Question

Enough of a monoprotic acid is dissolved in water to produce a 0.0190 M solution. The pH of the resulting solution is 2.33. Calculate the Ka for the acid.

Answer 1

pH = -log (H^+)
2.33 = -log(H^+)
(H^+) = about 0.005 which is only an estimate.
..........HA ==> H^+ + A^-
I......0.0190....0......0
C.......-x.......x......x
E...0.0190-x.....x......x

Ka = (H^+)(A^-)/(HA)
You know the value of x, substitute those values into the Ka expression and solve for Ka.

Answer 2

I know how to do the ice chart and set up i just don't know how to get the x value that you are saying that i know...