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 ion in the solution and its pH.

Explanation

This is weak acid , So, following equilibrium established in `CH_3COOH` solution.
Dissociation degree = `alpha` So,
`[H^+] = [CH_3COO^-] = 0.05 alpha`
`{:(,CH_3COOH_((aq)) hArr, H_((aq))^(+) + , CH_3COO_((aq))^(-)),("Molarity in initial", 0.05,0,0),("Change in equili.",-0.05 alpha,+0.05 alpha,+0.05alpha),("M at equilibrium",(0.05-0.05alpha),0.05alpha,0.05alpha),(,=0.05(1-alpha) approx 0.05M,,):}`
`K_a=([H^+][CH_3COO^-])/([CH_3COOH])=1.74xx10^(-5)`
`therefore ((0.05alpha)(0.05alpha))/(0.05) =1.74xx10^(-5)`
`therefore 0.05 alpha^2 = 1.74xx10^(-5)`
`therefore alpha=sqrt((1.74xx10^(-5))/0.05)=sqrt(3.48xx10^(-4))`
`therefore alpha`= Degree of dissociation
`=1.865xx10^(-2)`
=0.01865
`[H^+]=[CH_3COO^-]`
`=0.05 xx alpha= 0.05 xx 0.01865 = 9.327xx10^(-4)` M
pH=-log `[H^-]`
=-log `(9.327xx10^(-4))`
=-(0.9697-4)=-(-3.03)=3.03