If 45.5 mL lead(II) nitrate solution reacts completely with excess sodium iodide solution to yield 0.540 g precipitate, what is the molarity of lead(II) ion in the original solution?
2007-09-27 07:51:40 · 2 answers · asked by ronnie c 1 in Science & Mathematics ➔ Chemistry
Not at all difficult:
Pb+2 + 2 I- ===> PbI2 (s)
If you get 100% reaction, then the precipitate will consist of lead iodide -- PbI2
The molecular weight of lead iodide is 461.00 grams/mole
So, the precipitate is 0.540 g/461.00 g/mole = 0.00117 moles of lead iodide
Which means that you have 0.00117 moles of lead in the original solution.
(0.00117 moles)/[(45.5 ml)*(1 liter/1000 ml)] = 0.0257 moles/liter
So your original solution is 0.0257 M = 2.57 x 10^-2 M
2007-09-27 08:02:48 · answer #1 · answered by Dave_Stark 7 · 0⤊ 0⤋
Pb(NO3)2 + 2 NaI >> PbI2 + 2 NaNO3
Molar mass PbI2 = 460.99 g/mol
Moles PbI2 = 0.540 / 460.99 = 0.00117 = moles Pb2+ in
Pb(NO3)3
M = moles / L = 0.00117 / 0.0455 L = 0.0257 M
2007-09-27 15:00:53 · answer #2 · answered by Dr.A 7 · 0⤊ 0⤋