Students in a chemistry class were required to design a procedure to determine gravimetrically the concentration of lead(II) in a sample of hair dye, Pb(CH₃COO)₂ - VCE - SSCE Chemistry - Question 7 - 2012 - Paper 1

The filter paper and precipitate were heated to ensure that any remaining moisture was removed. Heating helps to eliminate water, which can affect the mass measurement by adding to the weight of the precipitate. Weighing several times after heating ensures that the mass is consistent and has stabilized, indicating that the precipitate is dry.

To find the mass of lead(II) iodide formed, we first calculate the mass of the precipitate:

1. Mass of filter paper with precipitate after the first heating = 0.4831 g
2. Mass of filter paper = 0.3120 g
3. Mass of lead(II) iodide = Mass of filter paper plus precipitate - Mass of filter paper

Thus,

$m(PbI_{2}) = 0.4831 ext{ g} - 0.3120 ext{ g} = 0.1711 ext{ g}$

Using the concentration of lead ethanoate in the hair dye, we can calculate the mass present in the solution. First, from the previous information, if there are 0.1711 g of PbI₂ formed:

1. The molar mass of lead(II) ethanoate, Pb(CH₃COO)₂ = 325.3 g/mol
2. Calculate moles of lead(II) iodide:

$n(PbI_2) = \frac{0.1711 ext{ g}}{461 ext{ g/mol}} = 0.0003706 ext{ mol}$

Using the stoichiometry from the balanced equations, we conclude that 1 mole of Pb(CH₃COO)₂ yields 1 mole of PbI₂:

3. Therefore, the mass of lead(II) ethanoate in 100 mL of solution is:

$m(Pb(CH₃COO)₂) = n(PbI_2) \times 325.3 ext{ g/mol} = 0.0003706 imes 325.3 = 0.1218 ext{ g}$

No precipitate of lead(II) nitrate formed because the compounds used in the reaction did not result in exceeding the solubility product for lead(II) nitrate in the solution. Lead(II) nitrate remains soluble, and thus does not precipitate under the given conditions.