a 0.250 mol sample of HX is dissolved in enough water to form 655mL of solution. If the pH of the solution is 3.44, what is the Ka of HX.

Answer 1

(HX) = moles/L = 0.250/0.655 = approx 0.4 but you need a better answer than that.
.............HX ==> H^+ + X^-
initial.....0.4M.....0......0
change.......-x......x......x
equi.....0.4-x.......x.......x

Convert pH 3.44 to H^+ from
pH = 3.44 = -log(H^+)
That will be x, substitute into the Ka expression and solve for Ka.

Answer 2

[HX] = .250moles/(655/1000L)
=.38M
HX<=====>H+ + x-
I .38 0 0
C -X +x +x
E .38-x x x
pH = 3.44 to find H+
-log[H+] = pH = H= 3.63x 10^-4
Ka = [H+][x-]/[HX]
= [3.63x10^-4]^2/3.8x10^-1
= 3.5x10^-9