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 77.9 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 3.451 g. What was the concentration of the original lead(II) nitrate solution?

Answer 1

The balanced equation is :

Pb(NO3)2 + 2NaCl --------------------> PbCl2(s) + 2NaNO3.

The molecular weight of PbCl2 = 278.1 g/mol

moles of PbCl2 has = 3.451 / 278.1

                                = 0.01241

from the balanced equation :

1 mol of Pb(NO3)2 gives ------------- 1 mol of PbCl2

moles of Pb(NO3)2 in 2 mL = 0.01241 moles

moles of Pb(NO3)2 in 77.9 mL = 77.9 x 0.01241 / 2 = 0.4834

when Pb(NO3)2 solution is evaporated only water is evaporated.and the number of moles of Pb(NO3)2 will remain same.

moles of Pb(NO3)2 in 100 mL = moles of Pb(NO3)2 in 77.9 mL

moles of Pb(NO3)2 = 0.4834 moles

volume of original solution = 100 mL = 0.1 L

Molarity = moles / volume = 0.4834 / 0.1 = 4.834

Molarity of Pb(NO3)2 = 4.834 M