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Chemistry - desperately need help with buffer calculation.

  1. Dec 10, 2012 #1
    1. A buffer is prepared by mixing a 100.0ml of a 0.100M NH3 solution with a 0.200M solution of NH4Cl and making the total volume up to 1.000L of water. What is the volume of the NH4Cl solution required to achieve a buffer at pH=9.5
    Ka of NH4+ = 5.6x10^-10

    The textbook tell me that the answer is 28.1mL, can anyone please help me to figure this out? i've been trying for the past few hours and just about ready to give up!

    I'm almost certain you need to use hendersen hasselbach but i cannot get the correct answer.
    Thank you :)
     
  2. jcsd
  3. Dec 11, 2012 #2

    Borek

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    Staff: Mentor

    HH equation will give a ratio of concentrations of NH3 and NH4+. You are given amount of ammonia and the final volume, so you can easily calculate ammonia concentration in the final solution. Use this number and the ratio to calculate concentration of NH4+ in the final solution, then calculate how much NH4+ in the form of 0.200M solution is needed.

    Actually dilution calculations can be ignored, but I am afraid explaining why will just confuse you more.
     
  4. Dec 11, 2012 #3
    oh ok, thank you for your feedback. is there any chance you would mind showing me the working for the question?
    im just a bit confused about which ratio you're talking about, i've completed the question and gotten an answer of 0.2823L which i know isn't right.
     
    Last edited: Dec 11, 2012
  5. Dec 11, 2012 #4

    Borek

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    Staff: Mentor

    Show what you did and how.
     
  6. Dec 11, 2012 #5
    okay, so i began by finding the moles of ammonia in 1L,
    first found moles in 100mL
    c=n/v so n=cV
    n=0.100M/0.1
    = 0.01

    then using henderson hasselbach,

    pH=pKa+ log (NH4+/NH3)
    9.5=9.25+log(NH4+/0.01)
    0.25=log(NH4/0.01)
    10^-0.25= (NH4/0.01)
    NH4= 0.564 x 0.01
    n of NH4 = 0.00565

    then from moles find volume of NH4 required,
    c=n/v
    v=n/c
    v= 0.00564/0.02M
    v= 0.2823L
    v= 282.3mL

    i know my working isn't right but i'm at a loss about what to do. i have an exam on this on friday so i really appreciate your help.
     
  7. Dec 11, 2012 #6

    Borek

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    Why 0.02M?
     
  8. Dec 11, 2012 #7
    i've just had a look at the question again,

    i think i've calcualted the number of moles incorrectly,

    moles of ammonia should actually be,
    n= 0.1M/1.0L
    n=0.1

    is that correct?
     
  9. Dec 11, 2012 #8
    oh that should be 0.2M not 0.02. my mistake
     
  10. Dec 11, 2012 #9
    so then if i sub that in, so..

    0.00564 M/0.2M
    =0.0282L, is that then correct?
     
  11. Dec 12, 2012 #10

    Borek

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    Looks OK to me, and fits the book answer, doesn't it?
     
  12. Dec 12, 2012 #11

    Borek

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    No.

    First, it is C*V, not C/V (actually you wrote it wrong in the original solution as well, but then you calculated right).

    Second, in the final solution concentration of ammonia is not 0.1M, it was diluted tenfold, to 0.01M.

    Your original calculation (0.1M*0.1L=0.01 moles) was OK.
     
  13. Dec 13, 2012 #12
    oh okay thank you very much for your help :) i appreciate it!
     
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