What is [H^(+)] in mol/L of a solution that is 0.20 M in CH_(3)CO ON aand 0.10M in CH_(3)CO OH ? K_(a) for CH_(3)CO OH = 1.8 xx 10^(-5).

What is [H^(+)] in mol/L of a solution that is 0.20 M in CH_(3)CO ON a

1.	What is [H^(+)] in mol/L of a solution that is 0.20 M in CH_(3)CO ON aand 0.10M in CH_(3)CO OH ? K_(a) for CH_(3)CO OH = 1.8 xx 10^(-5).
Answer» `9.0xx10^(-6)` `3.5xx10^(-4)` `1.1xx 10^(-5)` `1.8xx10^(-5)` Solution :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`