Question

A 1.379-g sample of commercial KOH contaminated K₂CO₃ by was dissolved in water, and the resulting solution was diluted to 500.0 mL. A 50.00-mL aliquot of this solution was treated with 40.00 mL of 0.05304 M HCl and boiled to remove CO₂. The excess acid consumed 4.55 mL of 0.04925 M (phenolphthalein indicator). An excess of neutral BaCl₂ was added to another 50.00-mL aliquot to precipitate the carbonate as BaCO₃. The solution was then titrated with 28.56 mL of the acid to a phenolphthalein end point. Calculate the percentage KOH,K₂CO₃ , and H₂O in the sample, assuming that these are the only compounds present.

Answer 1

For the first titration

Amount of HCl consumed = mmol NaOH + mmol of KOH + (2 x mmol K2CO3)

mmol KOH + ( 2 x mmol K2CO3) = (0.05304 mmol/mL HCl x 40 mL HCl) - (0.04925 mmol/mL NaOH x 4.55 mL) = 1.897 mmol

For the second titration

Amount of HCl = mmol KOH = 0.05304 mmol/mL HCl x 28.56 mL = 1.514 mmol KOH

amount of K2CO2 = (1.897 mmol - 1.514 mmol)/2 = 0.191 mmol

Percentage KOH = [1.514 x (56.1g KOH/1000 mmol)]/[1.379 x (50 mL/500 mL)] x 100 = 61.58%

Percentage K2CO3 =  [0.191 x (138.21g K2CO3/1000 mmol)]/[1.379 x (50 mL/500 mL)] x 100 = 19.13%

100-(61.58 + 19.13) = 19.28% water