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!
4 answers
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.
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
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
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) = ?
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) = ?
I how you answers this but what is "c" HAc? NaAc?
1 answer