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Question

An impure sample of barium hydroxide of mass 1.6524g was allowed to react with 100cm3 of HCL of concentration 0.200M.When the excess acid was titrated against NaOH, 10.9cm3 of NaOH solution was required.25.0cm3 of the NaOH required 28.5cm3 of the HCL in a separate titration.Calculate the percentage purity of the Ba(OH)2 sample.

Answer 1

First calculate the concentration of the NaOH:

From the second titration with NaOH/HCl

28.5ml(0.2MHCl) = 25ml (xM NaOH) ==> xM NaOH = 0.228 M

Now start at the beginning:

Ba(OH)2 + 2HCl --> BaCl2 + H2O

each mole of Ba(OH)2 will take 2 moles of HCl in your sample

Ba(OH)2 MW = 171.342 g/mole

the excess acid left required 10.9 ml NaOH
and you started with 0.100L(0.2M) = 0.02 moles HCl

moles excess acid = 0.0109L*(0.228M) = 0.0024852 moles HCl

acid used = 0.02 - 0.0024852 = 0.0175 moles

(REMEMBER ONLY 1/2 THIS REPRESENTS THE MOLES OF Ba(OH)2)

Ba(OH)2 present = 0.0175/2 = 0.00875 moles

0.00875 mole * 171.342g/mole = 1.4992g

% purity = 1.499/1.6524*100 = 90.73 %

Answer 2

Molarity of NaOH soln = 28.5*0.2/25 = 0.228 M

Moles of HCl left over = 0.228 * 10.9 = 2.49 mmol

Moles of Ba(OH)2 = (20-2.49)/2 mmol = 8.755 mmol

Mass of Ba(OH)2 = 1.497 g

Thus %age purity = 1.497*100/1.6524 = 90.59 %