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Topic: Find the pH of Aspirin Solution  (Read 24173 times)

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kladd56

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Find the pH of Aspirin Solution
« on: July 10, 2006, 01:01:23 PM »
Acetylsalicylic acid (HC9H7O4) is the active ingredient in aspirin.  Two tablets each containing 500 mg of acetylsalicylic acid are dissolved in 500 mL of water.  What is the pH of this solution, if the Ka of the acid is 3.3 x 10-4
« Last Edit: July 17, 2006, 03:41:04 PM by geodome »

Offline Will

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Re: can you help with this problem
« Reply #1 on: July 10, 2006, 01:28:03 PM »
Acetylsalicylic acid (HC9H7O4) is the active ingredient in aspirin.  Two tablets each containing 500 mg of acetylsalicylic acid are dissolved in 500 mL of water.  What is the pH of this solution, if the Ka of the acid is 3.3 x 10-4

pH = -log[H+]
                                                        
[H+] = _/Ka x [C6H4(COOH)(OAc)]

You just have to work out what the concentration of the acid is (moles is mass/Mr and concentration is moles/volume).

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Re: can you help with this problem
« Reply #2 on: July 10, 2006, 01:32:01 PM »
Use the previously posted equations, here is the procedure I would use

1. Determine relative molecular mass of the acid
2. Determine number of moles of the acid in two 500 mg tablets
3. Determine the concentration of the acid in the solution
4. using the pKa, determine the concentration of H+ ions in solution, [H+]
5. Determine pH
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Offline Borek

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Re: can you help with this problem
« Reply #3 on: July 10, 2006, 01:53:07 PM »
                                                        
[H+] = _/Ka x [C6H4(COOH)(OAc)]

Wrong. This equation (8.13 in terms of the lecture pointed to below) holds only for dissociation percentage below 5%.

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Borek

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Re: can you help with this problem
« Reply #4 on: July 10, 2006, 01:53:43 PM »
Acetylsalicylic acid (HC9H7O4) is the active ingredient in aspirin.  Two tablets each containing 500 mg of acetylsalicylic acid are dissolved in 500 mL of water.  What is the pH of this solution, if the Ka of the acid is 3.3 x 10-4

Please read forum rules, as Will suggested.
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Offline Will

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Re: can you help with this problem
« Reply #5 on: July 10, 2006, 02:58:46 PM »
                                                        
[H+] = _/Ka x [C6H4(COOH)(OAc)]

Wrong. This equation (8.13 in terms of the lecture pointed to below) holds only for dissociation percentage below 5%.

http://www.chembuddy.com/?left=pH-calculation&right=pH-weak-acid-base

Woops! That is proof of just how cr*ppy the A-Level syllabus is.

If I did it my way I would've gotten an inaccurate pH of 3.019085045, when in fact the real pH is 3.093596791 (I think!).

Offline Borek

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Re: can you help with this problem
« Reply #6 on: July 10, 2006, 03:34:03 PM »
Woops! That is proof of just how cr*ppy the A-Level syllabus is.

If I did it my way I would've gotten an inaccurate pH of 3.019085045, when in fact the real pH is 3.093596791 (I think!).

Hmm, either your syllabus is even more crappy or my BATE isn't worth money I am asking for ;)

2.75 here (2.8 to be correct in terms of significant digits).
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Offline Will

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Re: can you help with this problem
« Reply #7 on: July 10, 2006, 03:35:32 PM »
Woops! That is proof of just how cr*ppy the A-Level syllabus is.

If I did it my way I would've gotten an inaccurate pH of 3.019085045, when in fact the real pH is 3.093596791 (I think!).

Hmm, either your syllabus is even more crappy or my BATE isn't worth money I am asking for ;)

2.75 here (2.8 to be correct in terms of significant digits).

It's probably me, my room is boiling hot and I can't think properly! ;D

Edit: my calculator messed me up on the concentration, it showed me 0.002775... instead of 0.0111014... >:(
« Last Edit: July 10, 2006, 03:41:50 PM by Will »

Offline Will

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Re: can you help with this problem
« Reply #8 on: July 10, 2006, 03:49:16 PM »
The pH I get now is 2.755446731 or 2.76 to 2dp, probably due to the small difference in concentration (I used 180.15742 as Mr) and I would've gotten a pH of 2.718054109 if I used the inaccurate equation.

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Re: can you help with this problem
« Reply #9 on: July 10, 2006, 05:31:53 PM »
Other probable source of difference is Ka value - BATE requires logarithmic input, so it got 3.48 for pKa, instead of 3.3*10-4. But 2.75 or 2.76 is perfectly acceptable here.

Note that BATE uses accurate value for molar mass, it just displays only two decimal digits.
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