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The ionization constant of propionic acid is `1.32xx10^(-5)`. Calculate the degree of ionization of the acid in its `0.05`M solution and also its pH. What will be its degree of ionization in the solution of `0.01N HCI` ?

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`CH_(3)underset(C(1-alpha))(CH_(2)COO)HhArrCH_(3)underset(Calpha)(CH_(2))COO^(-)+underset(Calpha)(H^(+))`
`K_(a)=1.32xx10^(-5)`
`=[(Ch_(3)CH_(2)COO^(-)][H^(+)])/([CH_(3)CH_(2)COOH])`
`(CalphaxxC alpha)/(C(-alpha))=Calpha^(2)" "(therefore1-alpha~~)`
`:. 0.05xxalpha^(2)=1.32xx10^(-5)`
`alpha = 1.63xx10^(-2)`
`pH= -Iog [H^(+)]= - Iog(C alpha)`
`= - Iog(0.05xx1.63xx10^(-2))=3.09`
In `0.01 NHCI: [H^(+)]= 0.01` and thus
`1.32xx10^(-5)= (Calphaxx0.01)/(C(1-alpha))= alphaxx0.1 , (.: 1-alpha ~~1)`
`alpha = 1.32xx10^(-3)`

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