Question

The ionization constant of acetic acid `1.74xx10^(-5)`. Calculate the degree of dissociation of acetic acid in its `0.05 M` solution. Calculate the concentration of acetate ion in the solution and its `pH`.

Answer 1

`CH_(3)COOH hArr CH_(3)COO^(-)+H^(+)`
`K_(a) [[CH_(3)COO^(-)][H^(+)]]/[[CH_(3)COOH]] =[[H^(+)]^(2)]/[[CH_(3)COOH]]`
`[H^(+)] =(K_a)xx [CH_(3)COOH])^(1//2)`
`=(1.74 xx 10^(-5) xx 5 xx 10^(-2))^(1//2) =9.33 xx 10^(-4)M`
`pH =- log [H^(+)]=- log (9.33 xx 10^(-4))`
`=(4 - log 9.33 ) = 4- 0.97 =3.03`