Question

The solutions at the two Pb electrodes of a concentration cell were prepared as follows:

Cell A: A mixture of 1.00 mL of 0.0500 M Pb(NO3)2 with 4.00 mL of 0.0500 M KX (the soluble potassium salt of an unspecified monovalent ion X-). Some PbX2(s) precipitates.

Cell B: 5.00 mL of 0.0500 M Pb(NO3)2.

The cell potential was measured to be 0.05100 V at 25 °C.

1.)By use of the Nernst equation, determine the concentration (M) of Pb2+ in the solution of Cell A.

2.)In Cell A, how many moles of X- have reacted with Pb2+?

3.)What is the concentration (M) of X- in the solution of cell A.

4.) Calculate Ksp of PbX2.

Answer 1

A) At Anode :

Pb(A) (s) →Pb2+(A) + 2e–     E°(oxidation) = 0.13 V

At Cathode:

Pb2+(B) + 2e → Pb(B) (s) E°(red uction)= - 0.13 V

E°cell = E red-Eox=0

The Nernst Eqn to calculate [Pb2+(A)]

E° = 0

E = E°(cell) - (0.0591/n) log([Pb2+(A)/[Pb2+(B)])

0.05100= 0 - (0.0591/2) log([Pb2+(A)/[Pb2+(B)])

2(0.051) / - 0.0591 = - 1.726 = log([Pb2+(A)/[Pb2+(B)])

10^(- 1.726)= log([Pb2+(A)/[Pb2+(B)])

0.0188= log([Pb2+(A)/[Pb2+(B)])

[Pb2+(A)] = 0.0188 (0.05 M)

[Pb2+(A)] = 0.00094 M

B) reaction

Pb2+ + 2X– →PbX2(s)

In Cell A

Initial moles of Pb2+= (0.00100 L) *(0.0500 M =5.0 *10^-5 moles , Pb(NO3)2dissolved completely

Equilibrium concentration of [Pb2+] = 0.00027 M

Moles of pb2+ (0.00027 M)*(0.00500 L) = 1.35* 10^(-6) moles =0.135*10^-5

Amount of Pb2+ precipitated as PbX2=5.0 *10^-5 moles-1.35* 10^(-6) moles =4.86*10^-5

Moles of X-=2 *moles of pb2+ in pbx2=2*4.86*10^-5=9.73*10^-5

C) Moles of X-=2 *moles of pb2+ in pbx2=2*4.86*10^-5=9.73*10^-5 moles

But initial conc of X-(KX)=0.004L *0.05 M=2.0*10^-4 moles

Moles of X- still in solution= 2.0*10^-4 moles-9.73*10^-5 moles=1.027*10^-4 moles

[X-]=moles/total volume of solution=1.027*10^-4 moles/0.005L=0.02054M

D)

PbX2→ Pb2+ + 2X-

Ksp = [Pb2+] [X-]^2 = (0.00027 M)( 0.02054 M)^2 = 5.5* 10^-6