You made 100.0 mL of a lead(II) nitrate solution for lab but forgot to cap it. The next lab session you noticed that there was only 80.7 mL left (the rest had evaporated). In addition, you forgot the initial concentration of the solution. You decide to take 2.00 mL of the solution and add an excess of a concentrated sodium chloride solution. You obtain a solid with a mass of 3.264 g. What was the concentration of the original lead(II) nitrate solution?

Question

Steve · Answer

If the reaction is

Pb(NO3)2 + 2NaCl = PbCl2 + 2NaNO3

then each mole of PbCl2 uses one mole of Pb(NO3)2.

So, with 3.264/278.11 = 0.0117 moles of PbCl2 produced, the 2.00mL was .0117/.002 = 5.868M

5.868*80.7/100 = 4.735M original concentration