calculate concentration?

Question

Ok, so i did this equilibrium problem, but the answer doesnt seem right!

If a given weak acid HX has a Ka = 8.3x10^-3. what is the H+ concentration of a solution of HX that has a concentration of 0.0057 mol/L?
*what is that pH of the solution of HX?
*what is the percent ionization of HX?

So i did the first part out, and i got 0.0068 M [H+]
which doesnt make sense because it is bigger than the original concetnration of HX?
is there a mistake in the problem?
and then when i get to the % ionization, it would be 0.0068/0.0057 and then the answer is like 102%!!
Its a weak acid???
is this problem wack??

Answer 1

.. .. .. .. .. .. .. .. HX <=> H+ + X-
Initial .. .. .. .. .. C
React .. .. .. .. .. y
Produce .. .. .. .. .. .. .. ..y .. .. y
At Equil. .. .. .. C-y .. .. .y .. .. y

Ka= [H+][X-] /[HX] = y^2/(C-y)
C=0.0057 =5.7*10^-3 M, so
y^2 / (5.7*10^-3 -y) = 8.3*10^-3

Let's do the assumption (which I am pretty sure that it will fail) that y<< 5.7*10^-3 so that 5.7*10^-3 -y=5.7*10^-3
then y^2/ 5.7*10^-3 = 8.3*10^-3 =>
y= squareroot (8.3*5.7*10^-6) = 0.00687 =0.0069 (be careful how you round the numbers) which is bigger than the initial amount and of course it is wrong. This simply means that the assumption is not correct and we can find the correct answer by solving the quadratic.
Note that I expected that the assumption would not work by looking at C and Ka values. The assumption works for high C (>0.01) and very low Ka (<10^-4) values.

y^2 / (5.7*10^-3-y) = 8.3*10^-3 =>
y^2 = 4.371*10^-5 -0.0083y =>
y^2 +0.0083y -4.371*10^-5=0
y1= 0.0036559 = 0.0037
y2= -0.011956<0 rejected

Thus pH= -logy= -log(0.0037) = 2.43

a=y/C = 0.0037/0.0057 =0.649 =65%