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Homework Help: How do you calculate the pH of 35.0-mL of 0.15M acetic acid?

  1. Nov 10, 2015 #1
    1. The problem statement, all variables and given/known data
    "A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) os titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added: (a) 0 mL, (b) 17.5 mL, (c) 34.5 mL, (d) 35.0 mL, (f) 50.0 mL."

    2. Relevant equations
    Ka = [H+][A-]/[HA]
    Ka = 1.8 ⋅ 10-5
    x = ½(-b ± √(b2 - 4ac)
    pH = -log[H+]
    pH = 2.78 (from back of book)

    3. The attempt at a solution
    0.15 mol/L ⋅ 1 L/1000 mL ⋅ 35.0 mL = 5.25 ⋅ 10-3 mol HAc
    1.8 ⋅ 10-5 = (x)(x)/(5.25 ⋅ 10-3 - x)
    x2 + 1.8 ⋅ 10-5x - 9.45 ⋅ 10-8 = 0
    x = ½(-1.8 ⋅ 10-5 ± √3.24 ⋅ 10-10 - 37.8 ⋅ 10-8)
    x = 2.99 ⋅ 10-4 mol H+

    [H+] = (2.99 ⋅ 10-4 mol H+)/(0.0350 L) = 8.54 ⋅ 10-3 M
    -log(8.54 ⋅ 10-3) = 2.069 ≠ 2.78

    This is rather an unorthodox, and thus, unreliable method of finding the initial pH of the acetic acid solution. But I couldn't find any other way to do it... Could someone tell me if I'm doing this problem right?
  2. jcsd
  3. Nov 11, 2015 #2


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    Staff: Mentor

    You can't do equilibrium calculations using Ka (which is a concentration constant) and ignoring the concentrations. Why do you convert to number of moles, instead of using concentrations throughout the calculations?
  4. Nov 11, 2015 #3


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    Homework Helper
    Gold Member

    I don't think the approach is unorthodox except as below. I think there is only an arithmetical/algebraic mistake. Your square root is written without right bracketing and you have got wrong sign in the -4ac term. Probably you will get the right answer if you correct that.

    Where it is not wrong but unorthodox is there is a bit of unnecessary calculation.

    Orthodox would be to say that [H+] is very much smaller than total acid, and therefore you can approximate the denominator of your equilibrium equation by total acid concentration [HAc] = 0.15. Then your calculation reduces to taking a square root rather than solving a quadratic. Do it both ways and you will see they are close. You can if you wish, with benefit, look into the algebra and see why.

    The point is this: all this family of problems (the most frequent single one on this part of this forum) are solved using a handful of principles: mass conservation, charge conservation, equilibrium laws, I think that's all... plus use of approximations. Don't think of it as Important Laws and then, oh yes, also get rough answer quicker if approximate. No, the appropriate approximations, understanding them, and how to use them and why and when and how good and sufficient they may be depending on situations, is an essential part of what you have to think about, grasp and learn!

    In most cases if you have a quadratic or higher equation you have missed a possibility of approximation. (Of course if the physical nature of the problem requires it, e.g. some equilibria with two protonation etc. polynomials may be of the essence.)
  5. Nov 11, 2015 #4


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    Staff: Mentor

    I think he will not - units are wrong. Using number of moles instead of concentrations works when the volumes cancel out, like [itex]\frac {C_1}{C_2} = \frac {n_1}{n_2}[/itex], but here we have [itex]\frac {C_1^2}{C_2} = \frac {n_1^2}{V\times n_2} \neq \frac {n_1^2}{n_2}[/itex].

    I admit I have not checked numbers, but the logic looks faulty.
  6. Nov 11, 2015 #5
    Okay, I'll try using concentration instead of moles.
  7. Nov 11, 2015 #6
    So I thought the answer was more complicated than it seemed. I just used the initial concentration, and used the "x-is-small" approximation.

    1.8 ⋅ 10-5 = (x)(x)/(0.15 - x)
    1.8 ⋅ 10-5 = (x)(x)/(0.15)
    2.7 ⋅ 10-6 = x2
    x = 0.00164
    -log[H+] = 2.78

    Now to figure out the rest of the problem...
  8. Nov 11, 2015 #7


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    Staff: Mentor

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