Asked by Emily
Calculate the approximate pH of a 0.1 M solution of ethanoic acid at 298 K, given that the
dissociation constant, Ka, is 1.75 x 10-5 mol/L at 298 K.Ethanoic acid dissociates in aqueous solution as follows:
CH3COOH(aq) + H2O H3O+(aq) + CH3COO-(aq)
dissociation constant, Ka, is 1.75 x 10-5 mol/L at 298 K.Ethanoic acid dissociates in aqueous solution as follows:
CH3COOH(aq) + H2O H3O+(aq) + CH3COO-(aq)
Answers
Answered by
DrBob222
You need to find the arrow key on your keyboard and use it. I will use acetic acid (ethanoic acid) as HAc.
.........HAc + H2O ==> H3O^+ + Ac^-
I........0.1...........0........0
C.........-x...........x........x
E.......0.1-x..........x........x
Substitute the E line into Ka expression and solve for x = (H3O^+) and convert that to pH.
.........HAc + H2O ==> H3O^+ + Ac^-
I........0.1...........0........0
C.........-x...........x........x
E.......0.1-x..........x........x
Substitute the E line into Ka expression and solve for x = (H3O^+) and convert that to pH.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.