[stmoney]

I recently went through the majority of the olympiad questions and wasn't really stumped or surprised by any of the answers, but when I looked at the most recent math-based questions I felt the writers really increased the difficulty. Just curious someone could give feed back on to how to solve a few problems.

http://www.acs.org/content/dam/acsorg/education/students/highschool/olympiad/pastexams/2015-usnco-local-exam.pdf

35. Copper (II) hydroxide, Cu(OH)₂, has a K_sp = 2.2x10^-20. For the reaction below, K_eq = 4.0x10^-7. What is K_f for Cu(NH₃)₄²⁺?
Cu(OH)₂(s) + 4NH₃(aq)  Cu(NH₃)₄²⁺(aq) + 2OH^-(aq)

The answer is 1.8x10¹³. I tired starting with Ksp and finding the OH- concentration, then tried to use Keq = [Cu(NH₃)₄²⁺][OH^-]² / [NH₃]⁴  but this didn't lead to anything. I tried to flip my products and reactants for Keq, but this also didn't lead me to the correct answer. I'm missing a step some where.
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The second one was question 41, which asks...

The reduction of O₂ to H₂O in acidic solution has a standard reduction potential of +1.23 V. What is the effect on the half-cell potential at 25ºC when the pH of the solution is increased by one unit?

O₂(g) + 4H⁺(aq) + e^-  2H₂O(l)

The answer is A. the half-cell potential decreases by 59 mV

I thought this one was going to be pretty easy by using the Nerst equation. Since the pH is increasing by a unit I set K to [1]/[10] being that as the pH goes up the concentration of H⁺ would be decreasing by a factor of 10. My equation was E = 1.23 - (0.0592/4 e^-) log(1/10). Nope, that didn't work. Tried to play with numbers and did log(10/1). I knew it wouldn't work, but I tried it anyway and of course it didn't work.

Any suggestions on how to fix, or is there someone that has posted work on how to solve these two problems? Thanks!

[Vidya]

Q35 is not a difficult math ...
It is based on simple concepts of addition of two reactions ..when you add two reactions to get the net reactions than their equilibrium constants get multiplied...look at this example
If I need K for the equation
A--->C
and given equations are
B---> A K1
B---->C K2
To get the desired reaction
you need to flip equation 1
A--->B   1/K1
B--->C   K2
Net reaction A--->C
K = 1/K1 * K2
In your question two reactions are there ...one is given and other is ksp equation
see how can you add these equations to get the desired reaction which is

Cu2+ (aq) 4NH3(aq)<---> [Cu(NH3)4]^2+
If if you do it rightly you will get the answer

[Vidya]

in electrochemistry problem
you are right that we will use Nernst equation ...but the conc of H+ changes by 1/10 so it will become log 10 and if rest of the things are same than it reduces by -0.0591log10 = 0.0591 V = 0.59mV
You need to calculate the change ...

[mjc123]

Well, your equation doesn't balance, which doesn't help for a start. There should be 4 electrons on the LHS, as well as 4 H⁺. Assuming constant standard pressure of oxygen, the Nernst equation will be
E = E° + 0.059/4*log([H⁺]⁴) = E° + 0.059/4*4log[H⁺] = E° + 0.059 log[H⁺]

[CalcCrunch]

What is Kf? And how do we use the given Ksp?

[Vidya]

What is Kf? And how do we use the given Ksp?

Do you know how to write ksp expression and k eq expressions...
so write ksp for Cu (OH)2 ,keq for the reaction

http://www.acs.org/content/dam/acsorg/education/students/highschool/olympiad/pastexams/2015-usnco-local-exam.pdf

35. Copper (II) hydroxide, Cu(OH)₂, has a K_sp = 2.2x10^-20. For the reaction below, K_eq = 4.0x10^-7. What is K_f for Cu(NH₃)₄²⁺?
Cu(OH)₂(s) + 4NH₃(aq)  Cu(NH₃)₄²⁺(aq) + 2OH^-(aq)

write kf expression ..
if you reverse equation of ksp and add it to keq equation  you will get expression of kf or its equation..