Chemistry

Consider the titration of 100.0 mL of 0.100 M HC2H3O2 by 0.100 M KOH for the next five questions (Ka for HC2H3O2 = 1.8 x 10-5). Calculate all pH values to two decimal places.

1) Calculate the pH after 0.0 mL of KOH has been added.
2) Calculate the pH after 50.0 mL of KOH has been added.
3) Calculate the pH after 75.0 mL of KOH has been added.
4) Calculate the pH at the equivalence point.
5) Calculate the pH after 150.0 mL of KOH has been added.

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  1. I think you already know how to do these. What you must do to know how to calculate each is to recognize what you have in the solution. Here is the equation and the millimoles added. Start with 100 mL x 0.100 M = 10 millimoles HC2H3O2.
    ...................HC2H3O2 + KOH ==> KC2H3O2 + H2O
    I.....................10................0.................0................0
    add...................................x.................................
    C...................-x...............-x...................+x
    E....................10-x.............0.....................x
    1) Calculate the pH after 0.0 mL of KOH has been added.
    2) Calculate the pH after 50.0 mL of KOH has been added.
    3) Calculate the pH after 75.0 mL of KOH has been added.
    4) Calculate the pH at the equivalence point.
    5) Calculate the pH after 150.0 mL of KOH has been added.

    1) No KOH is added so the pH is determined by the ionization of the 0.1 M acid.
    2) Adding 50 mL KOH addes 50 x 0.1 = 5 mmoles KOH which forms 5 mmoles KC2H3O2. You should recognize this as a buffered solution and solve accordingly.
    3) 75.00 mL KOH adds 7.5 mmoles KOH. Just a rerun of (2).
    4) The equivalence point. You should know that you have no HC2H3O2 and no NaOH. You have a solution of the salt so the pH will be determined by the hydrolysis of the salt. I've worked some of those problems for you. The pH of NaC2H3O2 solution are in the neighborhood of 8.2 or 8.3 so you will know if you're in the ball park or not.
    5). After 150 mL NaOH that's 150 x 0.100 = 15 mmoles NaOH which is more than enough to neutralize all of the acid and have some NaOH left.
    The (OH^-) = (NaOH) and pH from there
    Post your work if you get stuck on any of these.

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    DrBob222

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