What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous..?

at equivalence point the mmoles of acid and base are equal so the mmoles of base consumed by the formic acid are 29.8mL x 0.0567 M NaOH = 1.6897 mmoles since you have completely reacted all the base with acid  you have only sodium formate and water at the EP ::
   HCOOH + NaOH ------ Na+  CHO2^-1   + H2O as formate is the conjugate base of  weak acid ( the Ka of HCOOH is 1.8 X 10^-4 ) then formate is a pretty strong base ( Lowry/Bronsted principle ) and will hydrolyze in water  CHO2^-1 + H2O ------->  HCOOH  + OH^-1 according to its hydrolysis constant which is Kh= Kw/Ka

so the concentration  of sodium formate  ( M ) present at EP is 1.6897 mmoles/25mL + 29.8 mL ( the total volume )= 0.0308M  a small amount of this,X,  will hydrolyze  to yield X amounts of HCOOH and X amounts of OH^-1

using the Kh=  1 x 10^-14/1.8 x 10^-4 = 5.56 X 10^-11
then 5.56 X 10^-11 =

• /0.0308M -x

assuming x is small compared to 0.0308  then the equation reduces to
x^2 = 0.0308 x 5.56 X 10^-11 :::  x^2 = 1.71 X 10^-12;  x = OH = 1.308 X 10^-6 ( so X IS small compared to 0.0308 ). Since [OH^-1] = 1.308 X 10^-6 ,, the pOH is -log ( 1.308 X 10^-6 ) = 5.88  pH is thn 14- pOH
= 14- 5.88 = 8.12