Asked by fenerbahce

acetic acid: 0.1 M 30ml
hydrochloric acid 0.1 M 5 ml
sodium hydroxide 0.1 M 15 ml

Buffer + 5ml of: 1.91 ph
ph (measured): 4:52 ph

HOW DO I CALCULATE PH(CALCULATED)?

AND ALSO IT SAYS

SHOW YOUR CALCULATION FOR THE PH OF THE BUFFER BEFIRE AND AFTER THE ADDITION OF HCL?

COULD YOU PLEASE HELP ME THANK YOU!

Answers

Answered by DrBob222
I could help but your post is too disjointed and some information seems to be missing. Please repost the problem as it appears in your text. If this is a lab experiment, show how the buffer is made, what it consists of, etc.
Answered by fenerbahce
Okay let me write it this way.

I mixed 30 mL of 0.1 M Acetic Acid and 15 mL of 0.1 M Sodium Hydroxide and measured the pH. Then I added 5 mL of 0.1 M NaOH to this buffer and measured the pH.

It's asking me:

Show the calculations for the pH of the buffer before and after the addition of HCL?

Thank You
Answered by DrBob222
OK. Sorry for the delay in getting back to you but I had a concert to attend.
30 mL x 0.1M HAc = 3 millimoles.
15 mL x 0.1M NaOH = 1.5 millimols.

......NaOH + HAc ==> NaAc + H2O
I.....1.5......3.0....0.......0
C.....-1.5.....-1.5...1.5......0
E.......0......1.5.....1.5
Thus, the buffer consists of 1.5 mmols Ac^- and 1.5 mmols HAc. The pH of this buffer is pH = pKa + log(base)/(acid) = pKa + log(1.5/1.5) = pKa + log 1 = pKa + 0 = pKa = 4.74 approximately but you need to look up Ka in your tables and use that value. I used 1.8E-5 to obtain 4.74.

If we add 5 mL of 0.1M HCl we add 0.5 mmols H^+.
..........Ac^- + H^+ ==> HAc
I........1.5......0.......1.5
add ..............0.5..........
C.......-0.5.....-0.5.....+0.5
E.......1.0........0......2.0

Then recalculate the pH with the H-H equation.
pH = pKa + log(1/2) = ?
Answered by fenerbahce
I how you answers this but what is "c" HAc? NaAc?
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