If `50ml` of `0.2 M KOH` is added to `40 ml` of `0.5 M HCOOH`, the `pH` of the resulting solution is (`K_(a)=1.8xx10^(-4)`)

Options

A. `3.4`
B. `7.5`
C. `5.6`
D. `3.75`

Answer

D

Explanation

`{:(HCOOH,+,KOH,hArr,HCOOK,+,H_(2)O,),(40xx0.5,,50xx0.2,,0,,0,"Intial"),(20-10,,10-10,,10,,10,"Final"):}`
Moles of HCOOK = Moles of HCOOH
[salt] = [acid]
`pH = pK_(a)`
`= - log (1 .8 xx 10^(-4))`
`= 4 - log 1.8 = 3.75`