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What is the final concentration of Pb2+ in μG/100 mL blood?
A child has a dangerous blood lead level of 120 μg/100 mL. If the child is administered 100.0 mL of 0.10 M Na2Ca(EDTA), assuming the exchange reaction and excretion process are 100% efficient, what is the final concentration of Pb2+ in μG/100 mL blood? (Total blood volume is 1.5 L.)
1 Answer
- JOHLv 68 years agoFavorite Answer
Not 100% sure on this...
Na2Ca(EDTA)(aq) + Pb^2+(aq) ----> PbCa(EDTA)(aq) + 2Na^+(aq)
n(Na2Ca(EDTA) = (0.100 L)(0.10 M) = 0.010 mol
n(Pb^2+) = (120 μg/ 100 mL)(1000 mL / 1 L)(1 g / 10^6 μg)(1.5 L) / 207.20 g/mol = 8.69 μmol
Na2Ca(EDTA) is in excess.
As the reaction is 100% efficient, all the lead (II) ions are consumed in the reaction and excreted.
Hence,
[Pb^2+] = 0 μg / mL
[Na2Ca(EDTA)] = (0.010 - 8.69x10^-6)mol / 1.5 L = 0.00666 M = 6.66 mM
