A student dissolves 125.0 g of lead(II) nitrate in enough water to make 250.0 mL of solution. What is the concentration of lead ions in solution in Eq/L?

Question

Make sure you have accounted for the charge on a lead(II) ion.

Answer 1

Lead nitrate = Pb(NO3)2

Pb(NO3)2(aq) ==> Pb2+(aq) + 2NO3-(aq) ... dissolution of lead nitrate in water

molar mass Pb(NO3)2 = 331.2 g / mole

moles Pb(NO3)2 used = 125.0 g x 1 mol / 331.2 g = 0.37742 moles

Volume of solution in liters = 250.0 ml x 1 L / 1000 ml = 0.2500 L

[Pb(NO3)2] = 0.37742 mols / 0.2500 L = 1.5097 moles/liter

Since Pb2+(aq) has 2 equivalents per mole, the concentration in Eq/L is...

1.5097 mols/L x 2 equivalents / mol = 3.019 Eq/L

P.S. Eq/L is an archaic unit of measurement and isn't used very much today.