Some progressive hair dyes marketed to men, such as Grecian Formula 16, contain

Question

Some progressive hair dyes marketed to men, such as Grecian Formula 16, contain lead acetate, Pb(CH3CO2)2. As the dye solution is rubbed on the hair, the Pb2+ ions react with the sulfur atoms in hair proteins to give lead(II) sulfide (PbS), which is black. A typical dye solution contains 0.3 mass % Pb(CH3CO2)2, and about 2 mL of dye solution is used per application.
   (a) Assuming that 30% of the Pb(CH3CO2)2 is converted to PbS, how many milligrams of PbS are formed per application of the dye?
   (b) Suppose the hair is washed with shampoo and water that has pH = 5.50. How many washings would be required to remove 50% of the black color? Assume that 3 gal of water is used per washing and that the water becomes saturated with PbS.
   (c) Does the calculated number of washings look reasonable, given that frequent application of the dye is recommended? What process(es) in addition to dissolution might    contribute to the loss of color?

Explanation / Answer

A typical dye solution contains 0.3 mass % Pb(CH3CO2)2,

2 mL of dye solution is used per application

So the mass of lead acetate in 2mL = 0.3 % X 2mL = 0.006 grams

out of this 30% is converted to PbS = 30 % of 0.006 grams = 0.0018 grams of lead acetate

Pb(CH3COO)2 + S-2 ---> PbS(s)

So one mole lead acetate will give one mole of PbS

or 325.28 grams of lead acetate will give 239.26 grams of PbS

one gram will give = 239.26 / 325.28 grams

0.0018 grams will give = 0.0018 X 239.26 / 325.28 = 0.00132 grams = 1.32 milligrams will be formed of PbS

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