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 concentrationof acetate ion in the solution and its pH.

The ionization constant of acetic acid is 1.74xx10^(-5) . Calculate th

1.	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 concentrationof acetate ion in the solution and its pH.
Answer» Solution :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