Question

A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) , until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 10.47 g PbCl2(s) 10.47 g PbCl 2 ( s ) is obtained from 200.0 mL 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq ) solution. concentration: M

Answer 1

Balanced equation: Pb(NO3)2(aq) + 2NaCl(aq) ----> PbCl2(s) + 2NaNO3(aq)

The moles of PbCl2 = 10.47 g/(278.1 g.mol^-1) = 0.03765 mol

From the balanced equation, the mole ratio Pb(NO3)2:PbCl2 = 1:1

i.e. The molarity of Pb(NO3)2 solution = 0.03765 mol/0.2 L = 0.18824 M

Note: 200 mL = 200/1000 = 0.2 L