The pH of 0.1 M solution of weak mono protic acid is 4.0. Calculate its `[H^(+)]` and `Ka`.

Explanation

Given `pH=4.0`
`: [H^(+)]=?`
`pH=-log[H^(+)]`
`:. [H^(+)]=10^(-pH)`
`=10^(-4)`
`:. [H^(+)]=C xx alpha`
`alpha=(10^(-4))/(0.1)`
`=10^(-3)`
`K_(a)=C xx alpha^(2)`
`=0.1xx(10^(-3))^(2)`
`=0.1xx10^(-6)`
`=10^(-7)`