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K(a) of 0.02M CH^(3)COOH is 1.8xx10^(-5)...

`K_(a)` of `0.02M CH^(3)COOH` is `1.8xx10^(-5)` Calculate
a. `[H_(3)O^(+)]` b. % ionization c. pH

Text Solution

Verified by Experts

`d^(2)=(k_(a))/(C)`
`=(1.8xx10^(-5))/(0.02)`
`alpha^(2)=9xx10^(-4)`
`alpha=3xx10^(-2)`
`:. alpha=3%`
% of ionisation
`[H_(3)O^(+)]=C xx alpha`
`=0.2xx3xx10^(-2)=6xx10^(-4)`
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