[43256488]

Sorry if you find this post long...I tried my best to comply within the forum rules!  This was a question on my midterm which I didn't understand.  The prof is not releasing the answers so if you can help me out on any part I would appreciate it!

The question:

Imagine that 24.00 mL of a solution of 0.0200F Mg²⁺ is being titrated with 0.0250 M EDTA at a pH = 9.0.  In order to follow the progress of the titration we will use Hg²⁺ as a "reporter" so enough HgY was added to give 1.0*10^-4 F HgY at the start of the titration.  Data (K_HgY = 5.0*10²¹;K_MgY = 6.17 * 10⁸; E⁰ for Hg²⁺ + e^-   Hg _(l) is +0.852V).  A pic of the titration setup is attached.

a) Describe how the titration of the magnesium with EDTA can be monitored using the voltage shown on the voltmeter.  That is, describe why the voltage changes as [Mg²⁺] changes.

b)Write a mass balance equation that would use as a first step in calculating [Y'] prior to the start of the titration. Y' is the symbol for unbound EDTA in all it's forms. Hint: account for all sources of [Y'].  Express your equation in terms of concentration of species and equilibrium constants.  Don't do any calculations.

c) If the titration was done in the opposite fashion, with 0.0200 M Mg²⁺ in the buret and 0.0250 F EDTA and 1.0*10^-4 F HgY in the beaker at the start of the titration, what would the voltage be on the voltmeter?  Assume the reference electrode is a standard hydrogen electrode.

d) Should the value calculated above be expected to be accurate?  Explain your answer.

My attempt:

For a I wrote the voltage changes because adding EDTA reduces the [Hg2+], which will affect the Hg cell.  But I don't know how Mg has anything to do with reducing the [Hg2+].

For b I wrote

[HgY^2-] = 1.0*10^-4

Hg²⁺ + Y⁴⁺   HgY^2- so k_HgY^2- = [HgY^2-]/([Hg²⁺][Y']α6) = 5.0*10²¹. Where α6 is the fraction of EDTA in Y⁴⁺ form.

So [Y']=(1.0*10^-4)/{[Hg²⁺](5.0*10²¹)(α6)}.  I think I have to add the other forms of Y', but I can't see how.

For c I wrote

E = E⁰ -(0.05916/2)log(1/[Hg²⁺)

[HgY]/([Hg²⁺][Y']α6) = 5.0*10²¹
[Hg²⁺] = [HgY]/((5.0*10²¹)[Y']α6)
α6 = 5.21*10^-2 (from table), Y' = 0.0250, [HgY]=1.0*10^-4, E⁰ = 0.852V
so [Hg²⁺] = 1.54*10^-23 M
E = 0.852 -(0.05916/2)log(1/(1.54*10^-23 M)) = 0.177V

For d I wrote

I don't think the value given for c should be accurate as the electrode has to detect such a small concentration of Hg²⁺ that it is likely smaller than it's detection limit.  As well, the equation does not take into account activity coefficients, which further reduces it's accuracy.

[Borek]

Mg does react with EDTA, lowering its concentration, doesn't it?

[43256488]

That's true, but how does that affect the voltage?  The electrode is based on the concentration of Hg, so that must be changing.

[Borek]

The electrode is based on the concentration of Hg, so that must be changing.

Very good point, but the answer should be obvious. Write formula for HgY stability constant. What happens when [Y] changes?

[43256488]

Ok, I think I have it.  As you add EDTA into the solution, it will react with Mg²⁺ to form MgY.  This species is in equilibrium with the free form:

MgY  Mg²⁺ + Y^4-

This equilibrium is the only source of Y' because the formation constant of HgY is very large. So [Mg]=[Y']. 

Adding EDTA decreases [Y'] as MgY is produced, which increases [Hg2+] as HgY dissociates.  Right?

[Borek]

Yep.

[43256488]

How would you go about finding [Y'] before the titration begins?

[Borek]

Initially there is only one source of Y - dissociating HgY.

[thecookie]

Hi, I'm in the same class as the original poster. I'm very confused. For part b), you say that the only source of Y' is the HgY and I agree, but there have been suggestions that some of the Y that dissociates from HgY2- goes to complex with Mg2+ and hence there's a source of Y' from MgY2- as well.

I am also confused by the fact that apparently (this coming from the prof) there are two ways of writing the concentration of Y' - initial concentration or equilibrium. If we're talking about initial, there's still SOME equilibrium in that initial - you will never have a solution of pure HgY2-, no matter how hard you try...

Any hints?
Thanks!

[Borek]

Good point, I forgot about Mg²⁺, sorry about that. However, BEFORE titration there is only one source of Y - and that's HgY. Mass balance must account for all three Y containing compounds present (two complexes and free ion), but their sum will equal initial concentration of HgY.

[thecookie]

Ok, that makes more sense now, thank you!

As for part c), the answer is correct as written by the original poster? I ask because that's how I'd do it too.

[Borek]

As for part c), the answer is correct as written by the original poster? I ask because that's how I'd do it too.

I just skimmed - approach looks OK, no idea about numbers.

As you already know I am not infallible.