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Topic: Explain qualitatively the effect of adding 0.01M hydrochloric acid to the aspiri  (Read 8574 times)

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sci0x

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Question: Aspirin is a weak acid.

(a) Calculate pH of 0.2M solution of aspirin at 25 degrees celsius (Ka = 3.0 x 10^-4 at 25 degrees celcius).

(b) Determine the percent ionisation

(c) Explain qualitatively the effect of adding 0.01M hydrochloric acid to the aspirin solution.

(d) Calculate the pH of the resulting solution

(e) Determine the percent ionisation of the aspirin under these conditions
« Last Edit: January 21, 2006, 01:42:22 PM by Mitch »

Offline plu

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Re:Acids, pH,
« Reply #1 on: January 21, 2006, 11:17:33 AM »
Give us a hand here, mate  :-\  Write some equations!

sci0x

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Re:Acids, pH,
« Reply #2 on: January 21, 2006, 11:20:21 AM »
Thats all the information I am given in the question

Offline Borek

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Re:Acids, pH,
« Reply #3 on: January 21, 2006, 12:20:54 PM »
Do they really told you to do the calculation without introducing you first to weak acids  and dissociation constants? Change school as fast as possible :P

Show us what you did.

Here is some theory for you - weak acid pH calculation.
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sci0x

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Re:Acids, pH,
« Reply #4 on: January 21, 2006, 12:48:22 PM »
Ok,

a) is it Ka = [H+]^2 / 0.2M

3.0 x 10^-4 = [H+]^2 / 0.2M
so, H+ = 7.7 x 10^-3
pH = -log[H+] = 2.11

b) pKa = -log10(Ka)
  = -log10(3.0 x 10^-4)
  = 3.522

Percent Ionisation = [H+]/M (original) x100
= (7.749 x 10^-3)/0.2  x 100
= 3.87%

c) Please explain this, i dont know this.

d) I'll be able to do this when i understand c

e) I'll be able to do this when i understand c

Offline Borek

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Re:Acids, pH,
« Reply #5 on: January 21, 2006, 12:58:17 PM »
a, b - results are correct (even if you use too much significant digits) - but are you sure you know why? You used some approximation - do you know why you used it and if it holds?

(c) Explain qualitatively the effect of adding 0.01M hydrochloric acid to the aspirin solution.

Dissociated aspirin is a weak base.
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st3v3n

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Re:Acids, pH,
« Reply #6 on: January 21, 2006, 01:25:03 PM »
well, for the third qn..., you are asked to explain qualitatively... meaning not using any numbers or any calculations...

I'll assume you know Le Chatelier's principle... So I guess you can say that when HCI is added to the aspirin acid solution, due to the common ion effect,, the equilibrium wil shift to the left so as to remove the hydrogen ions.... Meaning less acid dissociates...

As for part D, pH of the resultant solution... Here you will have to add the H+ concentration from HCI and the aspirin acid and then find the new pH... ans will be 1.75 to 3 s.f

For the last part, you will have to use the equation form earlier by sciox...
3.0 x 10^-4 = [H+]^2 / 0.2M
but now, you substitute in the total H+ concentration from part D and find the amount of dissociation for the conjugate base of the aspirin acid, which will be equal to the amount of dissociated H+ concentration under the new conditions... and find the percent ionisation like you did earlier... bingo... my ans here is 1.69%
 ;D

sci0x

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"As for part D, pH of the resultant solution... Here you will have to add the H+ concentration from HCI and the aspirin acid and then find the new pH... ans will be 1.75 to 3 s.f"

I cant work this out.
I assume its not simply add 0.2M to 0.01M and get a new H+ and a new pH. I cant get 1.75 through this method. What did you do exactly?

Offline Borek

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d is difficult, to be honest, I have no idea what kind of solution I can propose - I mean, I know how to do it, but not in an easy way.

Once d is done, e is simple - you may use Henderson-Hasselbalch equation.
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sci0x

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Any chance you could write out how to do it? Even if it is hard, and i'll try and understand when I see your method.

Offline Borek

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http://www.chembuddy.com/?left=pH-calculation&right=general-pH-calculation

You may try by yourself - start with all equations described in the lecture.

Assumptions to try:

water autodissociation can be neglected (ie [OH- ] = 0)
concentration of undissociated aspirin is equal to its analytical concentration
hydrocholoric acid is 100% dissociated

I did it and it is much easier than I expected. You will have second degree polynomial to solve.
« Last Edit: January 21, 2006, 08:41:05 PM by Borek »
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Offline Mitch

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Most Common Suggestions I Make on the Forums.
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Offline Borek

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OK, here goes the solution for d (hopefully sci0x tried it in the meantime).

HA <-> H+ + A-

What equations do we have?

dissociation constant:
Ka = [A- ][H+ ]/[HA]

water autodissociation:

Kw = [OH- ][H+ ]

charge balance:
[H+ ] = [Cl- ] + [OH- ] + [A- ]

and acid mass balance:
CA = [HA] + [A- ]

We know solution is acidic (pH less then 2 - that is granted by 0.01M HCl). If so, concentration of [OH- ] can be neglected (must be below 10-12) in charge balance equation (concentration of Cl- is 1010 times higher). So charge balance equation takes form:

[H+ ] = [Cl- ] + [A- ]

that allows us to remove water autodissociation from the set of equations.

Now, aspirin is a weak acid in acidified solution - so we can assume it is almost not dissociated (we already this this assumption in point a - it worked then it will work even better now). If so, we can write acid mass balance equation in form

[HA] = CA

substitute it into acid dissociation constant:
Ka = [A- ][H+ ]/CA

solve for [A- ]:

[A- ] = CA Ka / [H+ ]

and finally put this [A- ] into already simplified charge balance equation:

[H+ ] = [Cl- ] + CA Ka / [H+ ]

This is quadratic equation in [H+ ], and [H+ ] is the only unknown - all other values are given.

pH = 1.84
« Last Edit: January 22, 2006, 07:26:43 AM by Borek »
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