Question
A 0.035 M solution of a weak acid (HA) has a pH of 4.88. What is the Ka of the acid?
HA ==>H^+ + A^-
Ka = (H^+)(A^-)/(HA)
pH = -log(H^+)
You know pH. Convert that to (H^+).
(H^+)=(A^-) so plug those into the expression for Ka.
(HA) = 0.035 M - (H^+). Plug that in.
Calculate Ka.
Post your work if you get stuck.
THANK YOU SOO MUCH!!! I don't know why I cant get the simplest problems!
HA ==>H^+ + A^-
Ka = (H^+)(A^-)/(HA)
pH = -log(H^+)
You know pH. Convert that to (H^+).
(H^+)=(A^-) so plug those into the expression for Ka.
(HA) = 0.035 M - (H^+). Plug that in.
Calculate Ka.
Post your work if you get stuck.
THANK YOU SOO MUCH!!! I don't know why I cant get the simplest problems!