`CH_(3)CH_(2)COOH ltimplies CH_(3)CH_(2)COO^(-) + H^(+)`
`K_(alpha) = ([CH_(3)CH_(2)COO^(-)][H^(+)])/([CH_(3)CH_(2)COOH])`
`1.32 xx 10^(-5) = (Calpha xx Calpha)/(C(1-alpha)) = Calpha^(2) (therefore 1-alpha ~~ 1)`
`therefore alpha^(2) = (1.32 xx 10^(_5))/(0.05) = 2.64 xx 10^(-4)`
`alpha= 1.63 xx 10^(-2)`
`pH = -log_(10)[H^(+)]=-log_(10)(Calpha)`
`=-log_(10)(0.05 xx 1.63 xx 10^(-2)) = 3.09`
`0.01 M HCl` में `[H^(+)] = 0.01` अतः
`1.32 xx 10^(-5) = (Calpha xx 0.01)/(C(1-alpha)) = alpha xx 0.01`
`therefore alpha = (1.32 xx 10^(-5))/(0.01) = 1.32 xx 10^(-3)`