How many mL 0.1 M NaOH could you add to a buffer
formed from 0.1 moles NaHCOO and 0.1 moles HCOOH
without the pH going above 4.74?
Ka (formic acid = 1.8 x 10-4)
0.1 mol = 100 millimoles.
........HCOOH + OH^- ==> HCOO^- + H2O
I........100....0.........100
add.............x..............
C.....100-x....-x.........100+x
4.74 = 3.74 + log (100+x)/(100-x)
Solve for x = millimols OH to be added. I get approximately 82 millimols and
M = millimols/mL, You know M and mmoles, solve for mL.
Quite a large number when you think about it.