Question

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.

Answer 1

Step I : Get the average titre of the acid

$\dfrac{23.26+23.29+23.34}{3} =\dfrac{69.89}{3}=23.29cm3$

Step II : Calculate moles of HCl in this volume

We know that, concentration of the acid is 0.0987M

$\therefore 0.0987moles =1000cm3$

$? =23.29cm$

$=\dfrac{0.0987 x 23.29}{1000}$

$=0.0022999mol$

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) $\to$ BaCl_2(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

$\therefore moles of barium hydroxide= \dfrac{1}{2} x moles of HCl$$Moles of barium hydroxide = \dfrac{1}{2}x0.0022999=0.00114969 moles$

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;

$=\dfrac{0.00114969 x 250}{25} = 0.0114969 moles$

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; $Moles = \dfrac{mass}{RFM}$ we get,

$RFM = \dfrac{mass}{moles}$

$RFM = \dfrac{3.632}{0.0114969} = 315.9111083 amu$

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

$x=\dfrac{144.61}{18} = 8.034$

The value of x is approximately 8

The formula of hydrated barium hydroxide is therefore Ba(OH)₂ . 8H₂O