Post a New Question

chemistry

posted by on .

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 - ,

    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 - ,

    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

  • chemistry - ,

    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) = ?

  • chemistry - ,

    I how you answers this but what is "c" HAc? NaAc?

Answer This Question

First Name:
School Subject:
Answer:

Related Questions

More Related Questions

Post a New Question