[antimatter101]

Hello, everyone

I tried calculating the pH of a 1M solution of fluoroantimonic acid (strongest acid in the world). Its pKa is 10²⁵, which is a huge amount.

First, can somebody tell me how to insert a picture?

[antimatter101]

NOw i know how to upload pictures

[Arkcon]

Now that you have the basics down, you can add value to the images by cropping out white space or worse, the program's border.  And definitely crop out the edges of your screen wallpaper -- some people don't do that.  And maybe clean up the image -- reduce colors to just black and white.

[magician4]

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 c₀ of acid HX and purred it into water, so it would partially *⁾ dissociate:
c₀(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] = c₀(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 c₀(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 HSbF₆ , HI  or HClO₄ 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: 10²⁵) is all about