Question

4. a solution that is 0.025M in HCOOH and 0.025M in NaCOOH

Answer 1

KOH added : 7.5 ml

Further explanation

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]

$\tt \dfrac{x}{x+100~ml}=3\dfrac{10-x}{x+100}\\\\x=3(10-x)\\\\x=30-3x\\\\4x=30\rightarrow x=7.5~ml$