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Buffer solution preparation

  1. Sep 29, 2016 #1
    1. The problem statement, all variables and given/known data
    So I have prepare a buffer solution of 100 mL with ph of 4.00 .I have to use acetic acid and NaAc (sodium acetate ) to do so . The sum of the concentration of weak acid and its conjugate base is equal to 0.10 M
    ca + cb = 0.10 M
    How much volume of acetic acid and NaAc should I use to make buffersolution with ph of 4.00
    Pka of acetic acid is = 4.76
    Also information given :
    NaAc (molar weight = 82.08g/mol ) and 100 % m
    HAc (mw= 60.05) (density = 1.05 ) (100 % m)

    2. Relevant equations
    ph = pka - log (HAc/Ac-)

    3. The attempt at a solution
    I tried solving it but i'm not sure if its correct
    4.00 =4.76 - log(HAc)/Ac-)
    4.00-4.76 = - log ( Hac/ Ac-)
    [HAc]/ [Ac-] = 10^0.76 = 5.754
    we know ca +cb = 0.10 M x 0.1 L = 0.01 mol
    nAc- + 5.754 x n Ac- = 0.01 mol
    6.754 nAc- = 0.01 mol
    so nAc- = 0.01/ 6.754 = 0.00148 mol nAc-
    HAc = 0.01 - 0.00148 =0.008519 mol HAc
    SO we have to use 0.01 mol of HAc and 0.001480 mol of NaAc
    HAc
    c=(100% x 10 x1.05 g/ml)/ 60.05 g/mol = 17.458 M
    c= n/v
    17.458 M= 0.01 mol/ V
    V= 5.71 x10^-4 L => 0.57 ml We gonna add 0.57 ml of HAc
    and for NaAc = 0.001480 mol x Mw ( 82.08) = 0.121520 g NaAc
    Is this correct ??Please explain if its not ??
     
  2. jcsd
  3. Sep 29, 2016 #2

    Borek

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    Staff: Mentor

    0.008519 is not 0.01.

    But in general you are one the right track (and the mass of sodium acetate looks OK to me).
     
  4. Sep 29, 2016 #3
    But my teacher used 0.01 mol for HAc last year . Thats why i 'm confused
     
  5. Sep 29, 2016 #4
    He used the total mol
     
  6. Sep 29, 2016 #5

    Borek

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    Staff: Mentor

    Not sure what you mean.

    You have correctly calculated you need 0.008519 moles of the acid, but in the next line 0.008519 miraculously became 0.01 moles. Why?
     
  7. Sep 29, 2016 #6
    The following buffer problem is solved by my teacher .
    300 ml ph =9.90 c= ca+cb
    We're going to use NaHCO3 and NaOH
    ph = 9.90 = pka - log nHCO3-/ n CO32-
    pka = 10.30
    nHCo3-/ n co32- = 2.51
    n HCO3- +n co32- = 0.500 x 300 ml = 150 mmol
    2.51 n co32- + n co32- = 150
    nco32- = 150/ 3.51 =42.7 mmol
    How much NahCo and NAoh do we have to use to make that buffer solution
    His answer was 150 mmol NaHCO3 and 42.7 mmol NaOH
    Our teacher wrote this answer . Why did he use total mol ( 150 mmol) for NaHCO3 why not 150-42.7 mmol ??
     
    Last edited by a moderator: Sep 29, 2016
  8. Sep 29, 2016 #7

    Borek

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    Staff: Mentor

    These are different cases. In one you mix a conjugate acid and conjugate base, in the other you mix an acid with a base to neutralize it - and produce the conjugate base.

    I guess your mistake is based on a common misunderstanding of the buffers. Can you tell what are the conjugate acid and base in each case?
     
  9. Sep 29, 2016 #8
    Ac- is conjugate base and HAc weak acid
    on second one
    HC03- = acid
    c032- = conjugate base
     
  10. Sep 29, 2016 #9

    Borek

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    Staff: Mentor

    OK.

    Where does the CO32- (base) come from?

    Where does the Ac- (base) come from?

    Do you see why these are different problems?
     
  11. Sep 29, 2016 #10
    Yes Ac- came from Sodium acetate but on second case we added strong base NaoH . Naoh completely deprotonate weak acids .
     
  12. Sep 29, 2016 #11

    Borek

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    Staff: Mentor

    So is it clear now why these questions are solved a bit differently, and why your solution is incorrect?
     
  13. Sep 29, 2016 #12
    Yeah thank you very much
     
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