1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: I have my acids and bases test tomorrow, and i need help with a few types of problems

  1. Feb 12, 2007 #1
    im looking at some review problems my teacher gave me today, and i can pretty much do most of them. however, there is this one i really have no idea what to do. Its a 2 part problem.

    1a.What is the pH of a 1000 ml of a 0.1 M acetic acid solution, given pKa = 4.8?
    1b. what is the pH of 1000ml of 0.1 M acetic acid solution after you have added 5 ml of 10.0 M NaOH, assuming net volume change is negligible

    so, for the first part, i think i got it wrong. But here is what i did.
    First i found the Ka with
    pKa = 4.8 = -log[Ka]
    10^-4.8 = Ka
    Ka = 1.58 x 10^5

    so then i wrote out the equilibrium reaction.
    C2H4O2 + H20 > C2H3O2 + H30
    where everything is aqueous except H20, and the products on the right side are - and + ions, respectively.
    so then
    Ka = [x][x]/0.1 - x, where is x is the concentration of H30 and C2H3O2.
    solving for x gives 1.25 * 10^-3

    so, [Hplus] = 1.25 * 10^-3 M
    then, i did this step
    (1.25 x 10^-3 Hplus ions/L) * 1000ml * 1L/1000ml gives 1.25 x 10^-3 Hplus ions

    pH = -log[Hplus] = 2.903

    i really need help, im pretty sure i made some erroneous steps.

    and for part b, i really didnt have any idea how to start. I just started by finding the number of mols of OHminus, but i dont think that does anything.
    So, can someone help me do both parts?
  2. jcsd
  3. Feb 12, 2007 #2


    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    Hi, Pakmingki,
    You essentially have the right idea. In part A, you obtain about the same result whether a more full quadratic form is used or the simplification. In part B, the concentration of acid is essentially cut in half.

    This is a BASIC program which can give the results that you want; if you want a more full explanation, write a private message:
    You get a pH of about 3 for the addition of caustic soda, a pH of a bit less than 3 for just the acid before the addition of caustic soda.
  4. Feb 12, 2007 #3


    User Avatar
    Homework Helper
    Education Advisor
    Gold Member

    Excuse me, but what I wrote for part B may be inaccurate. Better, to take this form instead for the addition of NaOH:

    if x= molarity of hydronium,
    Fs = formal salt concentration
    Fa = formal acid concentration

    x(Fs + x)/(Fa - x) = Ka

    Now, just find x; this is a quadratic equation, so no real trouble. My little BASIC program is fine for little or no salt, but is not necessarily so good when a large portion of salt (neutralized acid) is present.
  5. Feb 13, 2007 #4


    User Avatar

    Staff: Mentor

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook