How do I find the moles and ksp?

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 0.06700 V at 25 °C.

By use of the Nernst equation, determine the concentration (M) of Pb2+ in the solution of Cell A.

In Cell A, how many moles of X- have reacted with Pb2+?

What is the concentration (M) of X- in the solution of cell A.

Calculate Ksp of PbX2.
please explain :) I will give 10 points!

    1 day ago
    - 3 days left to answer.

Additional Details
okay after ltrying this out for about 10 hours I finally figured out the first part!!! my concentration o Pb2+ is 0.000592M.
However, I am still stuck on the other questions.
for question 2 I thought I would just multiply the original moles times volume and then minus the concentration I found by the volume.
(0.05M*.001L)-(.000592 M*.005L) = 0.00005 mol - 0.00000296 mol = 0.00004704 mol but that is the wrong answer.

Anonymous · Accepted Answer

By use of the Nernst equation, determine the concentration (M) of Pb2+ in the solution of Cell A.

Anode	    Pb(A) (s) = Pb(A) ++ + 2e–	E°ox = 0.13 V (from table)

Cathode	    Pb(B)++ + 2e– = Pb(B) (s) 	E°red =  - 0.13 V 

E°tot =  0

		
Nernst Eqn.: To calculate [Pb(A) ++]
E° = 0 for a Concentration Cell as shown above
E = E° - (0.0591/n) log([Pb(A) ++]/[Pb(B)++])
0.06700 = 0 - (0.0591/2) log([Pb(A) ++]/[Pb(B)++]) 
2(0.067) / - 0.0591 = - 2.267 = log([Pb(A) ++]/[Pb(B)++]) 
0.00540 = [Pb(A) ++]/[Pb(B)++]
[Pb(A) ++] = 0.00540 (0.05 M)
[Pb(A) ++] = 0.00027 M 

In Cell A, how many moles of X- have reacted with Pb2+ ?
Pb++ + 2X–   = PbX2(s)
When Cell A was being prepared (0.00100 L)(0.0500 M Pb(NO3)2) = 5.00 x 10^(-5) moles of Pb++ dissolved, as Pb(NO3)2 is completely soluble in water.
Since the above calculation shows that [Pb(A) ++] = 0.00027 M, then at equilibrium (before the hookup to Cell B) there are (0.00027 M)(0.00500 L) = 1.35 x 10^(-6) moles of Pb++ still in solution.
So 5.00 x 10^(-5) - 1.35 x 10^(-6) = 4.865 x 10^(-5) moles of  Pb++  have precipitated out as PbX2.
Since twice as many moles of X- are tied up in PbX2 as moles of Pb++, then 9.73 x 10^(-5) moles of X- have reacted with Pb++. 

What is the concentration (M) of X- in the solution of cell A ?
From the last calculation, 9.73 x 10^(-5) moles of X- have reacted with Pb++ to form PbX2.
When Cell A was being prepared (0.00400 L)(0.0500 M KX) = 2.00 x 10^(-4) moles of X- dissolved, as KX is completely soluble in water.
So 2.00 x 10^(-4) – 9.73 x 10^(-5) = 1.027 x 10^(-4) moles of  X- are still in solution.
So the concentration (M) of X- in the solution of cell A = (1.027 x 10^(-4) moles)/(0.00500 L) = 0.02054 M

Calculate Ksp of PbX2(s):
PbX2(s)  = Pb++ + 2X–
Ksp = [Pb++] [X-]^2 = (0.00027 M)( 0.02054 M)^2 = 5.5 x 10^(-6)

? · Answer

The benzene-toluene question, or others very like it, have been asked before. Search the archives. The first question doesn't make any sense. Does all the solid dissolve? If so, you probably don't have a saturated solution, so you certainly can't calculate Ksp. If the 2g is excess, so that you do have a saturated solution, use the depression of freezing point to find out the concentration of solute. (At these low concentrations, you don't need to worry about the difference between molarity and molality.) Half of this solute is M+ and the other half is X-, and, by definition, Ksp = [M+][X-]. This of course only tells you Ksp at the freezing point.

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How do I find the moles and ksp?

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