Calculate pH of 0.1 M acetic acid having `K_(a)=1.8xx10^(-5)`

Explanation

`CH_(3)COOH+H_(2)O hArr CH_(3)COO^(-)+H_(3)O^(+)`
`{:("Initial", 0.1,O,O),("At equilibrium",(0.1-0.1alpha),0.1alpha,0.1alpha):}`
`K_(a)=C.alpha^(2)`
`1.8xx10^(-5)=0.1xxalpha^(2)`
`1.8xx10^(-5)= alpha^(2)`
`alpha=8.4xx10^(-4)`
`[H_(3)O^(+)]=0.1xx10^(-4)xx8.4`
`pH=-log[H_(3)O^(+)]`
`:. pH=-log[8.4xx10^(-5)]`
`=2.9`