Question

A student mixes 39.0 mL of 2.88 M Pb(NO3)2(aq) with 20.0 mL of 0.00197 M Na2C2O4(aq). How many moles of PbC2O4(s) precipitate from the resulting solution? What are the values of [Pb2 ], [C2O42–], [NO3–], and [Na ] after the solution has reached equilibrium at 25 °C?

Answer 1

Pb (NO3)2 --- > Pb +2 + 2 NO3-

Here 39.0 ml = 0.0390 L

Number of moles of Pb2+ :

0.0390 liters x 2.88 moles Pb+2/1 liter = 0.11232 moles Pb+2

Number of moles of C2O4-2
0.0200 liters x 0.00197moles I-/1 liter = 3.94 x 10^-5 moles C2O4-2


Pb+2 + C2O4-2 ===⇒ Pb C2O4

Limiting reagent is the C2O4-2

3.94x 10^-5 moles C2O4-2x 1mole PbI2/1 moles C2O4-2

= 3.94 x 10^-5 moles of Pb C2O4 precipitated

Final volume of solution = 0.059 liters ; (39ml+20 ml)

Moles of Pb+2 remaining in solution = 0.11232 – 3.94 x 10^-5

=0.1123 moles of Pb+2 remain

Molarity = number of moles / volumes in L
[Pb+2] = 0.1123/0.059 liters = 1.90 M

Ksp = [Pb+2][ C2O4-2] =8.5 x 10^-9

[1.90][ C2O4-2]= 8.5 x 10^-9
[C2O4-2] = 4.47 x 10^-9 M


moles NO3^-1

= 0.039 liters x 2.88 moles Pb(NO3)2/1 liter x 2 moles NO3^-1/1 mole Pb(NO3)2 = 0.22464 moles NO3^-1

[NO3^-1] = 0.22464moles/0.059 liters = 3.81M

Number of moles of Na+
0.0200 liters x 0.00197moles I-/1 liter *2

= 7.88x 10^-5 moles Na+

[Na+] = 7.88x 10^-5 moles Na+/0.059 liters = 1.34*10^3 M