Micro chemistry

Question

According to Chamot and Mason’s Handbook of Chemical Microscopy, Vol. II, a dilute solution of hydrochloric acid can be used to test for the presence of lead ion. To prepare a positive control solution, from which the test drop will be aliquoted, 50.0 mg of lead nitrate is added to 10.0 mL of distilled H2O which has been acidified with nitric acid. A dilute drop of hydrochloric acid is added to a 40.0 μL test drop using Method I. You may use the following information to answer the questions which follow: Atomic weights: H = 1.008 grams/mole, N = 14.01 grams/mole, O = 16.00 grams/mole, Cl = 35.453 grams/mole, Pb = 207.2 grams/mole. 

d. For this chemical test, only 75.0% of the lead in the test drop is expected to react with the reagent to form the crystalline product. How many micrograms of the lead crystalline product will form from the 40.0 μL test drop? Show each step clearly as you carry out the calculation. List any assumptions that you make. Pay particular attention to significant figures, units, and unit cancellations

Answer 1

lead nitrate = Pb(NO₃)₂

molar mass lead nitrate = 331.21 g/mol

mass % Pb in Pb(NO₃)₂ = 207.2 g / 331.21 g (x100%) = 62.56%

mass of Pb in 50.0 mg Pb(NO₃)₂ = 0.6256 x 50.0 mg = 26.28 mg Pb

26.28 mg Pb / 10 ml = 2.628 mg / ml

2.628 mg / ml x 1 ml / 1000 ul x 40 ul = 0.1051 mg Pb in the 40.0 ul sample

75% reacts so ... 0.1051 mg x 0.75 = 0.0788 mg Pb reacts

0.0788 mg Pb x 1000 ug Pb/mg Pb x 1 umol Pb / 207.2 ug = 0.3803 umol Pb

Assuming the crystalline product is lead chloride, PbCl₂, there will be 0.3803 umol PbCl₂

molar mass PbCl₂ = 278.1 g/mol

mass PbCl₂ in the 40.0 ul test drop = 0.3803 umol x 278.1 ug/umol = 106 ug (3 sig. figs.)