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Buffer Calculation using the Henderson-Hasselbalch Equation

  1. Sep 28, 2012 #1
    1. The problem statement, all variables and given/known data
    Prepare 750 mL of 0.5 M phosphate buffer, pH 6.7 and pKa 7.21. NaH2PO4 (119.96 g/mol) and Na2HPO4 (141.96 g/mol) is available to use.


    2. Relevant equations

    pH=pKa+log(B/A)

    3. The attempt at a solution

    I have the solution from my teacher, but I have no idea how he got the numbers. Could someone please explain how to use this equation? I'm trying to work backwards (I'm terrible at math, I learn the methods from solved examples) but I can't figure out where the numbers are coming from:

    6.7=7.21 +log(B/A)
    B=0.3090A (How did he get that???)

    0.75L x 0.5 mol/L = 0.375 mol = B+A

    1.3090A = 0.375 (Why is it 1.3090 and not 0.3090A like above?)

    34.37g = 119.96 x (He put a symbol here that looks like <=) A=0.2865 mol
    12.57g = 141.96 x <= B=0.885 mol

    If someone would explain the process above, I would be extremely grateful. Thanks
     
  2. jcsd
  3. Sep 28, 2012 #2

    Borek

    User Avatar

    Staff: Mentor

    That's where you need to know the math that you don't know. I can show you how to do it, but if you don't know properties of logarithms it won't help you much:

    [tex]6.7-7.21=-0.51=log\frac{B}{A}[/tex]

    Antilogarithm (both sides used as exponents):

    [tex]10^{-0.51}=\frac{B}{A}[/tex]

    [tex]B=10^{-0.51}A=0.3090A[/tex]

    [tex]B+A=0.375[/tex]

    But we know B=0.3090A, so we can substitute it:

    [tex]0.3090A + A = 1.3090A = 0.375[/tex]

    [/quote]34.37g = 119.96 x (He put a symbol here that looks like <=) A=0.2865 mol
    12.57g = 141.96 x <= B=0.885 mol[/QUOTE]

    Number of moles of A and B were calculated from the equations above, here it is just a conversion of moles to mass.
     
  4. Sep 28, 2012 #3
    OHH I was confusing antilog with natural log. I did say I was terrible at math. :blushing: Thanks for your help!
     
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