what ration of sodium benzoate to benzoic acid is required..

Question

is required to prepare a buffer with a pH of 4.28 at 25 degrees C? pKa for benzoic acid is 4.20 at 25 degrees C.

The answer has to be as a decimal, ratio with a denominator =1

 

I really don't understand how to do this problem.

Please help.

Answer 1

pH = pK_a + log(benzoate/benzoic acid)    (from Henderson Hasselbalch equation

4.28 = 4.20 + log(benzoate/benzoic acid)  -->  log(benzoate/benzoic acid) = 0.08  -->  benzoate/benzoic acid = 10^0.08 = 1.20