I.2. Titration of lead in paint Flakes of white paint (m=5.00g) are dissolved in
ID: 967715 • Letter: I
Question
I.2. Titration of lead in paint Flakes of white paint (m=5.00g) are dissolved in 50 mL of concentrated nitric acid. The formation of a gas is also observed. To 10.0 mL of the obtained solution are added 10.0 mL of a solution of chromium (III) nitrate (concentration 0.10 mol.L-1) and 10.0 mL of a solution of potassium bromate (KBrO3, concetration 0.20 mol. L-1). The pH is adjusted to 4.5 by addition of a few drops of concentrated sodium hydroxide solution and the solution is boiled for one minute. A precipitate of lead chromate PbCrO4 is obtained. The precipitate is filtrated, washed and redissolved in 200 mL of chlorhydric acid to give solution S.
a. Give the balaced equations for all reacitons taking place. 10.0 mL of solution S are titrated by a solution of Fe2+ (concentration 0.10 mol. L-1) in the presence of barium diphenylsulfonate as color indicator. Equivalence is observed at a volume of titrating solution Veq=12.0mL.
b. Give the balanced equation of the titration reaction. Calculate its equilibrium constant. Is the reaction quantitative?
c. Calculate the mass percentage of lead in the paint.
Explanation / Answer
Ans. (a) Pb based white paints are made by using lead (II) carbonate (PbCO3).
When paint flakes of white paint are dissolved in 50 mL of concentrated nitric acid, the following reaction take place:
PbCO3 + 2HNO3 Pb(NO3)2 + CO2 + H2O
Thus the gas formed is CO2.
To 10.0 mL of the obtained solution, 10.0 mL of a solution of chromium (III) nitrate and 10.0 mL of a solution of potassium bromate are added. The pH is adjusted to 4.5 by addition of a few drops of concentrated sodium hydroxide solution and the solution is boiled for one minute.
3 Cr(NO3)2 + 2 KBrO3 + 6 H2O 2 KBr + 6 HNO3 + 3 H2CrO4
Pb(NO3)2 + H2CrO4 = PbCrO4 + 2 HNO3