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
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. .
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.
Why did the math book go to the therapist? Because it had too many problems.
To determine the number of moles of PbC2O4(s) precipitate, we need to use stoichiometry. The balanced equation for the reaction between Pb(NO3)2(aq) and Na2C2O4(aq) is:
Pb(NO3)2(aq) + Na2C2O4(aq) -> PbC2O4(s) + 2NaNO3(aq)
From the balanced equation, we can see that one mole of Pb(NO3)2 reacts with one mole of Na2C2O4 to produce one mole of PbC2O4. Therefore, the number of moles of PbC2O4(s) precipitate is equal to the number of moles of Pb(NO3)2 used.
To find the number of moles of Pb(NO3)2 used, we can use the formula:
moles = concentration × volume (in L)
For Pb(NO3)2:
moles of Pb(NO3)2 = concentration of Pb(NO3)2 × volume of Pb(NO3)2(aq)
= 2.82 M × 0.0350 L
= 0.0987 moles Pb(NO3)2
Now, let's move on to the second part of the question. To find the values of [Pb2+], [C2O4^2-], [NO3^-], and [Na+], we need to take into account the stoichiometry of the reaction and the concept of solubility product (Ksp).
Since we have used all of the Pb(NO3)2 to form PbC2O4(s), the concentration of Pb2+ ions after precipitation is 0 M.
As for the C2O4^2- ions from Na2C2O4(aq), we need to determine how much of it is left in solution. To do that, we need to calculate the initial moles of C2O4^2- and see if it forms a precipitate.
moles of C2O4^2- = concentration of C2O4^2- × volume of Na2C2O4(aq)
= 0.00221 M × 0.0200 L
= 4.42 × 10^-5 moles C2O4^2-
Since the Ksp for PbC2O4(s) is given as 8.5 × 10^-9, we compare it to the initial concentration of C2O4^2- to see if a precipitate forms. If the initial concentration is higher than the Ksp, a precipitate will form.
Initial concentration of C2O4^2- = 4.42 × 10^-5 moles C2O4^2- / (0.0350 L + 0.0200 L) = 7.5 × 10^-4 M
Since the initial concentration of C2O4^2- is higher than the Ksp, a precipitate of PbC2O4(s) forms, and the concentration of C2O4^2- after precipitation is 0 M.
For Na+ and NO3^- ions, since they are spectator ions and do not contribute to the formation of the precipitate, their concentrations remain unchanged: [Na+] = 0.00221 M and [NO3^-] = 2 × 0.00221 M = 0.00442 M.
To sum up, after the solution reaches equilibrium at 25 °C, the values of the concentrations are:
[Pb2+] = 0 M
[C2O4^2-] = 0 M
[NO3^-] = 0.00442 M
[Na+] = 0.00221 M (Remain the same throughout the reaction)
I hope this answers your question and adds a dose of humor to your chemistry journey!
To determine the number of moles of PbC2O4(s) precipitate from the resulting solution, we need to use stoichiometry. Here's the step-by-step process:
Step 1: Calculate the moles of Pb(NO3)2(aq) and Na2C2O4(aq) used:
Moles of Pb(NO3)2(aq) = volume (L) x concentration (M)
Moles of Pb(NO3)2(aq) = 0.035 L x 2.82 M = 0.0987 moles
Moles of Na2C2O4(aq) = volume (L) x concentration (M)
Moles of Na2C2O4(aq) = 0.02 L x 0.00221 M = 4.42 x 10^-5 moles
Step 2: Use stoichiometry to determine the moles of PbC2O4(s) precipitate formed:
From the balanced equation:
Pb(NO3)2(aq) + Na2C2O4(aq) -> PbC2O4(s) + 2NaNO3(aq)
The stoichiometry shows that 1 mole of Pb(NO3)2 reacts with 1 mole of Na2C2O4 to form 1 mole of PbC2O4.
Therefore, the moles of PbC2O4(s) precipitate formed would also be 0.0987 moles.
Step 3: Calculate the concentrations of Pb2+, C2O42–, NO3–, and Na+ in the solution at equilibrium:
Since PbC2O4(s) is an ionic solid, it will dissociate into Pb2+ and C2O42– ions.
[Pb2+] = [C2O42–] = (moles of PbC2O4(s) / volume of solution)
[Pb2+] = [C2O42–] = (0.0987 moles / (0.035 L + 0.02 L))
[Pb2+] = [C2O42–] = 1.47 M
Since Na2C2O4 dissociates into 2 Na+ ions and 1 C2O42– ion:
[Na+] = 2 x (moles of Na2C2O4(aq) / volume of solution)
[Na+] = 2 x (4.42 x 10^-5 moles / (0.035 L + 0.02 L))
[Na+] = 2.59 x 10^-3 M
[NO3–] remains unchanged, as it does not participate in the precipitation reaction.
Therefore, the values of [Pb2+], [C2O42–], [NO3–], and [Na+] after the solution reaches equilibrium at 25 °C are:
[Pb2+] = [C2O42–] = 1.47 M
[NO3–] remains unchanged
[Na+] = 2.59 x 10^-3 M
To determine the number of moles of PbC2O4(s) precipitated, you need to first determine the limiting reactant.
1. Convert the volume of the Pb(NO3)2 solution to moles:
- 35.0 mL = 0.0350 L
- Moles of Pb(NO3)2 = volume (L) x concentration (M) = 0.0350 L x 2.82 M = 0.0987 moles
2. Convert the volume of the Na2C2O4 solution to moles:
- 20.0 mL = 0.0200 L
- Moles of Na2C2O4 = volume (L) x concentration (M) = 0.0200 L x 0.00221 M = 0.0000442 moles
3. Calculate the stoichiometric ratio between Pb(NO3)2 and PbC2O4:
- Balanced equation: Pb(NO3)2 + Na2C2O4 -> PbC2O4(s) + 2NaNO3
- From the balanced equation, the molar ratio between Pb(NO3)2 and PbC2O4 is 1:1.
4. Compare the number of moles of each reactant to determine the limiting reactant:
- The moles of Pb(NO3)2 are 0.0987 moles, and the moles of Na2C2O4 are 0.0000442 moles.
- Since the mole ratio between Pb(NO3)2 and PbC2O4 is 1:1, Pb(NO3)2 is the limiting reactant as it has fewer moles.
5. Determine the moles of PbC2O4(s) precipitated:
- Since Pb(NO3)2 is the limiting reactant, the moles of PbC2O4(s) formed will be equal to the moles of Pb(NO3)2 used.
- Therefore, the number of moles of PbC2O4(s) precipitated is 0.0987 moles.
To determine the values of [Pb2+], [C2O4^2-], [NO3^-], and [Na+] after the solution reaches equilibrium at 25 °C, you need to consider the solubility product constant (Ksp) of PbC2O4.
The Ksp expression for PbC2O4(s) is:
Ksp = [Pb2+][C2O4^2-]
Given:
Ksp(PbC2O4(s)) = 8.5 x 10^-9
At equilibrium, the concentrations of [Pb2+] and [C2O4^2-] will be equal to each other (x) since the molar ratio between them is 1:1.
Therefore:
Ksp = [Pb2+][C2O4^2-]
8.5 x 10^-9 = (x)(x)
x = sqrt(8.5 x 10^-9)
x ≈ 9.22 x 10^-5 M
So, after reaching equilibrium at 25°C:
[Pb2+] ≈ [C2O4^2-] ≈ 9.22 x 10^-5 M
[NO3^-] will be equal to the initial concentration of 2.82 M for Pb(NO3)2
[Na+] will be equal to the initial concentration of 0.00221 M for Na2C2O4