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Consider a 20.0mL sample of 0.105M HC2H3O2 is titrated with 0.125M NaOH.
Ka=1.8x10^-5. Determine each of the following:

a) the initial pH

b) the pH at 5.0mL of added base

c) the pH at one-half of the equivalence point

d) the pH at the equivalence point

  • chemistry -

    The secret to these titration problems is to know where you are on the titration curve.
    a. at the beginning of the titration; therefore, you have a solution of acetic acid. Set up and ICE chart, substitute and solve for H^+ then convert to pH.

    b. use the Henderson-Hasselbalch equation.
    c. same as b but I can tell you the answer is pH = pKa.

    d. The equivalence point pH is determined by the hydrolysis of the salt.
    Set up the hydrolysis equation, write the K expression, set it equal to Kb for acetate (which is Kb = Kw/Ka) and solve for OH^-, convert to pOH, then to pH.
    Post your work if you get stuck.

  • chemistry -

    I understand everything except for part d.
    Could you show me what you mean by writing the k expression.

  • chemistry -

    Let's call acetate, Ac^- just to save some typing.
    Ac^- + HOH ==> HAc (acetic acid) + OH^-

    Kb for acetate = (Kw/Ka) where Ka is the acid constant for acetic acid.
    Kb = (Kw/Ka) = (HAc)(OH^-)/(Ac^-)
    Let HAc = x = OH, then
    (Kw/Ka) = x^2/Ac^-.
    Kw you know. Ka you know. Ac^- is the concn of acetate ion at the equivalence point which is M x L of the acid or base and that divided by the total volume. Solve for x, convert to pOH and pH.

  • chemistry -

    Here is the K expression (for Kb).
    Kb = (Kw/Ka) = (HAc)(OH^-)/(Ac^-)

  • chemistry -

    is Ac^- equal to to 20 ml x the concentration? if so then I am getting the wrong answer according to my answer sheet. By the way what would my total volume be?

  • chemistry -

    nvm i got it. Thanks for the help

  • chemistry -

    for part d, how do you know the total volume at the equiv point?

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