Barium hydroxide crystallises as on elf several possible hydrates, Ba(OH)2. xH2O. A sample of barium hydroxide is analysed in order to determine its formula. 3.632g of the sample was dissolved in water to make 250.0mL of solution. 25.00mL of this solution was titrated with 0.0987 M HCl, using methyl orange indicator. T.26, 23.29 and 23.34 mL were obtained. Determine the formula (that is find the value of x) of the barium hydroxide sample.
The reaction is Ba(OH)2 + 2HCl = BaCl2 + 2H2O.
The average litres = 1/3(23.26 + 23.29 + 23.34)= 23.30mL. So, number of moles in 23.30mL of 0.0987M HCl is = (23.30mL/1000mL) * 0.0987= 0.0023mol.
Based on the equation, the mole of barium hydroxidewould be the half of 0.0023(HCl moles): n(Ba(OH)2)= 1/2(0.0023mol)= 0.00115mol in 25 mL. In 250mL we would have (250 * 0.00115)/25= 0.0115mol.
The molar mass of Ba(OH)2 = 137g/mol +2 * 16g/mol+2 * 1g/mol = 171 g/mol, so it's mass in the dissolved sample of hydrate would be` m= n/M = 0.0115mol /171g/mol = 1.967g.
The mass of the sample (that contains barium hydroxide and water) is 3.632g, and 1.967 g from it is barium hydroxide. So, the mass of the water would be = 0.0115mol * x * 18g/mol = 3.632g - 1.967g = 1.665g.