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What is `[H^(+)]` in mol/L of a solution that is 0.20 M in `CH_(3)CO ON a ` and 0.10M in `CH_(3)CO OH` ? `K_(a)` for `CH_(3)CO OH = 1.8 xx 10^(-5)`.A. `9.0xx10^(-6)`B. `3.5xx10^(-4)`C. `1.1xx 10^(-5)`D. `1.8xx10^(-5)`

Answer» Correct Answer - A
The given solution is an acidic buffer. Hence, its pH will be
`pH = pK_(a) +log. (["Salt"])/(["Acid"])`
`=-log.(1.8xx10^(-5))+log.(0.20)/(0.10)`
or `-log(h^(+)]=log.(2)/(1.8xx10^(-5))`
or `log.(1)/([H^(+)])=log.(2)/(1.8xx10^(-5))`
or `[H^(+)]=(1.8xx10^(-5))/(2) = 0.9xx10^(-5)M`
`=9xx10^(-6)M`


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