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Titration help

  1. Apr 23, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider the titration of 20.0mL of a 0.100 M solution of HBrO, a weak acid (Ka=2.5x10-9) with 0.200 M KOH. Calculate the pH of the following volumes of KOH.

    a. 0.00mL
    b. 5.00mL
    c. 10.00mL
    d. 30.00mL



    2. Relevant equations
    pH=pKa +log [base]/[acid]


    3. The attempt at a solution

    Im having trouble with C when the pH is at its equivalance point.

    so,

    10.00mL (.200 mmol KOH/mL) = 2.00 mmol OH-

    So stoichiometricaly my OH- and HBrO are used to completely to give me a conjugate base of 2.00 mmol BrO-

    [BrO-]=2.00mmol/30.00ml=.0667 M

    I set up my ice chart:

    BrO-<=====>OH-+HBrO
    I .0667 0 0
    C -x +x +x
    E .0667-x x x


    Where

    Kb=[OH-][HBrO]/[BrO-]=x2/.0667-x=4.0x10-6

    Assume x is small to find x=[OH-]=5.2x10-4

    pOH=3.28
    pH= 14-pOH=10.72

    What doesn't make sense to me is in Part A pH=4.80, part B pH=8.5, Part C=10.72 and part D pH=8.20.

    Now shouldnt the pH in part D be greater than part C since I have excess OH-?

    I'll type of my work for Part D in a bit. I have to leave right now. But if somebody can help my checking my answers I'd appreciate it. Thanks.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Apr 23, 2009 #2

    symbolipoint

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    Homework Helper
    Education Advisor
    Gold Member

    When your solution is far past the endpoint, the pH only depends on the excess hydroxide. How much or what is the concentration of the excess hydroxide? This helps to give you your pH calculation. pH + pOH = 14.
     
  4. Apr 23, 2009 #3
    Ohhhhhhhh!

    I see what I forgot to do. I knew that once past the end point the only thing affecting pH was excess hydroxides. I just forgot to properly calculate the concentration. I tried using the henderson-hassalbach(sp?) equation instead. So since I have an excess of 4.00 mmol of hydroxide and a total volume of 50.00mL, [OH-]=.0800 M. So leaving me with a final pH of 12.903 for part D. Thank you very much!
     
  5. Apr 24, 2009 #4

    Borek

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    Staff: Mentor

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