Question 7 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, in a sample of hair dye - VCE - SSCE Chemistry - Question 7 - 2012 - Paper 1

The filter paper and precipitate were heated to ensure that all moisture was removed, which is crucial for accurate mass measurements. The presence of water would add extra weight, leading to an erroneous calculation of the precipitate's mass. Repeated heating and weighing help ensure that a constant mass is achieved, confirming that all water has been evaporated.

To find the mass of lead(II) iodide formed, we can use the mass measurements provided. The mass of the filter paper plus precipitate after first heating is 0.4831 g, and the mass of the filter paper is 0.3120 g:

$m_{PbI₂} = m_{filter\, paper + precipitate} - m_{filter\, paper} = 0.4831 g - 0.3120 g = 0.1711 g$

This is the mass of lead(II) iodide formed.

Using the provided solubility data, we can calculate the mass of lead(II) ethanoate in 100.0 mL of hair dye solution. The solubility of lead(II) ethanoate at 25 °C is given as 55.0 g per 100 g of solution. Since the density of the hair dye is approximately the same as water, we can assume:

$ext{Mass of solution} \approx 100 ext{ g (for 100 mL)}$

Thus, the mass of lead(II) ethanoate in 100.0 mL is:

$\text{Mass}_{Pb(CH₃COO)₂} = 55.0 g$