Question

1. a.) A student mixes 47.0 mL of 3.30 M Pb(NO3)2(aq) with 20.0 mL of 0.00183 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? Ksp of PbC2O4 (s) is 8.5e^-9 M^2 b.) What are the values of [Pb^2+], [C2O4^2-], [NO3^-], and [Na^+] after the solution has reached equilibrium?

Answer 1

Volume of Pb(NO₃)₂ = 47 mL =0.047 L

Molarity of Pb(NO₃)₂ = 3.30 M

moles of Pb(NO₃)₂ =3.30*0.047 =0.155 mol (formula for moles = volume * molarity)

Volume of Na₂C₂O₄ = 20 mL=0.020 L

Molarity of Na₂C₂O₄ = 0.00183 M

moles of Na₂C₂O₄ = 0.00183*0.020 L=3,66*10^-5 mol

Pb(NO₃)₂(aq) + Na₂C₂O₄(aq) -------> PbC₂O₄(s) + 2NaNO₃(aq)

Here limiting reactant is Na₂C₂O₄

Moles of PbC₂O₄ formed =

PbC₂O₄ contains Pb²⁺ and C₂O₄^2- ions

PbC₂O₄ <=====> Pb²⁺ + C₂O₄^2--

Concentration of [Pb²⁺] = moles / volume of the solution

Final volume of the solution = 0.047 L + 0.020 L = 0.067 L

moles of Pb2+ remained in the solution = 0.1551 mol -0.0000366 mol = 0.15506 mol

Concentration of Pb²⁺ = 0.15506 mol/0.067L = 4.63 M

moles of NO^3- =

Concentration of [NO₃^-] = 0.3102mol/0.067 L =2.315 M

Concentration of C2O4^2- =

Ksp = [Pb+2][C2O42-]

8.5*10^-9 = [2.31] [C2O42-]

[C2O42-] = 3.68*10^-9 M

Na₂C₂O₄ -----> 2Na⁺ + C₂O₄^2-

moles of Na+ =

Concentration of Na⁺ = 7.32*10^-5/0.067 = 1.09*10^-3M

Concentration of Na⁺=1.09*10^-3M