When 50 mL of 0.10M formic acid, HCHO2, is titrated with 0.10M sodium hydroxide, what is the pH at the equivalence point? (Be sure to take into account the change in volume during the titration.) Select a good indicator for this titration

Answer 1

HCOOH + NaOH ==> HCOONa + H2O
So you have 50 mL x 0.1M = 5 millimols HCOOH to start and you add 5 millimols NaOH so at the end you have 5 millimols HCOONa (sodium formate) in 100 mL or (HCOONa) = 5 mmols/100 mL = 0.05 M. The pH at the equivalence point then is determined by the hydrolysis of the formate part of sodium formate salt as follows:
.............HCOO^- + HOH ==> HCOOH + OH^-
I.............0.05M........................0...................0
C................-x...........................x....................x
E..............0.05-x......................x....................x
Kb for HCOO^- = (Kw/Ka for HCOOH) = (x)(x)/(0.05-x)
Solve for x = (OH^-) and convert that to pH.
Post your work if you get stuck.