Second Year Chem. pH problem?
Calculate the pH of a soln of acetic acid that is 3% ionized, Ka = 1.8 x 10^-5.
Since I don't have M of the solution, I can't solve by
% ionization = [H+}Equil / [HA]i
please help me understand what approach to take. Thank you in advance.
Answer:
The Ka is the (H+)(C2H3O2-) / HC2H3O2
You know that the HC2H3O2 is 0.03 and you have the Ka.
THe H+ and C2H3O2- are equal, so once you have the Ka times 0.03, take the square root and you have (H+). From there you should know how to calculate the pH using the negative log.
This might help:
Ka = [H+]*[A-] / [HA]
and % ionization is [H+] / ([A-] + [HA]) * 100%
Because most of the H+ is made from the dissociated HA, you may also make the approximation
[H+] ≈ [A-]
Making this approximation, you can eliminate [A-] from the other two equations, you get
P = [H+] / ([H+] + [HA])
or, more conveniently,
1/P = 1 + [HA]/[H+]
and Ka = [H+]^2 / [HA]
This second equation can be reformulated into
[HA] / [H+] = [H+] / Ka
from which we can substitute the left side to get
1/P = 1 + [H+] / Ka
From which [H+], hence pH, can be easily found.
I calculated [H+] to be 5.8e-4, which is much greater than the 1e-7 from the water, justifying our simplifying assumption. If the acid were weak enough to endanger our assumption that [H+] ≈ [A-], then P would probably be less than 0.01 and we could instead have assumed that [A-] ≈ [KA].
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