March 24, 2017

Post a New Question

Posted by on Sunday, May 2, 2010 at 3:50pm.

What is the pH of the solution created by combining 2.40 mL of the 0.10 M NaOH(aq) with 8.00 mL of the 0.10 M HC2H3O2(aq)?

Can somebody explain how to do this? I am stuck. I've tried solving for H+ concentration. I got .00024M and subtracted it from [OH-]. Then I solved for pOH and used that to get pH but still got the wrong answer. Somone plz explain this to me.

  • Chemistry - , Sunday, May 2, 2010 at 4:02pm

    I think it's a simple problem IF you recognize what you have. To simplify, let's call HC2H3O3, acetic acid, HAC. Then
    NaOH + HAc ==> NaAc + H2O

    moles NaOH initially = L x M = 0.00240 x 0.1 = 0.000240
    moles HAc = L x M = 0.008 x 0.1 M = 0.0008.
    Note the NaOH is the smaller amount; therefore, HAc will neutralize all of the NaOH and some HAc will be left over.
    Final mols.
    NaOH = 0
    HAc = 0.0008-0.00024 = 0.00056
    NaAc formed = 0.00056
    H2O formed but not worry with that.
    So what do you have when all of this is over? We have a weak acid (HAc) and its salt (NaAc) which makes a buffer. Use the Henderson-Hasselbalch equation to solve for pH.
    pH = pKa + log[(base)/(acid)]
    base = concn of Ac^- which is moles/L.
    acid = concn of HAc which is moles/L.
    Plug and chug. I'll post this, then repost with an approximate answer to help you check your answer.

  • Chemistry - , Sunday, May 2, 2010 at 4:04pm

    You should get close to 4.3 or so for pH.

  • Chemistry - , Sunday, May 2, 2010 at 9:17pm

    How do you find the concentration of the conjugate base?

  • Chemistry - , Sunday, May 2, 2010 at 10:33pm

    base = mols acetate/L
    moles acetate = 0.00056
    volume = 10.4 mL = 0.0104 L.
    concn = 0.00056/0.0104 = ??

  • Chemistry - , Friday, May 7, 2010 at 5:29pm

    How are we finding pKa for the Henderson-Hasselbalch equation?

  • Chemistry - , Friday, May 7, 2010 at 6:03pm

    Never-mind decided just to use a table of pKa values. By the way the NaAc formed wouldn't be 0.00056 it would be 0.00024 because you can only make as much product as your limiting reagent allows (in this case OH- ions) I seriously tried using these numbers a ton of times and it was the NaAc value that screwed everything up.

Answer This Question

First Name:
School Subject:

Related Questions

More Related Questions

Post a New Question