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Buffer Calculation using the Henderson-Hasselbalch Equation

  • Thread starter jibjab
  • Start date
  • #1
13
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Homework Statement


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.


Homework Equations



pH=pKa+log(B/A)

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
 

Answers and Replies

  • #2
Borek
Mentor
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6.7=7.21 +log(B/A)
B=0.3090A (How did he get that???)
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]

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?)
[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.
 
  • #3
13
0
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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