Chemistry 3/4 2013 Thread

[SocialRhubarb]

Insight paper asks: Name one method or change, other than altering the primary structure, which can alter an enzyme's catalytic ability.
I answered competitive inhibition, which isn't on the s.d. for chem and feels like more of a bio answer, but the answers Insight gave were 'alterations to temperature or pH'. Aren't alterations to temperature/pH just ways of altering the primary structure?

Most temperature and pH changes will only affect secondary and tertiary structure, not primary structure.

[Dismounted]

Aren't alterations to temperature/pH just ways of altering the primary structure?

Changes in temperature and pH alter the secondary and tertiary structures. The order of the amino acids remain the same (primary structure) due to the stronger covalent bonds, but we break some weaker bonds (e.g. H-bonds) and alter the protein's shape.

[Scooby]

I always thought pH was equivalent to [H⁺] = 10^-pH? 

I've attached their solution btw.

This is from Thushan. The values are different but it's the same principle


You forget that the system re-equilibriates. There is a reversible reaction going on:

2H2O <--> H3O+ + OH-

When the solution is pH = 10, [OH-] = 10^-4 M and [H+] = 10^-10 M, so there is a lot more OH- than H+ originally.

When you dilute the solution, [OH-] initially decreases 100fold to 10^-6 M, and so does [H+] to 10^-12 M.

However, the system has to reach equlibrium, with K = 10^-14 and the reaction quotient at that time Q = 10^-18 M.

Hence, the system will shift to the right (also consistent with Le Chatelier's principle, given we have a dilution and the system favours the reaction with more aqueous particles).

However, since there is 1000000 times more OH- than there is H+, the amount of OH- formed will be negligible compared to its initial diluted concentration 10^-6 M.

Hence, we can say that [H+] would increase from 10^-12 to about 10^-8 M. This is the same as the amount of OH- formed, which would be much smaller than 10^-6 M. So,

Final:
[OH-] = 10^-6 M
[H+] = 10^-8 M

[radl223]

Hello everyone

Not sure if this has been posted before, but can anyone please help me with this mcq?

A chemist prepares 0.10M aqueous solutions of each of the following acids. Which solution has the lowest pH?
A) CH3COOH
B) HNO2
C) HCN
D) HOCl
 
The answer is B and I don't understand how. Since they all have one H+ (except for option A) wouldn't that all yield the same pH?

Thanks in advance

[Kayte]

Hello everyone

Not sure if this has been posted before, but can anyone please help me with this mcq?

A chemist prepares 0.10M aqueous solutions of each of the following acids. Which solution has the lowest pH?
A) CH3COOH
B) HNO2
C) HCN
D) HOCl
 
The answer is B and I don't understand how. Since they all have one H+ (except for option A) wouldn't that all yield the same pH?

Thanks in advance

For this question you have to refer to the data booklet and see which one has the highest Ka value.

[radl223]

Ahh always forgetting what's in the databooklet... Thank you very much

[lzxnl]

A students has 10mL of a solution of KOH which has a pH = 10. The solution is diluted with deionized water to make up a litre solution. The pH of the diluted solution will be?

I used the formula:











However, this is incorrect. If at the start, you had worked with [OH^-] =10^-4, the pH would be calculated at 8. Why the difference?

Thanks

This is from Thushan. The values are different but it's the same principle


You forget that the system re-equilibriates. There is a reversible reaction going on:

2H2O <--> H3O+ + OH-

When the solution is pH = 10, [OH-] = 10^-4 M and [H+] = 10^-10 M, so there is a lot more OH- than H+ originally.

When you dilute the solution, [OH-] initially decreases 100fold to 10^-6 M, and so does [H+] to 10^-12 M.

However, the system has to reach equlibrium, with K = 10^-14 and the reaction quotient at that time Q = 10^-18 M.

Hence, the system will shift to the right (also consistent with Le Chatelier's principle, given we have a dilution and the system favours the reaction with more aqueous particles).

However, since there is 1000000 times more OH- than there is H+, the amount of OH- formed will be negligible compared to its initial diluted concentration 10^-6 M.

Hence, we can say that [H+] would increase from 10^-12 to about 10^-8 M. This is the same as the amount of OH- formed, which would be much smaller than 10^-6 M. So,

Final:
[OH-] = 10^-6 M
[H+] = 10^-8 M

And this is again from my previous reply to Thushan's post.

"Pure" water has [H+]=10^-7 M. Your initial H+ concentration is 10^-10 M, so the "dilution" of H+ isn't a dilution. You're adding a much more concentrated solution of H+. With hydroxide, you initially had 10^-4 M, and you're adding 10^-7 M. That change is small, so you can do the calculations from the hydroxide perspective.

How do you work this out (the attachment)? I get C, since surely even if [HOBr] increased by the amount [OH-] decreases by, the [H+] will still be higher since the change is only partially reversed. This would increase the value of the fraction, I thought.

The answers say D, yet surely adding H+ is going to change the value of Q. Is it just that they've confused equilibrium concentration and the equilibrium constant, or is my logic wrong?


Also, does the C-O-C in a glucose or other sugar count as an ether linkage?

Thanks

By definition, an ether group is exactly that; when you have an oxygen bonded to two carbons.

[Mafioso]

How do I work out MCQ 26 in the VCAA 2013 sample?
Are there answers to the short answer questions anywhere?

[Patches]

Note that the question tells us "...after a long period of time..." - this is telling us the system is at equilibrium.

Since we haven't changed the temperature of the system, the value of K remains the same. Thus, the value of the concentration fraction is the same as before we added the "strong acid" (Q = K). We assume the system was at equilibrium before the acid was added.

Ok, that makes perfect sense - thanks

By definition, an ether group is exactly that; when you have an oxygen bonded to two carbons.

Thanks - the solution was wrong.

[ahat]

And this is again from my previous reply to Thushan's post.

"Pure" water has [H+]=10^-7 M. Your initial H+ concentration is 10^-10 M, so the "dilution" of H+ isn't a dilution. You're adding a much more concentrated solution of H+. With hydroxide, you initially had 10^-4 M, and you're adding 10^-7 M. That change is small, so you can do the calculations from the hydroxide perspective.

By definition, an ether group is exactly that; when you have an oxygen bonded to two carbons.

Thank you Chemistry kings. That makes sense

[Patches]

How do I work out MCQ 26 in the VCAA 2013 sample?
Are there answers to the short answer questions anywhere?

One mole of methane will release 900kJ, converting one mole of liquid water to steam requires 44 kJ.

So 900kJ/mol divided by 44kJ is 20.45 mol.

That means you can convert 20.45 mol of water to steam.

20.45 mol * 18.0 g/mol
=368 g.

And the solutions are here
http://www.cea.asn.au/sites/default/files/sample_paper_answers_2013.pdf

[clıppy]

I just need to make sure.
In galvanic or electrolytic cells, ions flow through the electrolyte. Are cations attracted to the positive terminal and anions attracted to the negative terminal, or is it vice-versa?

[Jeggz]

I just need to make sure.
In galvanic or electrolytic cells, ions flow through the electrolyte. Are cations attracted to the positive terminal and anions attracted to the negative terminal, or is it vice-versa?

Cations are attracted to the cathode and anions are attracted to the anode.
And remember that in Galvanic Cells, the Cathode is Positive. And in Electrolytic Cells the Cathode is Negative.
Hope that helps

[Dismounted]

And remember that in Galvanic Cells, the Cathode is Positive. And in Electrolytic Cells the Cathode is Negative.

Mnemonic time! DANCAP: Discharging Anode Negative, Charging Anode Positive.

[clıppy]

Mnemonic time! DANCAP: Discharging Anode Negative, Charging Anode Positive.

Oh. My. God
That just made everything 100x easier. Dismounted I love you for that.