I will bet money big time that you omitted the negative sign (-) on three items. I KNOW Ka is 1.34 x 10^-5 and I KNOW Kw = 1.0 x 10^-14. Then I assume the concn of propionic acid is not 500 M but rather that it is 0.05M (that's 5 x 10^-2 M.) So let me call propionic acid (CH3CH2COOH) something much more simple such as HPr.
...................HPr ==> H^+ + Pr^-
I.................0.05M......0.........0
C................-x.............x.........x
E...............0.05-x........x.........x
Substitute the E line into the Ka expression of
Ka = 1.34E-5 = (H^+)(Pr^-)/(HPr) and solve for (H^+).
Then pH = -log(H^+) . Then knowing H^+ and Kw you can solve for OH^-
I'm not sure what you want for H3O^+ + H^+. You may have meant H3O^+ = (H^+)
Post your work if you get stuck or if you want me to check your answers.
The concentration of propionic acid (CH3CH2COOH) solution was found to be 5.0 x 10^2 M, at 25 degree Celsius. (Ka for propionic acid is; 1.34 x 10^5, Kw= [H+][OH-]= 1 x 10^14)
Calculate,
A) Hydrogen (H+) ion concentration, ([H3O+] + [H+])
B) Hydroxide (OH-) ion concentration
C) pH of propionic acid solution
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