January 10, 2025, 07:11:51 PM
Forum Rules: Read This Before Posting


Topic: Buffers, is my logic correct?  (Read 5687 times)

0 Members and 1 Guest are viewing this topic.

Offline Boxxxed

  • Full Member
  • ****
  • Posts: 203
  • Mole Snacks: +1/-0
Buffers, is my logic correct?
« on: March 11, 2011, 06:51:36 PM »
Calculate final pH of 25 ml buffer 0.1 M CH3COOH/CH3COO, pH=3.50

after the addition of 1.00 ml of 0.1 M HCl

H3O+ = 3.16x10-4

Do I sub in H3O+ for the acid, pka and ph to find the initial concentration of the base to start off?

Then with addition of HCl subtract 0.0001 from base and add 0.0001 to acid then use henderson formula again to solve for final pH?


Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27894
  • Mole Snacks: +1816/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Buffers, is my logic correct?
« Reply #1 on: March 12, 2011, 04:46:03 AM »
Acid + base = 0.1, and pH/pKa give you ratio of acid & base concentrations - thats two equations in two unknowns, so you can calculate both. And then yes, just assume base reacted quantitatively with HCl, like you wrote.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Boxxxed

  • Full Member
  • ****
  • Posts: 203
  • Mole Snacks: +1/-0
Re: Buffers, is my logic correct?
« Reply #2 on: March 12, 2011, 10:33:11 AM »
3.50=4.75+log(base/3.16x10-4)

Base = 1.78x10-5 M
Acid = 3.16x10-4 M


Mole base = 0.000712    - 0.0001 = 0.000612
Mole Acid = 0.01264      +0.0001  = 0.01274

Mole Acid added = 0.0001 Mole

ph=4.75+log(0.000612/0.01274)

ph = 2.43

Is this correct?
« Last Edit: March 12, 2011, 11:12:33 AM by Boxxxed »

Offline Boxxxed

  • Full Member
  • ****
  • Posts: 203
  • Mole Snacks: +1/-0
Re: Buffers, is my logic correct?
« Reply #3 on: March 12, 2011, 12:59:30 PM »
I have the ratio, 0.056 but I don't understand how to convert it to moles using the given volume.

What does the 0.1 M mean in the question? Is it concentration of base and acid together?

If so then total moles = 0.0025 moles
Ratio = 0.056.

How do I find out moles of acid and base?
« Last Edit: March 12, 2011, 01:10:30 PM by Boxxxed »

Offline Boxxxed

  • Full Member
  • ****
  • Posts: 203
  • Mole Snacks: +1/-0
Re: Buffers, is my logic correct?
« Reply #4 on: March 12, 2011, 01:40:01 PM »
I calculated the moles to be

I used 0.0025=17.8(base)+(base)
base = 0.00013
acid = 0.00237

Is it alright to sub in moles without converting them to concentrations?
ph = 2.88

« Last Edit: March 12, 2011, 02:06:55 PM by Boxxxed »

Offline Borek

  • Mr. pH
  • Administrator
  • Deity Member
  • *
  • Posts: 27894
  • Mole Snacks: +1816/-412
  • Gender: Male
  • I am known to be occasionally wrong.
    • Chembuddy
Re: Buffers, is my logic correct?
« Reply #5 on: March 12, 2011, 03:46:11 PM »
3.50=4.75+log(base/3.16x10-4)

3.16x10-4 is incorrect. As I wrote earlier you have to use the information that sum of concentrations of base and conjugate acid is 0.1M.
ChemBuddy chemical calculators - stoichiometry, pH, concentration, buffer preparation, titrations.info

Offline Boxxxed

  • Full Member
  • ****
  • Posts: 203
  • Mole Snacks: +1/-0
Re: Buffers, is my logic correct?
« Reply #6 on: March 12, 2011, 05:13:17 PM »
3.50=4.75+log(base/3.16x10-4)

3.16x10-4 is incorrect. As I wrote earlier you have to use the information that sum of concentrations of base and conjugate acid is 0.1M.

Yes, that post is wrong. I redid the question in the post after it using that method.

0.0025=17.8(base)+(base)

0.000133 mol base initially
0.00237 mol acid initially

Since the volume is the same for both of them I don't see why dividing both by the same number would change the answer.

Offline methic

  • Regular Member
  • ***
  • Posts: 27
  • Mole Snacks: +0/-1
Re: Buffers, is my logic correct?
« Reply #7 on: March 12, 2011, 06:22:25 PM »
Wait. Where'd that 17.8 come from?

Offline Boxxxed

  • Full Member
  • ****
  • Posts: 203
  • Mole Snacks: +1/-0
Re: Buffers, is my logic correct?
« Reply #8 on: March 12, 2011, 06:58:32 PM »
Ratio is 0.0562 base to acid

Acid is 17.8 times larger than the base.

1/0.0562 is 17.8

0.0025=17.8(base)+(base)

Or I could have done

0.0025=0.0562(acid)+(acid)

Sponsored Links