April 30, 2024, 11:28:13 AM
Forum Rules: Read This Before Posting


Topic: 2011 ChemOly Test Qs  (Read 9771 times)

0 Members and 1 Guest are viewing this topic.

Offline XGen

  • Full Member
  • ****
  • Posts: 127
  • Mole Snacks: +9/-4
2011 ChemOly Test Qs
« on: February 12, 2012, 09:10:19 PM »
Note: For many of the following questions not all 4 choices are listed because they are clearly incorrect.


34. Which one of the following mixtures creates a buffer solution?

b. 20.0 mL 0.10 M NaOH, 15.0 mL 0.10 M HF
c. 15.0mL 0.10 M NaOH, 20.0 mL 0.10 M HF

I was thinking that after the neutralization, there would be more NaOH and HF respectively. I wrote the buffer reaction as H+ + F- <-> HF. Is my answer of B incorrect because the protons would go through the neutralization reaction (H+ + OH- = H2O) before the buffer reaction? Would that invalidate it as a buffer solution?


35. The solubility of AgBrO3 in aqueous solution depends on the presence of other substances in solution. Relative to its solubility in H2O the solubility of AgBrO3 is higher in 0.10 M ____ and lower in 0.10 M ____.

a. NH3, KBrO3
b. KBrO3, NH3
c. HNO3, NH3
d. NH3, HNO3

For this question, I didn't understand how to approach it so I followed my gut in thinking that it would have to be based on the polarity of the solvents. The three solvents that are listed, NH3, KBrO3, and HNO3 are all polar, but HNO3 and KBrO3 appear to be more polar. After this, I did not really know where to progress.


41. Consider a voltaic cell in which the reaction below occurs in two half-cells connected by a salt bridge and an external circuit.
2Cr(s) + 3Sn2+(aq) -> 3Sn(s) + 2Cr3+(aq)       Eo = 0.603
Which change will cause the voltage to increase?

c. Diluting the solution in the anode compartment
d. Diluting the solution in the cathode compartment

Using the equilibrium portion of the Nerst equation, there is the expression [Cr3+]^2/[Sn2+]^3. If the denominator were to decrease, the value of the fraction would increase, which is not what is desired. If the numerator were to decrease, the value of the fraction would decrease. Am I correct in my thinking?


22. At the triple point of water how do the entropies of solid, liquid, and gas compare?

a. S solid < S liquid < S gas
d. S solid = S liquid = S gas

The triple point is the point at which the three phases coexist at the same temperature and pressure. Do they exist as different phases in this condition or do they sort of morph together?
« Last Edit: February 13, 2012, 07:06:32 AM by Borek »

Offline Arkcon

  • Retired Staff
  • Sr. Member
  • *
  • Posts: 7367
  • Mole Snacks: +533/-147
Re: 2011 ChemOly Test Qs
« Reply #1 on: February 12, 2012, 09:29:56 PM »
35. The solubility of AgBrO3 in aqueous solution depends on the presence of other substances in solution. Relative to its solubility in H2O the solubility of AgBrO3 is higher in 0.10 M ____ and lower in 0.10 M ____.

a. NH3, KBrO3
b. KBrO3, NH3
c. HNO3, NH3
d. NH3, HNO3

For this question, I didn't understand how to approach it so I followed my gut in thinking that it would have to be based on the polarity of the solvents. The three solvents that are listed, NH3, KBrO3, and HNO3 are all polar, but HNO3 and KBrO3 appear to be more polar. After this, I did not really know where to progress.

No, I don't think you've thought of this one right.  The options aren't the solvent, the solvent is water.  These are co-solutes, or maybe reactants, depending on the choice.  Try to look at this problem from that direction.
Hey, I'm not judging.  I just like to shoot straight.  I'm a man of science.

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: 2011 ChemOly Test Qs
« Reply #2 on: February 12, 2012, 10:18:55 PM »
Would that invalidate it as a buffer solution?
Yes, in the buffer you want a combination of HF and F- so it can buffer against added base and acid respectively.

35. The solubility of AgBrO3 in aqueous solution depends on the presence of other substances in solution. Relative to its solubility in H2O the solubility of AgBrO3 is higher in 0.10 M ____ and lower in 0.10 M ____.
Common ion effect, have you heard of it?

41. Consider a voltaic cell in which the reaction below occurs in two half-cells connected by a salt bridge and an external circuit.
2Cr(s) + 3Sn2+(aq) -> 3Sn(s) + 2Cr3+(aq)       Eo = 0.603

c. Diluting the solution in the anode compartment
d. Diluting the solution in the cathode compartment
I think you posted a portion of the question and forgot the rest?

Do they exist as different phases in this condition or do they sort of morph together?
Nah they don't morph together.

Offline XGen

  • Full Member
  • ****
  • Posts: 127
  • Mole Snacks: +9/-4
Re: 2011 ChemOly Test Qs
« Reply #3 on: February 12, 2012, 11:13:51 PM »
I'm sorry, but I can't find the edit button.

The end of that problem should actually have the question:
"Which change will cause the voltage to increase?"

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: 2011 ChemOly Test Qs
« Reply #4 on: February 13, 2012, 02:54:24 PM »
Your thinking with the Nernst equation is correct.

Offline XGen

  • Full Member
  • ****
  • Posts: 127
  • Mole Snacks: +9/-4
Re: 2011 ChemOly Test Qs
« Reply #5 on: February 13, 2012, 03:29:53 PM »
The solubility constant for AgBrO3 is represented by the expression:
Ksp = [Ag+][BrO3-].
Therefore, in KBrO3, it would make the concentration of the BrO3- larger, and so the solubility would be decreased as less bromate ions would be required from the solute AgBrO3.

I realize that this alone is enough to get the answer (option A has the only one stating KBrO3 would decrease the solubility) but how would ammonia affect the solubility? Ammonia would make the water more basic, but I feel to see any reaction between AgBrO3 and hydroxide ions that would increase the solubility.

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: 2011 ChemOly Test Qs
« Reply #6 on: February 13, 2012, 03:56:00 PM »
Think about complex formation between Ag+ and NH3 plus reaction of Ag+ with hydroxide ions to form silver oxide. These all remove free Ag+ ions from solution.

Offline XGen

  • Full Member
  • ****
  • Posts: 127
  • Mole Snacks: +9/-4
Re: 2011 ChemOly Test Qs
« Reply #7 on: February 13, 2012, 04:37:33 PM »
Ah, that makes sense.

Another question that I forgot to include, which I will summarize for convenience's sake, is:

What is the gas evolved when dilute nitric acid is combined with silver?

The answer is NO; however, I can not figure out the balanced reaction nor the reasoning for the life of me. Semi-related is, why do combustion reactions form CO in limited oxygen? Why can't the oxygen atoms simply form CO2, but less of it?

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: 2011 ChemOly Test Qs
« Reply #8 on: February 13, 2012, 09:33:08 PM »
I can not figure out the balanced reaction
Do you know how to balance redox reactions? In this case, something like NO3- :rarrow: NO in acidic conditions.

why do combustion reactions form CO in limited oxygen?
I am not too sure, maybe someone more knowledgeable can give a better answer. But I like to think of it as a two step reaction. During combustion, C is first oxidised to CO and then CO is oxidised to CO2. So if there is limited oxygen, only the first step can occur.

Offline XGen

  • Full Member
  • ****
  • Posts: 127
  • Mole Snacks: +9/-4
Re: 2011 ChemOly Test Qs
« Reply #9 on: February 13, 2012, 10:46:32 PM »
The reason I am unsure is the same as with combustion in limited oxygen; in dilute nitric acid, why can't it just be HNO3 + Ag(s) -> H2(g) + Ag+(aq) + NO3-(aq)?

Offline UG

  • Full Member
  • ****
  • Posts: 822
  • Mole Snacks: +134/-15
  • Gender: Male
Re: 2011 ChemOly Test Qs
« Reply #10 on: February 13, 2012, 11:19:50 PM »
Because silver cannot reduce H+ to H2 under standard conditions, have you seen their electrode potentials?

Sponsored Links