What did i do wrong here?
in my opinion, you somewhat lost directions whilst working the equations...
... mostly because there seems to be confusion between "initial value for acid" and "value of acid at equilibrium", it seems to me
my proposal would be:
let's set up the general
law of mass action expression for a 1-protonic acid HX
[tex] K_a \ = \ \frac {[X^-] \ \cdot \ [H^+]}{[HX]} [/tex]
now, [X
-] and [HX] are concentrations
at equilibrium.
how did we get there?
we started with an
initial concentration c
0 of acid HX and purred it into water, so it would
partially *
) dissociate:
c
0(HX) = [HX] + [X
-]
"the initial concentration of acid will be distributed between the equilibrium concentration of the named acid plus the equilibrium concentration of its corresponding anion"
form this, it follows directly that [HX] = c
0(HX) - [X
-]
furthermore , for each [X
-] generated, there will be a [H
+] generated, too, so we can argue [H
+] = [X
-]
let's introduce this to our initial expression:
[tex] K_a \ = \ \frac {[X^-] \ \cdot \ [H^+]}{[HX]} \ = \ \frac {[X^-] \ \cdot \ [H^+]}{c_0(HX) \ - [X^-]} \ = \ \frac {[H^+] \ \cdot \ [H^+]}{c_0(HX) \ - [H^+]}[/tex]
solving this expression for [H
+] results in:
[tex] \ [H^+] \ = \ - \frac {K_a}{2} \ + \ \sqrt { \left( \ - \frac {K_a}{2} \right)^2 \ + \ c_0 \ \cdot \ K_a} [/tex]
with the solution [H
+] = 0.9999999... [itex] \approx [/itex] 1 (if we inserted the values for your real problem here)
... leading to the revelation, that a strong acid for all practical purposes will be (sufficiently) dissociated to call it completely dissociated, with c
0(HX) becoming [H
+]
hence, even for the strongest acid around, you can't have more protons than your initial concentration of HX has to offer in total, and hence in effect there would be no difference with respect to the pH resulting thereof if we talked HSbF
6 , HI or HClO
4 or HCl , respectively, at same initial concentrations: all would result in pH = 0 , in this case
regards
Ingo
*
)note: even with next-to-complete dissociation, there still will be one or another molecule around, that didn't dissociate.
that's what a K
a-value [itex] \neq \infty [/itex] (in your case: 10
25) is all about