1.

What is [H^(+)] is mol/ L of a solution that is 0.20 M in CH_(3)CO ON a and 0.10 M in CH_(3)CO OH ? K_(a) for CH_(3)CO OH=1.8xx10^(-5)

Answer»

`3.5xx10^(-4)`
`1.1xx10^(-5)`
`1.8xx10^(-5)`
`9.0xx10^(-6)`

Solution :`pH=p K_(a)+LOG [("Salt")/("Acid")]`
`log[H^(+)]=log K_(a)-log [("Salt")/("Acid")]`
`log [H^(+)]=log K_(a)+log [("Acid")/("Salt")]`
`[H^(+)]=K_(a)[("Acid")/("Salt")]`
`=1.8xx10^(-5)xx(0.1)/(0.2)=9XX10^(-6)`


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