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February 22, 2026, 02:14:09 am

Author Topic: So you think Unit 4 is easy? Have a challenge. :)  (Read 6509 times)  Share 

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Mao

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So you think Unit 4 is easy? Have a challenge. :)
« on: August 10, 2010, 11:31:43 pm »
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So, a few of you think unit 4 is a joke compared to unit 3. To make it more interesting for you, and also for you to gain an undying interest for physical chemistry, have a challenge. :)

Note, mathematics recommended. Also, you will need to research these topics to find/work out the answer. Most of these should only require google, some light reading, very basic principles, and a fair bit of maths. Main things you'll need are literary references of Ka [or pKa], and some wikipedia pages for explanations of formulas.

Equilibrium

- Investigate the relationship between rate of reaction, and thus show how the expression for the equilibrium constant K can be derived from the rate of reaction. [wiki rate equation]
-- Follow on from that, using the Arrhenius equation, show the dependence of K on temperature, and thus verify how K changes with temperature and delta H.

- Prove Le Chatelier's principle, particularly the subsequent shift will only partially offset the initial change.

- Derive an expression for solving the pH of a weak acid without making the main assumption [HX]_equilibrium = [HX]_i
http://vcenotes.com/forum/index.php/topic,28290.msg291268.html#msg291268
http://vcenotes.com/forum/index.php/topic,28290.msg291979.html#msg291979

- How is pH affected when two weak acids of different strengths are mixed?

- Derive an expression for solving the pH of a weak diprotic acid, and a weak triprotic acid. Thus rank these polyprotic acids in order of their strength: sulfuric, phosphoric, carbonic

- Using a weak acid and its conjugate salt, derive an expression that gives the mole fractions of these two compounds to create a buffer of a specific pH

- Investigate the isoelectric point of amino acids, what do they mean? Try to derive an expression for the isoelectric point of simple amino acids. Try to derive an expression for the pH of an amino acid dissolved in water (start with simple amino acids).

This is all I can think of at this moment, maybe I'll add stuff in the future. GOOD LUCK. Winner(s) may or may not get a prize of arbitrary value. I also may or may not post solutions in the future. That is all.
« Last Edit: August 14, 2010, 11:24:50 am by Mao »
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happyhappyland

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #1 on: August 11, 2010, 08:51:01 pm »
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- Derive an expression for solving the pH of a weak acid without making the main assumption [HX]_equilibrium = [HX]_i

HX -> X- + H+

[X-][H+]/[HX]-[H+]

:S?
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appianway

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #2 on: August 11, 2010, 08:59:09 pm »
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Thanks Mao! The first question isn't too bad, but some of the others should provide some fun xD

qshyrn

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #3 on: August 11, 2010, 09:02:47 pm »
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isnt the second one quite simple by just looking at the K expression.   for example if concentration of products gets higher, then numerator gets higher and to maintain the k value, the denominator gets higher and numerator gets loweredd and stufff like that...

Russ

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #4 on: August 13, 2010, 04:30:03 pm »
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- Investigate the isoelectric point of amino acids, what do they mean? Try to derive an expression for the isoelectric point of simple amino acids. Try to derive an expression for the pH of an amino acid dissolved in water (start with simple amino acids).


Love this one lol.

Studyinghard

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #5 on: August 13, 2010, 07:01:49 pm »
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i bet hes just making you do his uni assignment.
"Your life is like a river, no matter what you just got to keep on going"

Mao

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #6 on: August 13, 2010, 08:16:23 pm »
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- Derive an expression for solving the pH of a weak acid without making the main assumption [HX]_equilibrium = [HX]_i

HX -> X- + H+

[X-][H+]/[HX]-[H+]

:S?

So far so good, but how do you keep going to solve for pH?
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pooshwaltzer

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #7 on: August 13, 2010, 08:40:36 pm »
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- Derive an expression for solving the pH of a weak acid without making the main assumption [HX]_equilibrium = [HX]_i

HX -> X- + H+

[X-][H+]/[HX]-[H+]

:S?

So far so good, but how do you keep going to solve for pH?

Logging helps

fady_22

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #8 on: August 14, 2010, 09:50:00 am »
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I can't be bothered typing out the whole working, but, from there:



By changing the expression into a quadratic, and solving using the quadratic formula.
« Last Edit: August 14, 2010, 12:16:44 pm by fady_22 »
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Mao

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #9 on: August 14, 2010, 11:23:19 am »
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I can't be bothered typing out the whole working, but, from there:



By changing the expression into a quadratic, and solving using the quadratic formula.

Very good. Just a small thing that it should be Ka instead of K.
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fady_22

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #10 on: August 14, 2010, 12:09:05 pm »
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I can't be bothered typing out the whole working, but, from there:



By changing the expression into a quadratic, and solving using the quadratic formula.

Very good. Just a small thing that it should be Ka instead of K.

I know, but I'm so bad at Latex, and so I couldn't be bothered typing that out. :)
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DuBistKomisch

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #11 on: August 16, 2010, 11:05:45 pm »
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- Using a weak acid and its conjugate salt, derive an expression that gives the mole fractions of these two compounds to create a buffer of a specific pH

Assuming a buffer represented by HA + H2O <-> A- + H3O+.
We know that K_a = [A-][H3O+]/[HA] and that [H3O+] = 10^(-pH).
Therefore the ratio [A-]/[HA] = K_a/10^(-pH) = K_a*10^pH.
I'm not sure how to interpret "mole fractions", but I guess if you want to express n(HA) and n(A-) as percentages of n(HA) + n(A-), then say the total is x, so [HA] + [A-] = x => [A-] = x - [HA], so:
(x-[HA])/[HA] = K_a*10^pH
x-[HA] = [HA]*K_a*10^pH
x = [HA](K_a*10^pH+1)
[HA] = x/(K_a*10^pH+1) --- notice how the denominator is always greater than 1, so [HA] < x and [HA] / x < 1 :D
Then %HA = [HA] / total = (x/(K_a*10^pH+1))/x = 1/(K_a*10^pH+1)
And %A- is simply 100% minus that.
[HA] <=> n(HA) and [A-] <=> n(A-) and such are always interchangeable because the volume is constant and n and c are proportional.
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charmanderp

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #12 on: November 21, 2012, 11:46:13 pm »
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I'm going to try and do all of these, in case anyone else is interested.
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nisha

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #13 on: November 21, 2012, 11:58:34 pm »
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I'm up for it as well. Looking for jobs is driving me bananas.
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Am taking in students for CHEMISTRY and MATHS METHODS tuition for 2014 as well as first year chemistry. If interested, pm me. Flexible with location.

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charmanderp

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Re: So you think Unit 4 is easy? Have a challenge. :)
« Reply #14 on: November 22, 2012, 12:02:47 am »
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I've started reading a uni chem textbook for LOLs. Started reading it before finishing my exams (during downtime) actually, that's how insane I'd gone.
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