Question

Calculate the degree of ionization of acetic acid in solution 1 through 3?

All I am giving is the pH...
pH of solution 1= 3
pH of solution 2= 3.5
pH of solution 3= 5

Then, I did this...

HC2H3O2 (aq) + H2O (l) equals H3O+ (aq) + C2H3O2- (aq)

Ka= 1.7 E -5 (according to chart in my book)

Ka= {H3O}{C2H3O2)/{HC2H3O2}=1.7 E -5

How do I figure the degree of ionization for solutions 1-3 from what I am given above? Any help would be greatly appreciated.

Answer 1

Nevermind. I got it. It would help if I looked at the concentration data that my lab book gives me complete the problem:)

Answer 2

1.75 x 10^-5=Ka= {H3O}{C2H3O2)/{HC2H3O2}


lets do the first one,


HC2H3O2 + H2O----> H3O+ + C2H3O2-

pH=3, so

H3O+=10^-(3)=1 x 10 ^-3 M


so the ratio of H3O+= C2H3O2-


So, Ka=1.75 x 10^-5= {H3O}{C2H3O2)/{HC2H3O2}

becomes

Ka=1.75 x 10^-5= [1 x 10 ^-3 M][1 x 10 ^-3 M]/{HC2H3O2},

solve for the molarity of HC2H3O2

1.75 x 10^-5/(1 x 10^-3)^2=HC2H3O2


(H3O+/HC2H3O2)*100, should give you the degree of ionization

Answer 3

Good to know.

Answer 4

Because the part should be,

(1 x 10^-3)^2/1.75 x 10^-5=HC2H3O2


(H3O+/HC2H3O2)*100, should give you the degree of ionization

Flipped it by accident.