Question

You want a buffer with a pH of 7.40.  How much 0.13 M NaOH would you have to add to 75 mL of 0.10 M phosphoric acid to make this buffer?

Answer 1

the reaction after addition of NaOH would be :

The pka of phosporic acid is missing it should be given to get the exact answer.

I am adding that data : pka1 = 2.16 , pka2 = 7.21 pka3 = 12.32

to get the desired pH = 7.4 , we need to make the buffer solution corresponding to the second dissociation constant so the reaction should be as follows

             H₃PO₄   + NaOH ------->   NaH₂PO₄      + 2H₂O

millimoles        7.5           0.13 x V₁             7.5

       v1 =   57.69 ml

                     NaH₂PO4   +     NaOH           ------->   Na2HPO₄      + 2H₂O

millimoles        7.5                  0.13 x V₂ added          

                      7.5 -0.13V₂                         0.13x V₂

applying henderson hasselbach equation

pH = Pka2 + log(salt/acid)

7.4   = 7.21 + log{ ( 0.13V₂)/(7.5 -0.13V₂ )}

0.19 = log{ ( 0.13V₂)/(7.5 -0.13V₂ )}

taking antilog

1.54 = { ( 0.13V₂)/(7.5 -0.13V₂ )}

on solving for V₂    = 34.9 78 ml

Total volume of NaOH added = V₁ + V₂   =   57.69 ml + 34.9 78 ml = 92,668 ml