1/25/2018 12:00 AM A 60/100 Gradebook Print Calculator Periodic Table Question 2

Question

1/25/2018 12:00 AM A 60/100 Gradebook Print Calculator Periodic Table Question 2 of 5 Map General Chemistry 4th Edition University Science Books presented by Sapling Leaming A student mixes 47.0 mL of 2.56 M Pb(NO3)2(aq) with 20.0 mL of 0.00237 M Na2C204(aq) How many moles of PbC204(s) precipitate from the resulting solution? Number mol What are the values of [Pb, [C202], [NO3], and [Na] after the solution has reached equilibrium at 25 °C? Number Number Number Number Previous Give Up & View Solution e Check Answer Next Exit Hint

Explanation / Answer

The balanced reaction

Pb(NO3)2 + Na2C2O4 = PbC2O4 + 2NaNO3

Initial Moles of Pb(NO3)2 = molarity x volume

= 2.56 mol/L x 0.047 L

= 0.12032 mol

Initial Moles of Na2C2O4 = molarity x volume

= 0.00237 mol/L x 0.020 L

= 0.0000474 mol

Ionic reaction
Pb+2 + C2O4 -2 = PbC2O4 (s)

We have

Moles of Pb+2 > moles of C2O4 -2

Limiting reagent = C2O4 -2

From the stoichiometry of the reaction

Moles of PbC2O4 precipitate = moles of Na2C2O4 consumed

= 0.0000474 mol

Total volume of solution = 47 + 20 = 67 mL = 0.067 L

Moles of Pb+2 remained = initial Moles - reacted moles

= 0.12032 - 0.0000474

= 0.1202726 mol

Equilibrium Concentration of [Pb+2] = 0.1202726 mol/0.067 L

= 1.7951134 M

From the equilibrium solubility product constant

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

1.7951134 x [C2O4 -2] = 8.5 x 10^-9

Equilibrium concentration of [C2O4 -2] = 4.735 x 10^-9 M

moles of Na+ =

= (0.0000474 mol C2O4 -2) x (2 Mol Na+/1 mol C2O4 -2)

= 0.0000948 moles Na+

Equilibrium concentration of [Na+] = 0.0000948 moles/0.067 L

= 0.0014149 M

Moles of NO3-

= 0.12032 mol x (2 moles NO3-/1 mole Pb(NO3)2)

= 0.24064 mol

Equilibrium concentration of

[NO3^-1] = 0.24064 moles/0.067 L

= 3.591 M

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