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nojha1 · Dec 3, 2009, 02:52 PM

Please help me with this part of problem/

Pb2+(aq) + EDTA2-(aq) -> PbEDTA(aq)
A 10.00 mL sample of the water from the holding tank requires 35.47 mL of 8.193 x 10-4 M EDTA to reach the end point.
(A) What will be the molar concentration of lead(II) ions if you added a 10% excess of the stoichiometric amount of potassium carbonate? (B) Is this enough potassium carbonate to “get the lead out” to meet the EPA limit? (E) Which principle is being applied to remove the lead(II) ions from solution?

Perito · Dec 3, 2009, 03:05 PM

Pb2+(aq) + EDTA2-(aq) -> PbEDTA(aq)

A 10.00 mL sample of the water from the holding tank requires 35.47 mL of 8.193 x 10-4 M EDTA to reach the end point.

(A) What will be the molar concentration of lead(II) ions if you added a 10% excess of the stoichiometric amount of potassium carbonate? (B) Is this enough potassium carbonate to “get the lead out” to meet the EPA limit? (E) Which principle is being applied to remove the lead(II) ions from solution?

A

$34.47\,mL \,\times\, 8.193\times10^{-4}\,\frac {mmoles}{mL} = 2.824\times10^{-2}\, mmoles = 2.848\times10^{-5}\,moles\,of\,EDTA$

Since the stoichiometry requires 1 mole of EDTA per mole of Pb(2+), that is also the number of moles of Pb. The concentration of Pb is

$\frac {2.848\times10^{-5}\,moles\,of\,EDTA}{10\,mL} = 2.848\times10^{-6}\,\frac {moles}{Liter}$

$Pb^{2+} + CO_3^{2-} \rightarrow PbCO_3(s)$

$K_{sp} = [Pb^{2+}][CO_3^{2-}$

The carbonate is 1.1 times the concentration of lead, so

$K_{sp} = [Pb^{2+}][2.848\times\10^{-6}$

To figure out the rest of this question, you need the solubility product of PbCO3 and then you solve for Pb(2+) and compare it with the EPA's best guess as to what's permissible.

B You need the answer in part A to solve this.
E If it's carbonate, I guess it's "precipitation". Or something like that.