Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Calculating pH for Strong Acid-Base Titration

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data
    The problem is as follows:
    100.0 ml of 0.100 M acetic acid is placed in a flask and is titrated with 0.100 M sodium hydroxide . An appropriate indicator is used. Ka for acetic acid is 1.7 x 10 -5. Calculate the pH in the flask at the following points in the titration.
    a. when no NaOH has been added.
    b. after 25.0 ml of NaOH is added
    c. after 50.0 ml of NaOH is added
    d. after 75.0 ml of NaOH is added
    e. after 100.0 ml of NaOH is added
    2. Relevant equations
    According to my text:
    (Liters of acetic acid solution)X(mol H+/1Liter of solution)=mol H+
    (Liters of sodium hydroxide solution)X(mol OH-/1Liter of solution)=mol OH-
    (Liters of acetic acid)+(Liters of hydroxide solution)=total Liters
    Concentration of H+=(moles H+/total Liters)
    pH=-log(concentration of H+ *above*)
    3. The attempt at a solution
    What I have so far is this, but my pH answers are nowhere near what they should be. What am I missing?
    "100.0 ml of 0.100 M acetic acid is placed in a flask and is titrated with 0.100 M sodium hydroxide . An appropriate indicator is used. Ka for acetic acid is 1.7 x 10 -5Calculate the pH in the flask at the following points in the titration."
    (0.1L soln)(0.100M H+/1L soln)=1.000 X 10-2
    a. "when no NaOH has been added" pH=-log(0.100)=1.000
    b. "after 25.0 ml of NaOH is added" (0.025 L soln)(0.100M OH-/1L soln)=2.5 X 10-3 mol OH-
    100.0mL+25.0mL=125.0mL=0.125L [H+]=(7.5x10-3)/(0.125L)=6.000X10-2M pH=-log(6.000X10-2M)=1.222

    c. "after 50.0 ml of NaOH is added" (0.050L soln)(0.100M H+/1L soln)=5.000 X 10-3 mol OH-
    100.0mL+50.0mL=150.0mL=0.150mL [H+]=(5.000 X 10-3)/(0.150L)=3.333 X 10-2 pH=-log(3.333 X 10-2)=1.477

    d. "after 75.0 ml of NaOH is added" (0.075 L soln)(0.100M H+/1L soln)=7.5X10-3 mol OH-
    100.0mL+75.0mL=175.0mL=0.175L [H+]=(2.5 X 10-3)/(0.175L)=1.429X10-2 pH=-log(1.429X10-2)=1.845

    e. "after 100.0 ml of NaOH is added" (0.1L soln)(0.100M H+/1L soln)=1.000 X 10-2 mol OH-
    100.0mL+100.0mL=200.0mL=0.200L [H+]=pH=7.0


    This is a table similar to that which is shown in my text, where I have filled in my answers.
    H+(aq) + OH-(aq) --> H2O(l)
    Before addition 1.000 X 10-2 0 ---------
    Addition 25.0 ml 2.5 X 10-3
    Addition 50.0 ml 5.000 X 10-3
    Addition 75.0 ml 7.5X10-3
    Addition100.0 ml 1.000 X 10-2
    After Adn25.0 ml 7.5x10-3 0 --------
    After Adn 50.0 ml 5.000 X 10-3 0 --------
    After Adn 75.0 ml 2.5 X 10-3 0 --------
    After And 100.0 ml 0 0 ---------
     
  2. jcsd
  3. Feb 26, 2010 #2

    Borek

    User Avatar

    Staff: Mentor

    One general problem - acetic acid is not a strong acid.

    Second general problem - when you add base to the solution it reacts with the acid.

    See acid-base titration curve calculation for details, scroll down the page - the most important information you is in the third paragraph from the end.

    --
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook