Question

A student mixes 31.0 mL of 3.06 M Pb(NO3)2(aq) with 20.0 mL of 0.00193 M Na2SO4(aq). How many moles of PbSO4(s) precipitate from the resulting solution? What are the values of [Pb2 ], [SO42–], [NO3–], and [Na ] after the solution has reached equilibrium at 25 °C?

.......mol .....Ksp={PbSo4}=2.5 * 10^-8

part 2:

what are the valuse of [Pb^+2], [SO4^-2], [NO3^+] after the solution has reached equilibrium at 25 C?

Answer 1

Pb(NO3)2 +Na2SO4→PbSO4+ 2 NaNO3

Moles of Pb(NO3)2 in the solution =31/1000L*3.06 M=31/1000 L*3.06 mol/L=0.095 mol

Moles of Na2SO4 in the solution=20 /1000L*0.00193M=0.0000386 moles=3.86*10^-5 moles

As Na2SO4 and Pb(NO3)2 react in 1:1 molar ratio ,so Na2SO4 is the limiting reagent here.

0.0000386 moles moles of both reactants will react to give 0.0000386 moles of PbSO4 in the solution.

1) Given ksp (PbSO4)=2.5*10^-8

Ksp=[Pb2+][SO4(2-)]

PbSO4 dissociates in the solution to give Pb2+ and SO4(2-) ions.If the solubility of PbSO4 is S

[Pb2+]=S, then [SO4(2-)]=S

2.5*10^-8=S*S

S^2=2.5*10^-8

S=1.6*10^-4

So 1.6*10^-4 mol/L is the concentration of PbSO4 in the solution

Total solution volume=31+20=51ml=51/1000

So moles of PbSO4 in the solution=1.6*10^-4 mol/L*51/1000L=81.6*10^-7 moles

Moles precipitated= moles of PbSO4 formed-moles in solution=3.83*10^-5 moles-81.6*10^-7 moles

                                                                                                                 =383 *10^-7-81.6*10^-7 moles

                                                                                                                   =301*10^-7 mol

Moles precipitated=301*10^-7 mol

2) at equilibrium =[Pb2+]=[SO4(2-)]=1.6*10^-4 mol/L

Moles of NaNO3 formed=2*3.86*10^-5 moles=7.72*10^-5 moles

Conc of NaNO3=7.72*10^-5 moles/total volume=7.72*10^-5 moles/51*10^-3 L=0.1513*10^-2 mol/L=0.00151M

It dissociates completely to form [Na+] and [NO3(-)] ions

[Na+] = [NO3(-)]=0.00151M