Pb(NO3)2 + Na2X2O4 ==>Pb(C2O4 + 2NaNO3
millimols Pb(NO3)2 = 35 x 2.82 = approx 98.7
mmols Na2C2O4 = 20 x 0.00221 = approx 0.0442
So 0.0442 mmols PbC2O4 will be formed.
(concn) = millimols/mL solution.
(Pb^2+) = essentially 98.6/55 = approx 1.8M
(NO3^-) = (2*98.7/55) = ?
(Na^+) = 2*0.0442/55( = ?
The (C2O4^2-) will be determined by the Ksp for the salt because the Pb^2+, acting as a common ion, decreases the solubility of the PbC2O4.
.......PbC2O4 ==> Pb^2+ + C2O4^2-
I......solid.......0.......0
C......solid.......x.......x
E......solid.......x.......x
(Pb^2+) = about 98.6/55 = about 1.79
Total (Pb^2+) = 1.79 + x
Ksp = (Pb^2+)(C2O4^2-) = 8.5E-9
(1.79+x)(x) = 8.5E-9
Assume 1.79+x = 1.79, then
1.79x = 8.5E-9
(x) = (C2O4^2-) = 8.5E-9 = about 4.7E-9M which is considerably less than what you might have guessed initially of 0.0442/55 = about 8E-4M
Check my work carefully. .
A student mixes 35.0 mL of 2.82 M Pb(NO3)2(aq) with 20.0 mL of 0.00221 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(PbC2O4(s)) = 8.5*10^-9
2 answers
After submitting the answer, this is what I received as feedback. The system is marking that the amount of moles of precipitate was incorrect, although when I did the problem originally I got the same number that you did. I am just posting this feedback for future reference, you DO NOT need to answer. BTW, THANK YOU SOOOOOOO MUCH!!!!
After mixing, find the new concentrations of the two ions that will combine to form PbC2O4(s). Then, using the relevant reaction equation, construct a Ksp expression. If we consider that the concentration of the cation is over 1000 times larger than the anion, it should be obvious that [C2O42–] can be considered negligible when compared to [Pb2 ]. With that in mind, use the equilibirum expression to calculate the final concentration of C2O42–(aq) in solution utilizing the concentration of Pb2 after mixing and the value of Ksp. The moles of C2O42– that have precipitated is the difference of the initial moles before mixing and the final moles after mixing.
Using the mole ratio present in the reaction equation, convert to moles of PbC2O4(s). The concentration of Pb2 is essentially constant throughout the reaction, as are the concentrations of NO3– and Na . The concentration of C2O42– was calculated to determine the moles of precipitate using the equilibrium expression.
After mixing, find the new concentrations of the two ions that will combine to form PbC2O4(s). Then, using the relevant reaction equation, construct a Ksp expression. If we consider that the concentration of the cation is over 1000 times larger than the anion, it should be obvious that [C2O42–] can be considered negligible when compared to [Pb2 ]. With that in mind, use the equilibirum expression to calculate the final concentration of C2O42–(aq) in solution utilizing the concentration of Pb2 after mixing and the value of Ksp. The moles of C2O42– that have precipitated is the difference of the initial moles before mixing and the final moles after mixing.
Using the mole ratio present in the reaction equation, convert to moles of PbC2O4(s). The concentration of Pb2 is essentially constant throughout the reaction, as are the concentrations of NO3– and Na . The concentration of C2O42– was calculated to determine the moles of precipitate using the equilibrium expression.
2 answers