A solution of NaCl(aq) is added slowly to Pb(NO3)2(aq) Pb ( NO 3 ) 2 ( aq )

Question

A solution of NaCl(aq) is added slowly to a solution of lead nitrate, Pb(NO3)2(aq), until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 14.19 g PbCl2(s) is obtained from 200.0 mL of the original solution. 

Calculate the molarity of the Pb(NO3)2(aq) solution.

Answer 1

The balanced equation is Pb(NO3)2(aq) + 2NaCl (aq) =====> PbCl2 (s) + 2NaNO3 (aq).

The conversion factor to get moles of Pb(NO3)2 from grams of PbCl2 is 1 mole Pb(NO3)2 / 278.1 g PbCl2 (molecular mass PbCl2) = 0.003596 moles Pb(NO3)2 / g PbCl2. Multiply this by the number grams PbCl2 recovered to get

0.003596 x 14.19 g PbCl2 = 0.0512 moles Pb(NO3)2.

Molarity = moles / L = 0.0512 moles x 1L / 200 mL = 0.255 Molar Pb(NO3)2