Chemistry...please help
posted by Kaili .
A galvanic cell is based on the following halfreactions:
Fe2+ + 2e > Fe(s) E= –0.440 V
2H+ + 2e > H2(g) E= 0.000 V
where the iron compartment contains an iron electrode and [Fe2+] = 1.00 x 103 M and the hydrogen compartment contains a platinum electrode, PH2 = 1.00 atm, and a weak acid HA at an initial concentration of 1.00 M. If the observed cell potential is 0.258 V at 25 degrees C, calculate the Ka value for the weak acid HA

I calculated Q from the nernst equation and got 4.45 X 10^3 but i go from here/how do i get the Ka from this

If you have Q you are half way home.
Q = 4.45E3 = pH2*(Fe^2+)/(Fe)(H^+)^2
Substitute 1.00 for pH2, 0.001 for (Fe^2+), y for (H^+) and solve for x.
Then ...........HA ==> H^+ + A^
initial.......1.00.....0.....0
change........y.......y.....y
equil.......1.00y......y.....y
Ka = (H^+)(A^)/(HA)
You know (H^+) which is the x you solved for, you know HA and A^ = x also. Solve for Ka. 
How did you get your q value?