Question

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

Answer 1

`{:(,CH_(3) CO OH,hArr,CH_(3)CO O^(-),+,H^(+),,),("Initial",C,,0,,0,,),("At eqm.",C-C alpha,,C alpha,,C alpha ,,):}`
`K_(a)=(C alpha. C alpha)/(C(1-alpha))=(C alpha^(2))/(1-alpha) ~= C alpha^(2)`
`alpha = sqrt((K_(a))/(C))` (or directly by Ostwald dilution law)
`=sqrt((1.74xx10^(-5))/(0.05))=sqrt(3.48xx10^(-4))=1.86xx10^(-2)`
`[CH_(3)CO O^(-)] = C alpha = 0.05 xx (1.86xx10^(-2))=9.3xx10^(-4)M`
`[H^(+)]=C alpha = 9.3 xx 10^(-4)`
`pH = - log [H^(+)]=- log (9.3xx10^(-4))=3.03`