Question

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)

Answer 1

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.