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Buffer (pH of a weak acid decreases when we add NaOH Yes, really

  1. Mar 2, 2013 #1
    Buffer (pH of a weak acid decreases when we add NaOH!! Yes, really..

    This is not homework, but just something that I find not following general rule. Just wondering why.
    We have the titration of 20.0 mL of .100M HF with .100M NaOH

    With 0mLof NaOH, the pH=2.08 (by calculating)

    But with 1.00mL of NaOH, the pH is 1.888 (by calculating)

    BUT then with 2.00ml of NaOH, the pH increases to 2.213
    After that, it keeps increasing and follow the rule

    given the Ka= 6.8e-4

    Can someone explain it?
  2. jcsd
  3. Mar 3, 2013 #2


    User Avatar

    Staff: Mentor

    Show how you got this result.
  4. Mar 3, 2013 #3
    For V of NaOH=0mL

    The equilibrium equation is
    HF <=> H+ + F-
    -x +x +x
    .1 x x

    (x^2)/.1 = Ka = 6.8e-4
    x= .008246 = [H+]
    pH = 2.083745544

    With V of Naoh=1mL

    HF + OH- <=> F- + H20
    (.02)(.1) (.001)(.1)
    .002mol .0001mol
    -.001mol -.001mol +.001mol
    = .0019mol 0 .0001mol

    => [HF]=.0019mol/(.021L) = .090476 M
    [F-]= .0001mol/(.021L) = .00476M

    Equilibrium equation:
    HF <=> H+ + F-
    .090476 M .00476M
    -x +x +x
    .090476 M x .00476M

    .00476x/.09476 = 6.8e-4
    [H+] = 1.86847 = x
    pH= 1.868472988
    Last edited: Mar 3, 2013
  5. Mar 3, 2013 #4


    User Avatar

    Staff: Mentor

    Check the last line again.
  6. Mar 3, 2013 #5
    Yes, I did, sir.

    My teacher also checked it but she had no clue why it's like that..

    SHe also did the same calculation and get the same pH.
  7. Mar 3, 2013 #6
    we're probably not supposed to ignore the x in this one, right?
  8. Mar 4, 2013 #7


    User Avatar

    Staff: Mentor

    Don't guess, try. The compare numbers you get with the numbers you got earlier.

    Simple check is always to plug concentrations you get into Ka and seeing if the calculated value agrees with the given one.
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