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Homework Help: Chemistry pH problem

  1. Apr 19, 2007 #1
    1. The problem statement, all variables and given/known data
    A solution was prepared by dissolving 0.02 moles of acetic acid (HOAc; pKa = 4.8) in water to give 1 liter of solution.
    a) What is the pH?
    b) 0.008 moles of concentrated sodium hydroxide (NaOH) was then added to this solution. What is the new pH? (In this problem, you may ignore changes in volume due to the addition of NaOH).
    c) An additional 0.012 moles of NaOH is then added. What is the pH?

    2. Relevant equations
    pH = -log [H+ (aq)]
    pH = pKa + log [A- (aq)]/[HA (aq)]

    3. The attempt at a solution
    a) pH = pKa + log [A- (aq)]/[HA (aq)] = 4.8 + log (0.02 mol/1 L) = 4.8 + log (0.02) = 4.8 - 0.301 = 4.699
    Can I get this problem checked? I also am not sure on how to approach parts b and c. Thanks.
    Last edited: Apr 19, 2007
  2. jcsd
  3. Apr 19, 2007 #2


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    Consider for part (a), you do not know pH or [H+]; you do not know [A-] and you only approximately know [HA]. Not a bad equation, but you might want to use a slightly better one:

    [tex] \[
    K_a = \frac{{[H][H]}}{{F_a - [H]}}
    [/tex] ...remember, that is for part (a) only.

    If you look at the Ka and also use charge balance and mass balance, and if you know hydronium is more significant than hydroxide (pH reasonably lower than 7) then you can use this:
    [tex] \[
    K_a = \frac{{[H][F_{salt} + H]}}{{F_a - [H]}}
    \] [/tex] Just be aware that F there stands for FORMALITY. This is based upon how the solution is prepared. If you take N moles of a substance and dissolve it in your 1 liter of solvent, and if any of it dissociate, the you will have less than N moles per liter of the substance.
  4. Apr 20, 2007 #3

    So in relation to the problem I am trying to solve, Fa would be mol/L, but what about Fsalt? And is there another equation(s) to use, since these equations you gave are not in my lab manual or textbook.
  5. Apr 20, 2007 #4


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    For the first question - use ICE table. For the second - assume neutralization is complete. You either have a buffer solution (Henderson-Hasselbalch equation) or there was excess base, and this excess is responsible for solution pH.

    Check out these pH calculation lectures, although they can be a little bit too advanced.
  6. Apr 20, 2007 #5


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    In relation to the message number #3, The "F", formality, is FORMULA UNITS PER LITER. It is like Molarity, but not exactly the same. This will generally depend on how a solution (liquid) is prepared.

    The formality of the salt ( parts "b" and "c") in your problem-description will depend on the amount of sodium hydroxide added. Each formula unit of sodium hydroxide will in effect create one formula unit of sodium acetate, UNTIL you add excess sodium hydroxide. Any excess sodium hydroxide IS NOT equivalent to any sodium acetate.

    Look back on post #2. There I gave two different formulas for acid dissociation constant expression. The first one is only for a solution of the acid; the second one is for a solution of acid and its salt. Neither of these formulas may be in your lab manual. Maybe Borek's posted links might help explain these a little more clearly. Remember these formulas for Ka, etc, are expressions for a CONSTANT.
  7. Apr 20, 2007 #6


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    Here is a little bit more of explanation:

    HAc <==> H+ + Ac-

    Each mole of HAc which dissociates gives one mole of H+ and one mole of Ac-.
    We can then say that [H] = [Ac-]. This gives for Ka,
    Ka=[H][H]/[HAc] in which one of the [H] is for hydronium and the other is same as acetate.

    IF ANY sodium salt is also present, then this changes the [Ac-] by the additional amount of this salt, so you show this with addition:
    Ka=[H](Fsalt + [H])/[HAc]

    So, some of the acetate comes from the formula units of sodium acetate, and some comes from the acetate dissociated from the acid.

    I still did not here explain the way I previously wrote the denominator; maybe you can trace this: (Facid - [H])
    Excuse me for not using TexAide for typsetting.
  8. Apr 22, 2007 #7


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    a)If it's purely acetic acid, then use the equation

    Ka=[A-][H3O+]/ [HA] = [x][x]/ [original concentration - x] and solve for x which should be the hydronium cation concentration ; then use this value to find the pH.

    for b) and c) use the Ka equation again, remember that for every acetic acid molecule consumed by the base, an equivalent amount of the conjugate is going to form. Use these new ratio values to solve again for "x" by incorporating "x" correctly into the Ka equation.
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