KOH added : 7.5 ml
Buffer solution of weak acid HCOOH and strong base KOH
Reaction
initial = 100 ml 0.1 M HCOOH = 10 ml mol HCOOH, and x mlmol of KOH
KOH + HCOOH ⇒ COOHK + H₂O
x 10
x x x x
- 10-x x x
[HCOO - ] = 3[HCOOH]
[tex]\tt \dfrac{x}{x+100~ml}=3\dfrac{10-x}{x+100}\\\\x=3(10-x)\\\\x=30-3x\\\\4x=30\rightarrow x=7.5~ml[/tex]