroxide crystallises as one of several possible hydrates. A sample of barium hydroxide is analysed in order to determine its formula. 3.632 g of the sample was dissolved in water to make 250.0 mL of solution. 25.00 mL of this solution was titrated with 0.0987 M HCl, using methyl orange indicator. Titres of 23.26, 23.29 and 23.34 mL were obtained.

Determine the formula (that is, find the value of x) of the barium hydroxide sample.
Expert's answer

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

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