`200ml` of `1 M CH_(3)COOH(K_(a)=10^(-6))` is mixed with `200ml` of `0.1M HCOOH(K_(a)=10^(-5))`. The `pH` of the resulting mixture is

Answer

3

Explanation

`[CH_(3)COOH]=C_(1)=(200xx1)/400=0.5`
`[HCOOH]=C_(2)=(200xx0.1)/100=0.05`
`[H^(+)]=sqrt(K_(a_(1))C_(1)+K_(a_(2))C_(2))`
`=sqrt((10^(-6)xx0.5)+(10^(-5)xx0.05))`
`=10^(-3)`
`pH=-log 10^(-3)=3`