Question

A student mixes 33.0 mL of 2.54 M Pb(NO3)2(aq) with 20.0 mL of 0.00187 M Na2C2O4(aq). How many moles of PbC2O4(s) precipitate from the resulting solution?

Ksp[PBC2O4(s)]=8.5*10^-9

What are the values of [Pb2 ], [C2O42–], [NO3–], and [Na ] after the solution has reached equilibrium at 25 °C?

Answer 1

a)

mmol of Na2C2O4 = MV = 20*0.00187 = 0.0374 mmol

mmol of  Pb(NO3)2= MV = 2.54*33 = 83.82 mmol

Clearly, C2O4-2 will be limiting reaction

mmol of PbC2O4 = mmol of C2O4-2 = 0.0374 mmol

moles of precpitate = 0.0374 *10^-3 mol of PbC2O4 will form

b)

Vtotal = 33+20 = 53 mL

[Na+] = mmol of Na+ / Vtotal = 2*0.0374 /53 = 0.00141 M

[NO3-] = mmol of NO3- / Vtotal = 2*2.54*33/53 = 3.16301 M

[Pb+2] =mmol of PB+2 left / vtotal =  (83.82 -0.0374 ) / (53) = 1.580 M

finally, C2O4-2 must be calculated form equilbilrium

Ksp = [PB+2][C2O4-2]

8.5*10^-9 = (1.580 )([C2O4-2])

[C2O4-2] = (8.5*10^-9) / (1.580 ) = 5.379*10^-9 M