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.
4 answers
Nevermind. I got it. It would help if I looked at the concentration data that my lab book gives me complete the problem:)
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
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
Good to know.
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.
(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.
1 answer