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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
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... show full transcript
Worked Solution & Example Answer: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
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ii. Explain why the filter paper and precipitate were heated and weighed several times.
Answer
The filter paper and precipitate were heated and weighed several times to ensure that all residual moisture is removed. Heating helps to dry the precipitate thoroughly, allowing for an accurate measurement of the mass. If any water remains, it will contribute to the mass, leading to an incorrect calculation of the lead(II) iodide formed.
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iv. What is the mass, in grams, of lead(II) ethanoate that is present in 100.0 mL of hair dye solution?
Answer
To find the mass of lead(II) ethanoate present in 100.0 mL of hair dye solution, we use stoichiometry based on the reaction and lead(II) iodide formation:
Since we have determined the mass of lead(II) iodide formed:
From the molar mass, we can see that:
Calculating: ext{m}(Pb(CH₃COO)₂) = rac{0.1711}{461} imes 325.3 = 0.118 ext{ g} ext{ (approximately)}
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b. Explain why no precipitate of lead(II) nitrate formed.
Answer
No precipitate of lead(II) nitrate formed because lead(II) nitrate is soluble in water. The reaction of lead(II) ions with nitrate does not result in any insoluble compound, hence no precipitate is observed. This is in contrast to lead(II) iodide, which is insoluble and forms a precipitate.