Question

44.1 mL of aqueous 0.255 M Pb(NO3)2 is mixed with 38.5 mL of 0.415 M NaCl.

The equation for the precipitate reaction is: Pb(NO3)2 (aq) + 2 NaCl (aq) --> PbCl2 (s) + 2 NaNO3 (aq)

The concentration of Pb2+ ion in the solution is _____ M after the reaction is complete.

Answer 1

Concentration of Pb(NO3)2 = 0.255 M = 0.255 mol/L

Volume of Pb(NO3)2 = 44.1 ml = 44.1 L / 1000 = 0.0441 L

Number of moles of Pb(NO3)2 = 0.255 mol/L * 0.0441 L = 0.01124 mol

Concentration of NaCl = 0.415 M = 0.415 mol/L

Volume of NaCl solution = 38.5 ml = 38.5 L / 1000 = 0.0385 L

Number of moles of NaCl = 0.415 mol/L * 0.0385 L = 0.016 mol

From reaction, 2.0 mol of NaCl reacts with 1.0 mol of Pb(NO3)2 hence 0.016 mol of NaCl will react with 0.016 mol / 2.0 = 0.008 mol of Pb(NO3)2.

Number of moles of Pb(NO3)2 remaining = ( 0.01124 - 0.008) mol = 0.00324 mol

Total volume of solution = ( 0.0441 + 0.0385) L = 0.0826 L

In aqueous solution, [Pb^2+] = [Pb(NO3)2] = 0.00324 mol / 0.0826 L = 0.039 mol/L = 0.039 M

The Concentration of Pb^2+ ion in the solution after reaction = 0.039 M