Question

35.40 mL of a 1.20 M solution of KCl is mixed with 55.0 ML of a solution of Pb(NO3)2 with unknown Molarity. 2KCl(aq) + Pb(NO3)2(aq) -->2KNO3(aq) + PbCl2(s). The solid PbCl2 weights 5.29g. If the concerntration of lead (II) ions in solution after the complete precipitation is 0.3564 M, what was the original concentration of the lead(II) nitrate solution

Answer 1

Volume of KCl = 35.40 ml = 35.40 L / 1000 = 0.0354 L

Concentration of KCl = 1.20 M = 1.20 mol/L

Number of moles of KCl = 1.20 mol/L * 0.0354 L = 0.04248 mol

From reaction, 2.0 mol of KCl reacts with 1.0 mol of Pb(NO3)2 produces 1.0 mol of PbCl2 so 0.04248 mol of KCl will react with 1/2 * 0.04248 mol = 0.02124 mol of Pb(NO3)2 will produce 0.02 mol of PbCl2.

Molar mass of PbCl2 = 278.1 g/mol

Number of moles of PbCl2 produced = 5.29 g / 278.1 g/mol = 0.019 mol ~ 0.02 mol

From reaction, Number of moles of Pb(NO3)2 reacted = 0.02 mol

After the complete precipitation, concentration of lead(II) ions = 0.3564 M

Total volume of solution = (35.40 + 55.0)ml = 90.4 ml = 90.4 L/1000 = 0.0904 L

Number of moles of Lead(II) ions after precipitation = 0.0904 L * 0.3564 mol/L = 0.032 mol

Since number of moles of Pb(NO3)2 = Number of moles of lead(II) ions

Hence unreacted moles of Pb(NO3)2 = 0.032 mol

Total moles of Pb(NO3)2 = (0.032 + 0.02) mol = 0.052 mol

Original volume of Pb(NO3)2 solution = 55.0 ml = 0.055 L

Concentration of Pb(NO3)2 = 0.052 mol / 0.055 L = 0.945 mol/L = 0.945 M