Chemistry 3/4 2013 Thread

[barydos]

Could someone please explain to me VCAA 2007 EXAM 1 MULTIPLE CHOICE Q16?
Thanks in advance

I think this is the way to do it.
We have 1L of KOH with a pH of 12.
Therefore [OH] = 10^(-2)
We want to get rid of this first, because pH is power of hydrogen, we calculate pH with just [H+].
So to neutralise or get rid of all the OH we need 10^(-2) mol of H+ (via adding HCl).
Now we just need 10^(-2) mol of H+ to bring our overall [H+] = 10^(-2) since our V = 1.0 L.
So to make the pH of the solution 2.0, we have to add 10^(-2) + 10^(-2) = 0.02 mol of HCl.
Hope that helps

Edit: perhaps there's an alternative method, but this is how I see it.

[Edward21]

Could someone please explain to me VCAA 2007 EXAM 1 MULTIPLE CHOICE Q16?
Thanks in advance

I found this really confusing at first as well don't worry! If you have KOH at pH 12. There's [OH]=10^-2 as [H3O+]=10^-12 (using the Kw value for water at 25 degrees). VCAA have been lovely enough to have 1L so your conc. values equal the mole values as it's in mol/L.

So first to decrease pH you need to add more H+ to increase the [H3O+], first you have to neutralise your base which is 10^-2M at pH 12, which as we know for a 1L solution is actually 0.01mol of HCL needed. That's all cool and stuff that you've neutralised your base, but obviously pH 2 is somewhat ACIDIC. All we've done so far is neutralised the OH- ions from the KOH. We now need to get the pH to 2 by getting the [H3O+]=10^-2M and as 1L solution 0.01mol of H+ NEEDS to be present in our 1L solution for a pH of 2. So we add 0.01+0.01=0.02mol. You had to neutralise your OH- with 0.01mol of HCl, then add a further 0.01mol to actually get a H3O+/H+ concentration of 10^-2M for a pH of 2 ultimately. Just think it out logically, you've got a basic solution, that you want to make acidic. First you add acidic H+ to eliminate the OH- then you have to add even more H+ to get the pH so low that it's 2, which translates to 0.01mol of H+ in 1L of aqueous solution. 

Damn my post was beaten 

[Aelru]

By the way, anyone have solutions to the 2013 VCAA Trial exam? I'm sorry if it's already posted, but I can't seem to find it o-o

[barydos]

By the way, anyone have solutions to the 2013 VCAA Trial exam? I'm sorry if it's already posted, but I can't seem to find it o-o

The multiple choice questions aren't fully worked, but here: http://www.cea.asn.au/sites/default/files/sample_paper_answers_2013.pdf

[massachusetts8]

I think this is the way to do it.
We have 1L of KOH with a pH of 12.
Therefore [OH] = 10^(-2)
We want to get rid of this first, because pH is power of hydrogen, we calculate pH with just [H+].
So to neutralise or get rid of all the OH we need 10^(-2) mol of H+ (via adding HCl).
Now we just need 10^(-2) mol of H+ to bring our overall [H+] = 10^(-2) since our V = 1.0 L.
So to make the pH of the solution 2.0, we have to add 10^(-2) + 10^(-2) = 0.02 mol of HCl.
Hope that helps

Edit: perhaps there's an alternative method, but this is how I see it.

I found this really confusing at first as well don't worry! If you have KOH at pH 12. There's [OH]=10^-2 as [H3O+]=10^-12 (using the Kw value for water at 25 degrees). VCAA have been lovely enough to have 1L so your conc. values equal the mole values as it's in mol/L.

So first to decrease pH you need to add more H+ to increase the [H3O+], first you have to neutralise your base which is 10^-2M at pH 12, which as we know for a 1L solution is actually 0.01mol of HCL needed. That's all cool and stuff that you've neutralised your base, but obviously pH 2 is somewhat ACIDIC. All we've done so far is neutralised the OH- ions from the KOH. We now need to get the pH to 2 by getting the [H3O+]=10^-2M and as 1L solution 0.01mol of H+ NEEDS to be present in our 1L solution for a pH of 2. So we add 0.01+0.01=0.02mol. You had to neutralise your OH- with 0.01mol of HCl, then add a further 0.01mol to actually get a H3O+/H+ concentration of 10^-2M for a pH of 2 ultimately. Just think it out logically, you've got a basic solution, that you want to make acidic. First you add acidic H+ to eliminate the OH- then you have to add even more H+ to get the pH so low that it's 2, which translates to 0.01mol of H+ in 1L of aqueous solution. 

Damn my post was beaten 

Thank you both! That makes so much more sense, what's the bet there will be one on this year's exam haha
Thanks guys!

[barydos]

For the IR spectrum of Q4: http://www.vcaa.vic.edu.au/Documents/exams/chemistry/chem-specs-samp-w.pdf
Which represents the C-H and which represents the O-H (acids) part?

And is there not splitting in C-NMR?

And is the "residue" of amino acids everything but the carboxy and amino groups? (so including the H atom opposite the R group?)

[Edward21]

For the IR spectrum of Q4: http://www.vcaa.vic.edu.au/Documents/exams/chemistry/chem-specs-samp-w.pdf
Which represents the C-H and which represents the O-H (acids) part?

And is there not splitting in C-NMR?

And is the "residue" of amino acids everything but the carboxy and amino groups? (so including the H atom opposite the R group?)

C-H is basically in every organic molecule. Characteristically, the broad ones at that region of the O-H for the acids, the thinner ones are for the alkanol O-H group.

No splitting in C-NMR Only 1-H, using the n+1 rule

Not 100% sure on that, I'll wait until someone answers that, I don't want to mislead you for the sake of answering 

[barydos]

C-H is basically in every organic molecule. Characteristically, the broad ones at that region of the O-H for the acids, the thinner ones are for the alkanol O-H group.

No splitting in C-NMR Only 1-H, using the n+1 rule

Not 100% sure on that, I'll wait until someone answers that, I don't want to mislead you for the sake of answering 

Thanks for that!

[barydos]

Ahh I got another question that I'm concerned with in the sample exam:
http://www.vcaa.vic.edu.au/Documents/exams/chemistry/chem-specs-samp-w.pdf

Question 10b), the answer shows the molecule to have the Br's be substituted with OH.
But then how does that oxidise into the molecule with a carboxy group on both ends (did a carbon atom magically appear.. since there's now 2 carbons branching off the benzene on both ends)

Yeah, so I just need some clarification there for 10 b).

[lzxnl]

Yeah, the magic carbon atom appearing is a glitch.

[barydos]

Yeah, the magic carbon atom appearing is a glitch.

Well that was fun wasting 10 minutes of my life wondering how that was possible... Thanks again nliu haha

[Dismounted]

No splitting in C-NMR Only 1-H, using the n+1 rule

Not 100% sure on that, I'll wait until someone answers that, I don't want to mislead you for the sake of answering 

I can confirm. High-resolution NMR is only studied for protons (i.e. peak splitting). Low-resolution for both proton and carbon-13 NMR.

[Edward21]

I can confirm. High-resolution NMR is only studied for protons (i.e. peak splitting). Low-resolution for both proton and carbon-13 NMR.

Haha not that part, the part about the exact definition of an amino acid residue?

[Dismounted]

Haha not that part, the part about the exact definition of an amino acid residue?

Oh haha, my bad.

An amino acid residue is "whatever's left" of the amino acid after it undergoes condensation to form the polypeptide.

For example, if three NH₂-CHR-COOH molecules bond to form a polypeptide, each one will have a different residue (assume I've bonded them end to end with amide to the left and carboxyl to the right):

• Left molecule residue: NH₂-CHR-CO-
• Middle molecule residue: -NH-CHR-CO-
• Right molecule residue: -NH-CHR-COOH

Hope that makes sense!

EDIT: If that doesn't make sense, here's a definition straight from the horse's mouth (IUPAC): http://goldbook.iupac.org/A00279.html (may be more confusing than me!).

[Edward21]

Oh haha, my bad.

An amino acid residue is "whatever's left" of the amino acid after it undergoes condensation to form the polypeptide.

For example, if three NH₂-CHR-COOH molecules bond to form a polypeptide, each one will have a different residue (assume I've bonded them end to end with amide to the left and carboxyl to the right):

• Left molecule residue: NH₂-CHR-CO-
• Middle molecule residue: -NH-CHR-CO-
• Right molecule residue: -NH-CHR-COOH

Hope that makes sense!

EDIT: If that doesn't make sense, here's a definition straight from the horse's mouth (IUPAC): http://goldbook.iupac.org/A00279.html (may be more confusing than me!).

Oh I see so after the water's removed in the condensation reaction, whatever's left of it..