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 0.05600 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 K_sp of PbX₂.

Answer 1

with help of nernst equation

cell equation a is

anode : Pb(A) = Pb⁺²_(a) + 2e^- E⁰_{(OXI) =} 0.13v

Cathode : Pb_(B)⁺² + 2e^- = Pb(B) E⁰_{(RED)= -0.13v}

E^{0 =} E⁰_{(OXI) +}E⁰_(RED)

= 0.13-0.13

= 0

now from nernst equation

E=  E⁰-0.0591/n log[Pb(A)⁺²/Pb(B)⁺²]

0.067=-0.0591  log[Pb(A)⁺²/Pb(B)⁺²]

0.0054=Pb(A)⁺²/Pb(B)⁺²

Pb(A)⁺²=0.0054(0.05M)

Pb(A)⁺²=0.00027M --------------------------ANS (1)

Pb⁺² + 2X^- = PbX2

When we prepared cell A

Dissolved Pb⁺² is

(0.00100L)(0.0500M Pb(NO₃)₂=5*10^-5 moles of  Pb⁺² dissolved

when  Pb(NO₃)_{2 is dissolved in water we findout}

Pb(A)⁺² = 0.00027 As per Ans 1

so at equilibrium poimt moles of Pb^{+2 in solution is}

=0.00027M(0.00500L)-1.35*10^-6

4.865*10^-5 moles of PbX₂ precipitate

but moles of x is double than pb so

9.73*10^-5 moles reacted-----------------------ans(2)

From the second calculation moles of x=9.73*10^-5 moles

now prepare cell A

=(0.00400L)(0.0500M KX)

=2.00 * 10^{-4 moles}

so 2.00 * 10^-4 -9.73*10^-5

=1.027*10^-4 moles available in solution

so now concentration of solution

Cell A =1.027*10 / 0.00500L

=0.02054 M ------------------ans (3)

PbX_{2 = Pb}^{+2 +} 2x^-

dissociation constant

KSp= [_Pb⁺²][x^-]²

   ⁼0.00027M [0.002054M]²

   = 5.5 * 10^{-6 ----------------------ans (4)}