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bugmenomore bugmenomore
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11 years ago
I need to make a 0.1 M glycine buffer with a pH of 9.4. I have two buffers of 0.1 M glycine that have a pH of 9 and a pH of 10. How would I make 200mL of my pH 9.4 glycine buffer? Thank you in advance for your help!
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wrote...
11 years ago
Use the Henderson-Hasselbalch equation.

At ph 9 glycine is present in two forms, a -1 charged form and a 0 charged form.

the pka of the N-terminus of glycine is I beleive around 9.6.

9.0= 9.6 + log (b/a)
b/a= .25 which means that 20% is in the base form and 80% in the acid form at ph 9.

You want to get this to 9.4

9.4=9.6 + log (b/a)
b/a=.63 meaning you want 39% in the base form and 61% in the acid form.

You need to take your ph 9 glycine and convert some of the acid to the base form by adding something like NaOH.

Now, if you have 200 ml of the ph9 buffer this means .02 mols of total glycine, .004 will be in base form and .016 will be in acid form. you want to convert some of the .016 acid form mols to the base form so that you only have .0122 mols left in acid form so add .0038 mols of NaOH to it ( how much this is depends on the molarity).

Hope this helps.


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