Question

100.0 mL of a Pb(NO3)2 solution was concentrated to 80.0 mL. To 2.00 mL of the concentrated solution was added an excess of NaCl(aq), and 3.407 g of a solid was obtained. What was the molarity of the original Pb(NO3)2 solution? The "To 2.00mL of the concentrated...an excess of NaCl" confuses me.

Answer 1

YOUR CONSUSION --> this implies, we take 2 mL from the 80 mL solution. Note that the concentration of 80 and 2 mL solutions are the same....

The reaction:

Pb(NO3)2 + 2NaCl = PbCl2 + 2NaNO3

ratio is

1 mol of Pb per 2 mol of Cl-

solid obtained

m = 3.407 g of PbCl2

mol of PbCl2 = mass/MW = 3.407/278.1 = 0.01225 mol of PbCl2

mol of Pb+2 = 0.01225 mol

find [Pb+2] original:

from

2 mL solution --> 0.01225 mol are present

then,

80/2= 40x

0.01225*40 = 0.49 mol o in 80 mL

[Pb+2] = (80/100) * 0.49 = 0.392 M

[Pb(NO3)2] = 0.392 M