Chem problem?

Question

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 80.0 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 4.159 g. What was the concentration of the original lead(II) nitrate solution?

Answer 1

Here is the concept. You can do the math.

The solid obtained must be PbCl2. If you obtained 4.159g, calculate how many moles this is.

Now write the balanced equation

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

Do the stoichiometry to calculate how many moles of Pb(NO3)2 you started with. It is 1: 1 ratio, so the moles of PBCl2 = moles of Pb(NO3)2.

Now that you have the moles reacted, they were in 2.00 / 80.0 ml. of solutions. Multiply your moles by 40 and you have the total moles in the original vessel. Since it was originally in 100 ml of solution, multiply your moles by 10 and you will finally have the molarity.

Nice problem!

Answer 2

now its polluted..how ya gonna get an honest reading !