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PH for this Weak Acid-Strong Base titration.

  1. Dec 31, 2013 #1
    [itex]\:[/itex]1. The problem statement, all variables and given/known data
    What is the pH when 10 ml of 0.1 M [itex]NaOH[/itex] is added to 25 ml of 0.1 M [itex]H_{}C_{2}H_{3}O_{2}[/itex]?

    2. Relevant equations
    pH = -log[[itex]H_{3}O[/itex]]

    3. The attempt at a solution
    The main thing confusing me about this type of problem is that my teacher taught me a way which is different from the methods I see online. And I get a different answer.

    This is how I was taught to do it:

    moles of [itex]H_{}C_{2}H_{3}O_{2}[/itex] = c * v
    = 0.1 M * 0.025L
    = 0.0025 mol

    moles of [itex]NaOH[/itex] = c * v
    = 0.1M * 0.01L
    = 0.001 mol

    Ethanoic acid is in excess, so
    moles [itex]H_{}C_{2}H_{3}O_{2}[/itex] = 0.0025 mol - 0.001 mol
    = 0.0015 mol

    [[itex]H_{}C_{2}H_{3}O_{2}[/itex]] = [itex]\frac{n}{v}[/itex] = [itex]\frac{0.0015}{0.035}[/itex] = [itex]0.043[/itex]

    [itex]H_{}C_{2}H_{3}O_{2} + H_{2}O \leftrightharpoons H_{3}O + C_{2}H_{3}O_{2}^{-}[/itex]
    I [itex]\:\:[/itex] 0.043 [itex]\:\:\:\:\:\:[/itex] - [itex]\:\:\:\:\:\:\:\:\:\:\:\:[/itex] 0 [itex]\:\:\:\:\:\:\:\:\:\:\:\:[/itex] 0
    C [itex]\:\:\:\:\:\:[/itex] -x [itex]\:\:\:\:\:\:[/itex] - [itex]\:\:\:\:\:\:\:\:\:\:\:\:[/itex] +x [itex]\:\:\:\:\:\:\:\:\:\:\:\:[/itex] +x
    E [itex]\:\:\:[/itex] 0.043 - x [itex]\:\:\:\:\:\:[/itex] - [itex]\:\:\:\:\:\:\:\:\:\:\:\:[/itex] x [itex]\:\:\:\:\:\:\:\:\:\:\:\:[/itex] x

    ka = [itex]1.8 * 10^{-5}[/itex]

    [itex]1.8 * 10^{-5}[/itex] = [itex]\frac{x^{2}}{0.043}[/itex]
    [itex]x = 8.8 * 10^{-4} M[/itex]
    pH = [itex]-log[8.8 * 10^{-4}][/itex]
    = [itex]3.05[/itex]

    Is this the right approach?
  2. jcsd
  3. Dec 31, 2013 #2
    Your approach is correct but the concentration of ##C_2H_3O_2^-## at E is not x, it is x+something, can you figure out the "something" part? (Think about the reaction of NaOH with the given acid).

    What is other approach you talk about? To my knowledge, you have a buffer after reaction of NaOH with ##HC_2H_3O_2## so you might have seen the online resources using Henderson–Hasselbalch equation but that is actually obtained from the procedure you follow, there is really no need to look it up.
  4. Dec 31, 2013 #3
    Thanks for the reply. The other approach is where they use a mole chart
    like this

    I just want to know the situations of when to use a mole chart. This had been confusing me for weeks haha.
  5. Jan 1, 2014 #4
    You use a mole chart in every situation as you did in your attempt. Did you solve the problem?
  6. Jan 1, 2014 #5


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    Homework Helper
    Gold Member

    From moles and volume you have calculated the molarity of HAc. But you haven't calculated the molarity of Ac- that I can see.

    Write out the formula for Ka and you will see you you don't need to calculate either molarity. [HAc] and [Ac-] enter as a ratio, molarity ratio equals moles ratio. From the Ka formula and what you have, finding [H+] is staightforward enough.
  7. Jan 1, 2014 #6


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    Staff: Mentor

    You can assume neutralization went to completion, use this assumption to calculate [HAc] and [Ac-], plug into Henderson-Hasselbalch equation. Done.

    You can also start with above, assume dissociation went further and use ICE table to find the exact answer.

    The difference will be in most cases negligible, so the latter step can be safely ignored.

    These are about buffer pH, but it is exactly the same problem:


  8. Jan 1, 2014 #7

    Oh I see

    [itex]H_{}C_{2}H_{3}O_{2} + H_{2}O \leftrightharpoons H_{3}O + C_{2}H_{3}O_{2}^{-}[/itex]
    I [itex]\:\:[/itex] 0.043 [itex]\:\:\:\:\:\:[/itex] - [itex]\:\:\:\:\:\:\:\:\:\:\:\:[/itex] 0 [itex]\:\:\:\:\:\:\:\:\:\:\:\:[/itex] 0.029
    C [itex]\:\:\:\:\:\:[/itex] -x [itex]\:\:\:\:\:\:[/itex] - [itex]\:\:\:\:\:\:\:\:\:\:\:\:[/itex] +x [itex]\:\:\:\:\:\:\:\:\:\:\:\:[/itex] +x
    E [itex]\:\:\:[/itex] 0.043 - x [itex]\:\:\:\:\:\:[/itex] - [itex]\:\:\:\:\:\:\:\:\:\:\:\:[/itex] x [itex]\:\:\:\:\:\:\:\:\:\:\:\:[/itex] x + 0.029

    ka = [itex]1.8 * 10^{-5}[/itex]

    [itex]1.8 * 10^{-5}[/itex] = [itex]\frac{x * 0.029}{0.043}[/itex]
    [itex]x = 2.67 * 10^{-5} M[/itex]
    pH = [itex]-log[2.67 * 10^{-5}][/itex]
    = [itex]4.57[/itex]

    And to verify using henderson equation: [itex]pH = 4.74 + log(\frac{0.029}{0.043}) = 4.57[/itex]. Much easier.

    Thanks! It makes sense too, I guess my teacher taught me wrong, I'll talk to him about it.
    Last edited: Jan 1, 2014
  9. Jan 1, 2014 #8


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    Staff: Mentor

    Note that x (being 2.67×10-5) is three orders of magnitude smaller than initial concentrations (both being in the 10-2 range). Thats why you can ignore it.

    It wont be that way for very diluted solutions.
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