We have given,
40 ml 0.1 M Ba (HCOO)2
20 Ml 0.1 M HCOOH
Ka value of HCOOH = 10-5
Numbers of moles of HCOO- = 2 x 40 ml x 10-3 x 0.1 mole
= 8 millimole
Number of moles of HCOOH = 20 x 100-3 x 0.1 mole
= 2 millimole
After mixing, the volume of solution = (40 + 20) ml
= 60 ml
the concentration of HCOO- (salt) = \(\cfrac{8}{60}\)
= 0.133 M
the concentration of HCOOH (acid) = \(\cfrac2{60}\)
= 0.0 33 M
pKa at HCOOH = - log Ka
- log 10-5
= 5
Using Henderson Hesselbalch equation
pH = pKa + log \(\cfrac{[salt]}{[acid]}\)
5 + log \(\cfrac{0.133}{0.033}\)
= 5.6
Hence pH of solution will be 5.6