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
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To determine the number of moles of PbC2O4(s) that precipitate, we need to calculate the number of moles of Pb(NO3)2(aq) and Na2C2O4(aq) that react with each other.
First, we will calculate the moles of Pb(NO3)2(aq) using the given values:
Molarity of Pb(NO3)2(aq) = 2.82 M
Volume of Pb(NO3)2(aq) = 35.0 mL = 0.0350 L
Moles of Pb(NO3)2(aq) = Molarity x Volume
Moles of Pb(NO3)2(aq) = 2.82 M x 0.0350 L = 0.0987 moles
Next, we will calculate the moles of Na2C2O4(aq):
Molarity of Na2C2O4(aq) = 0.00221 M
Volume of Na2C2O4(aq) = 20.0 mL = 0.0200 L
Moles of Na2C2O4(aq) = Molarity x Volume
Moles of Na2C2O4(aq) = 0.00221 M x 0.0200 L = 4.42 x 10^-5 moles
The reaction between Pb(NO3)2 and Na2C2O4 is:
Pb(NO3)2(aq) + Na2C2O4(aq) → PbC2O4(s) + 2NaNO3(aq)
We can see from the balanced equation that the stoichiometric ratio between Pb(NO3)2 and PbC2O4 is 1:1. Therefore, the moles of PbC2O4(s) precipitated will be equal to the moles of Pb(NO3)2 used, which is 0.0987 moles.
To calculate the concentration of each species after equilibrium, we need to consider the volume of the resulting solution. Since we mixed 35.0 mL of Pb(NO3)2(aq) with 20.0 mL of Na2C2O4(aq), the total volume is 35.0 mL + 20.0 mL = 55.0 mL = 0.0550 L.
Using the balanced equation, we know that 1 mole of Pb(NO3)2 yields 1 mole of Pb2+, 1 mole of C2O42-, and 2 moles of NO3-. Therefore, the concentrations of each species are as follows:
[Pb2+] = [C2O42-] = 0.0987 moles / 0.0550 L ≈ 1.79 M
[NO3-] = 2 x 0.0987 moles / 0.0550 L ≈ 3.58 M
Since there is no reaction involving Na+ ions from Na2C2O4, the concentration of Na+ remains unchanged and is determined by the initial concentration:
[Na+] = 0.00221 M
Therefore, at equilibrium, the values of [Pb2+], [C2O42-], [NO3-], and [Na+] are approximately 1.79 M, 1.79 M, 3.58 M, and 0.00221 M, respectively.