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4. How would you prepare 250 mL of a buffer with pH = 3.6 using .100...

4. How would you prepare 250 mL of a buffer with pH = 3.6 using .100 M formic acid and .100 M sodium formate solutions?

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Answer #1

Apply buffer equation

pH = pKa + log(formate/Formic acid)

pKa = 3.75 for formic acid so

3.6 = 3.75 + log(formate/Formic acid)

(formate/Formic acid) = 10^(3.60-3.75) = 0.70794

note that

total molarity must be 0.1 M, or mmol = MV = 250*0.1 = 25 mmol

then

formic acid + formate = 25 mmol

(formate/Formic acid) = 0.70794

formate = 0.70794*acid

formic acid + formate = 25 mmol

acid +  0.70794*acid = 25

acid = 25/(1+0.70794) = 14.637 mmol of acid

formate= 25-14.637 = 10.363 mmol of formate

from the solutions:

Vacid required = 14.637 /0.1 = 146.37

Vformate required = 10.363 /0.1 = 103.63

Vtotal = 146.37+103.63 = 250 mL , and ratio is kept

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