A 2.9880g sample containing Cl- and ClO4- (as the potassium salts) was dissolved in sufficient water to give 250.0 mL of solution. a 50.00mL aliquot required 13.97mL of 0.08551M AgNO3 to titrate the Cl-. In a second 50.00mL aliquot, the ClO4- was reduced to Cl- with V2(SO4)3. Titration of the reduced sample required 40.12mL of the AgNO3 solution. Calculate the %KCl and %KClO4 in the sample.© BrainMass Inc. brainmass.com March 21, 2019, 6:57 pm ad1c9bdddf
2.9880g sample containing KCl and KClO4
Dissolve in 250mL
1) Take 50mL
13.97mL of a 0.08551 M AgNO3 solution = 0.0012 moles AgNO3
Since AgNO3 and KCl react 1 to 1 in stoichiometry, this means there is 0.0012 moles of KCl in the 50mL aliquot. So, 0.0012 x 5 = 0.06 moles of KCl in 250 mL solution ...
The expert examines stoicheometry of potassium salts. The %KCl and %KClO4 are calculated.