[Big-Daddy]

H₃AsO₄ and K₄Fe(CN)₆ are dissolved in water in a stoichiometric ratio. What will the [H₃AsO₄]/[H₃AsO₃] ratio be at equilibrium if pH = 2.00 is maintained? Given: Fe(CN)6]3–/[Fe(CN)6]4– has Eº₄ = +0.356 V, and H3AsO4 /H3AsO3 has Eº₅ = +0.560 V.

I'd tried asking the same question a while back and got no response so if you doubt my effort you may look at that thread: http://www.chemicalforums.com/index.php?topic=70105.msg252505#msg252505

Recently I looked at the solutions and saw a formula which, since the last time I attempted this problem, I have also seen elsewhere. It was the formula for E_eq here: http://www.titrations.info/potentiometric-titration-equivalence-point-calculation. However I don't consider the problem solved because I can't see why that formula is applicable to this situation or how to use it. And in any case I would like to understand the solution from first principles, and surely the problem did not require this formula to be known from beforehand.

[Borek]

The formula is derived and explained on the titration.info page, it doesn't appear from nowhere.

It applies here as you were told stoichiometric amounts were mixed. That's equivalent to reaching equivalence point (pun not intended).

[Big-Daddy]

Ok, maybe if I show a method I tried recently you can tell me where I'm going wrong. I first write out the redox equations for each couple:

H₃AsO₄ + 2H⁺ + 2e^- H₃AsO₃ + H₂O ... Rxn 1

[Fe(CN)₆]^3- + e^-  [Fe(CN)₆]^4- ... Rxn 2

Now the more negative E° reaction is going to be subtracted from the more positive E° reaction. Thus we perform Rxn 1 - 2 * Rxn 2:

H₃AsO₄ + 2H⁺ + 2[Fe(CN)₆]^4-  H₃AsO₃ + H₂O + 2[Fe(CN)₆]^3-

And this is the reaction going on in solution. It does not matter whether we are at equivalence point or not. Now we can find the E° of this reaction as +0.204.

Now maybe the fact comes in, that we're at equivalence point and a stoichiometric amount of the two substances has been added together. Again, this comes down to exactly how to define this "stoichiometric amount": [H₃AsO₄] = 1/2 * [[Fe(CN)₆]^3-] and [H₃AsO₃] = 1/2 * [[Fe(CN)₆]^4-] maybe?

Am I on the right track so far?

[Borek]

Again, this comes down to exactly how to define this "stoichiometric amount": [H₃AsO₄] = 1/2 * [[Fe(CN)₆]^3-]

OK

and [H₃AsO₃] = 1/2 * [[Fe(CN)₆]^4-] maybe?

Why do you include these? You are mixing REACTANTS in the stoichiometric ratio, at this stage products don't matter yet. (Doesn't mean this equation is wrong, it is just not part of the information about INITIAL concentrations).

You know several things:
- initial ratio of concentrations,
- potentials of both half reactions at equilibrium will be identical,
- reaction equation (which defines reaction stoichiometry).

That means several equations - just write them down and solve.

[Big-Daddy]

Is that formula actually correct? Since we are mixing reactants, shouldn't that mean that [H₃AsO₄]_eq = 1/2 * [[Fe(CN)₆]^4–]_eq, since these are the two substances we start with? And in any case, if this is the initial composition of the mixture we create by combining the two substances, how can we say that this equation is necessarily true at equilibrium?

[Borek]

Is that formula actually correct? Since we are mixing reactants, shouldn't that mean that [H₃AsO₄]_eq = 1/2 * [[Fe(CN)₆]^4–]_eq

Sorry, I thought that was what you wrote, my mistake.

[Big-Daddy]

Ok. But is it

[H₃AsO₄]_eq = 1/2 * [[Fe(CN)₆]^4–]_eq

that is true (along with correspondingly [H₃AsO₃]_eq = 1/2 * [[Fe(CN)₆]^3–]_eq), or is it

n₀(H₃AsO₄]) = 1/2 * n₀([Fe(CN)₆]^4–) where n₀(H₃AsO₄])/V = [H₃AsO₄]_eq + [H₃AsO₃]_eq and n₀([Fe(CN)₆]^4–)/V = [[Fe(CN)₆]^4–]_eq + [[Fe(CN)₆]^3–]_eq

Which one?

From the former, I used the calculated value of K for the reaction in post #3 and then expressed K as [H₃AsO₃]_eq³/([H₃AsO₄]_eq³ * [H⁺]²) leading to a value of 0.137 for the [H₃AsO₄]_eq/[H₃AsO_3eq ratio. This is in slight disagreement with the solutions' answer of 0.107 but those solutions are known for being inexact. Is this correct?

[Borek]

Ok. But is it

[H₃AsO₄]_eq = 1/2 * [[Fe(CN)₆]^4–]_eq

You are told about INITIAL concentrations being STOICHIOMETRIC, why do you try to apply it to equilibrium?

It may happen that this equation will hold at equilibrium, but you don't have to know it to begin finding the solution. Instead of trying to describe the system in the most simple and obvious way, you are trying to make it more and more confusing.

[Big-Daddy]

You are told about INITIAL concentrations being STOICHIOMETRIC, why do you try to apply it to equilibrium?

It may happen that this equation will hold at equilibrium, but you don't have to know it to begin finding the solution. Instead of trying to describe the system in the most simple and obvious way, you are trying to make it more and more confusing.

I am just looking for the method that gives the answer in the most straightforward possible way. Yes, you're right, this

n₀(H₃AsO₄) = 1/2 * n₀([Fe(CN)₆]^4–) where n₀(H₃AsO₄])/V = [H₃AsO₄]_eq + [H₃AsO₃]_eq and n₀([Fe(CN)₆]^4–)/V = [[Fe(CN)₆]^4–]_eq + [[Fe(CN)₆]^3–]_eq

seems probably better to me. But let's discuss the method for solving from there. The redox potentials of the two half-reactions are equal at equilibrium, thus I can set E (H₃AsO₄ Couple) = E ([Fe(CN)₆]^4– Couple). The method then is to replace the two cyanide-containing complex ion terms with equivalent expressions for the acid terms, so that we are left no longer with too many variables, and can solve the equation for the ratio. But I think one more equation is needed because if we try to substitute in for one of the iron complex terms (using rearrangement of the "stoichiometric amount" equation I wrote above), we'll end up with an expression that contains the other, leaving us unable with just this info to reduce the equation of all variables except the two acid terms.

[Big-Daddy]

Will it help me get a response to rephrase my comment?

My two equations are

[tex]c_{H3AsO4,eq} + c_{H3AsO3,eq} = \frac{1}{2} \cdot ( c_{Fe^4,eq} + c_{Fe^3,eq} )[/tex]

and

[tex]E°_{H3AsO4} - \frac{RT}{2F} \cdot ln(\frac{c_{H3AsO3,eq}}{c_{H3AsO4,eq}}) = E°_{Fe^4} - \frac{RT}{F} \cdot ln(\frac{c_{Fe^4,eq}}{c_{Fe^3,eq}})[/tex]

Where I have abbreviated the complex ion forms with their overall charges.

I don't see how a solution (requiring the Fe⁴ and Fe³ concentration terms to be eliminated) follows directly from this?