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In general buffer the Henderson/Hasselbach equation is valid to use:

pH = pKa -log [(Ca)/(Cb)]

In which:

* Ca = concentration of weak acid (benz. or formic.acid) and

* Cb = conc'n of their respective conjugated bases (benzoate and formiate)

From tables it is found: pKbenz.acid = 4.20 and pKformic.acid = 3.80

pHbenz.acid,sol'n = 4.20 - log[(475*0.200-25*2.00)/(25*2.00)] =

= equal pH values =

= pHformic.acid sol'n = 3.80 - log[(Va*0.200-Vb*2.00)]/(Vb*2.00)]

in which Va + Vb = 500 mL of the (new) HCOOH/HCOO- buffer

* Va = volume of the formic acid sol'n (HCOOH, in mL) and

* Vb = volume of the (2.00M) NaOH sol'n.

The rest is calculation. Outcome:

equal pH values: pH = 4.246

- log [(Va*0.200-Vb*2.00)]/(Vb*2.00)] = 4.246 - 3.80 = 0.446

[(Va*0.200-Vb*2.00)]/(Vb*2.00)] = 0.358

Va*0.200/Vb*2.00 - 1 = 0.358

0.100*Va/Vb = 1.358

Va = 13.58*Vb and with Va + Vb = 500 it comes to:

Va= 465.5 mL and Vb = 34.5 mL

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Q: What volumes of 0.200 m hcooh and 2.00 m naoh would make 500 ml of a buffer with the same pH as of one made from 475 ml of 0.200 m benzoic acid and 25 ml of 2.00 m naoh?
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