1.6gram sample with a weak acid HX (82g) is dissolved in 60ml of H2O, titrated with .250M NaOH. When 1/2 of the HX is neutralized the pH is 5.0; and at the equivalence point the pH is 9.0. Find the % of HX in the original sample and the Ka of HX.
Even with help i was only able to get the Ka=1.0E-5 (which is correct) but my % was 51.3% (which is wrong).
My steps:
HX --> H+ + X-
Ka=[H+] [X-] / [HX]
@ 1/2 neutralization
Ka= [H+][1/2X-] / [1/2HX]
Ka = [H+] and since pH=5 @ 1/2 neutralization
pH = -log [H+]
5 = -log [H+]
[H+] = 1E-5
Ka = 1E-5
then after this point my answers start going wrong. I found the Ka but don't know how to get the %HX.
Can anyone please help me on how to approach this. You don't even need to do the answer, just explain to me how to think going about the problem. Thanks