Posted by fenerbahce on Saturday, July 14, 2012 at 6:28pm.
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!
- chemistry - DrBob222, Saturday, July 14, 2012 at 7:31pm
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.
- chemistry - fenerbahce, Saturday, July 14, 2012 at 7:42pm
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?
- chemistry - DrBob222, Saturday, July 14, 2012 at 11:12pm
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
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
Then recalculate the pH with the H-H equation.
pH = pKa + log(1/2) = ?
- chemistry - fenerbahce, Sunday, July 15, 2012 at 12:36pm
I how you answers this but what is "c" HAc? NaAc?
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