Chem Homework help

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 10.93 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

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

From this balanced equation and the given mass of PbCl₂, we will calculate moles of PbCl₂ and then moles of Pb(NO₃)₂. Since we know the original volume of Pb(NO₃)₂, we can then find the molarity.

molar mass PbCl₂ = 278.1 g/mol

moles PbCl₂ present = 10.93 g x 1 mol/278.1 g = 0.03930 moles

moles Pb(NO₃)₂ originally present = 0.03930 mol PbCl₂ x 1 mol Pb(NO₃)₂ / mole PbCl₂ = 0.03930 moles

Original volume Pb(NO₃)₂ = 200.0 ml x 1 L/1000 ml = 0.2000 L

Molarity of Pb(NO₃)₂ solution = 0.03930 moles / 0.2000 L = 0.1965 M Pb(NO₃)₂