A galvanic cell is based on the following half-reactions:

Question

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 10-3 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
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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

Answer 1

If you have Q you are half way home.
Q = 4.45E-3 = 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.......1.00-y......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.

Answer 2

How did you get your q value?