What is the percent ionization of a solution prepared by dissolving 0.0286 mol of chloroacetic acid in 1.60 L of water? For chloroacetic acid ,Ka=1.4x10^-3

Answer 1

Chloroacetic acid's formula is too long to write. Let's just call it HAcl.
(HAcl) = 0.0286mol/1.60L = about 0.018 but you do it more accurately.
........HAcl ==> H^+ + Acl^-
I.......0.018.....0......0
C........-x.......x.......x
E......0.018-x....x......x

Ka = (H^+)(Acl^-)/(HAcl)
Substitute Ka and the E line into Ka expression and solve for x = (H^+). Then
%ionization = (H^+)/(0.018)
Don't forget to redo the 0.018 number.

Answer 2

What you mean redo 0.018?