Calculate `pH` of the solution which is `10^(-1)M` in `HCl` & `10^(-3)M` in `CH_(3)COOH[K_(a)=2xx10^(-5)]`.Also calculate `[H^(+)]` form `CH_(3)COOH`.
Calculate `pH` of the solution which is `10^(-1)M` in `HCl` & `10^(-3)M` in `CH_(3)COOH[K_(a)=2xx10^(-5)]`.Also calculate `[H^(+)]` form `CH_(3)COOH`.
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`underset(C(1-alpha))(CH_(3)COOH)hArr underset(Calpha)(CH_(3)COOH)+underset(10^(-1)+Calpha)(H^(+))`
` t=eq`
`H^(+)` ion can be considered completely form `HCl` due to less dissocation of `CH_(3)COOH` (because of common ion effect by `H^(+)` of `HCl,alpha ltlt1)` and its low conc.So `[H^(+)]=10^(-1)M:.pH=1`
from above equilibrium ,`2xx10^(-9)=(Calphaxx10^(-1))/(C)`
`alpha=2xx10^(-4)`
`[H^(+)]` form `CH_(3)COOH=Calpha=10^(-3)xx2xx10^(-4)=2xx10^(-7)M`
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