Question

3. A water supply was known to contain lead (II) ions. A 100.0 mL aliquot of the solution was pipetted. and its mass determined to be 100.53 g. The 100.0 mL aliquot was acidified, and 21.25 mL of a 0.01235 M potassium dichromate solution were required to oxidize all of the lead (I) ions, Calculate (1) the molarity of the lead (II) ions in the solution, and (2) the mass percent of the lead (II) ions in the solution.

Answer 1

Let weight of lead ions = w gm Pb^2 + rightarrow Pb^4 + + 2e^(-) (1) Cr_2O_7^2- + 6e^(-) rightarrow Cr^3 + 3 times Eq (1) + Eq (2) = > Net Reaction 3 Pb^2 + + Cr_2O_7^2- rightarrow 3 Pb^4 + + Cr^3 + (Net Reaction should not contain electrons) 1 mole of Cr_2O_7^2- reacts with 3 moles of Pb^2 + n_Pb^2 + = 3 times n_Cr_2O_7^2- n_Pb^2 + = 3 times (Molarity times volume (L))_Cr_2O_7^2- n_Pb^2 + = 3 times 0.01235 times 21.25/1000 n_Pb^2 + = 7.87 times 10^-4 molarity of lead = moles of lead/volume of solution (L) molarity of Pb^2 + = 7.87 times 10^-4/100 times 1000 = 7.87 times 10^-3 M weight of Pb^2 + = moles times molar mass = 7.87 times 10^-4 times 207 = 0.163 gm % weight = weight of lead/Total weight times 100 % weight = 0.163/100 times 100 = 0.163 %