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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?
    thanks
     
  2. jcsd
  3. Feb 12, 2007 #2

    symbolipoint

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    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

    symbolipoint

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    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

    Borek

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    Staff: Mentor

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