Question

Suppose we have a solution of lead nitrate, Pb(NO3)2(aq). A solution of NaCl(aq) is added slowly until no further precipitation occurs. The precipitate is collected by filtration, dried, and weighed. A total of 11.72 g of PbCl2(s) is obtained from 200.0 mL of the original solution. Calculate the molarity of the Pb(NO3)2(aq) solution.

Answer 1

Given that 11.72 g of PbCl2(s) is obtained from 200.0 mL of the original solution of Pb(NO3)2(aq).

Mass of PbCl2(s) = 11.72 g

Molar mass of PbCl2(s) = 278.1 g/mol

Moles of  PbCl2(s) = Mass of PbCl2(s) / Molar mass of PbCl2(s)

= 11.72 g/  278.1 g mol-1

= 0.042 mol

Moles of  PbCl2(s) =  0.042 mol

Volume of original Pb(NO3)2 solution = 200.0 mL = 0.20 L

Pb(NO3)2 (aq) + 2 NaCl (a) -------------> PbCl2 (s) + 2NaNO3

1 mol 1 mol

? 0.042 mol

? = [0.042 mol PbCl2 (s) / 1 mol PbCl2 (s)] x 1mol  Pb(NO3)2 (aq)

= 0.042 mol  Pb(NO3)2 (aq)

Hence ,

Moles of Pb(NO3)2 (aq) = 0.042 mol

Volume of original Pb(NO3)2 solution = 200.0 mL = 0.20 L

Molarity of Pb(NO3)2 (aq) = Moles of Pb(NO3)2 (aq) / Volume of Pb(NO3)2 solution

= 0.042 mol/  0.20 L

= 0.21 M

Molarity of Pb(NO3)2 (aq) = 0.21 M

Therefore, molarity of the Pb(NO3)2(aq) solution = 0.21 M