Question

If `50ml` of `0.2 M KOH` is added to `40 ml` of `0.05 M HCOOH`, the `pH` of the resulting solution is (`K_(a)=1.8xx10^(-4)`)
A. `3.4`
B. `7.5`
C. `5.6`
D. `3.75`

Answer 1

Correct Answer - A
`pH=-logK_(a)+log``(["Salt"])/([Acid])`
`["Salt"]=(0.2xx50)/(1000)=0.01`, `[Acid]=(0.5xx40)/(1000)=0.02`
`pH=-log(1.8xx10^(-4))+log``(0.01)/(0.02)`
`pH=4-log(1.8)+log 0.5`
`pH=4-log (1.8)-0.301`
`pH=3.4`