Talk:Inorganic nonaqueous solvent

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[edit] The molarity of pure water is NOT equal to one!

The expression for the acid dissociation constant of hydronium is [H⁺][H₂O]/[H₃O⁺], or, because [H⁺]=[H₃O⁺], simply [H₂O], the molarity of pure water. This is approx. 55.5 or 55.6, molarity is moles per liter, not moles per mole. Hence, pKa of hydronium is about -1.74, NOT zero. Because a strong acid must be stronger than the protonated solvent, an acid is only strong in the proper sense if its pKa is less than -1.74. Also, because K_a * K_b = K_w, the pKa of water is 14.00 - (-1.74) = 15.74, NOT 14.00! It is a very common error (I'll even confess, I used to make it frequently!) to assume that [H₂O] = 1 without even thinking about it. BE CAREFUL, Physchim62 and all others! -User: Nightvid

The problem is that pK_a values of acids in aqueous solutions are quoted according to the convention that a(H₂O) = 1. This causes a systematic difference with the "true" thermodynamic values, as you point out, but this difference is the same for all acids considered. What is not valid is to compare values calculated on the basis that a(H₂O) = 1 (that is the overwhelming majority of published values) with those calculated according to a different convention. You should also note that [H⁺] in aqueous solution is negligeably low, and that the acidity of aqueous strong acids is due to H₅O₂⁺. Physchim62 (talk) 16:16, 16 March 2006 (UTC)

The convention of omitting [H₂O] in the standard general expressions of K_a and K_b cannot be generalized in such a manner without leading to inconsistencies. Sure, the convention says that we omit [H₂O] when it appears in the general complete equilibrium constant expression (once in the denominator only in the case of K_a). But just as the K_a of an acid is not the same as the K_eq, which does NOT omit [H₂O], the K_a of hydronium (ok, to split hairs, this means "hydrated hydronium", [H₃O(H₂O)_n]⁺ for our purposes here) is not one, but rather the molarity of water. You could make a special rule and define it to be 1, but if you did, the comparison to another acid would become invalid, since you would be leaving out the [H₂O] on the numerator of the expression as well as the denominator, while on the other acid this doesn't happen. So sure, if you want to insist that pK_a of hydronium is zero, fine. But then you can't turn around and say that pK_a < 0 makes a strong acid because the omission of the [H₂O] in the numerator means you're comparing apples to oranges. The assumption that a lower pK_a means a stronger acid only is valid if the expressions are consistent with each other. An acid in aqueous solution is a better proton donor than hydrated hydronium ONLY IF pK_a < -1.74. Also, as mentioned earlier, the pK_w of water is 14.00 since both of the [H₂O] terms in the denominator of K_w are omitted. But by the general definition of pK_a and pK_b, which only omit ONE [H₂O] in the denominator of K_a or K_b, those must be 14.00 + log₁₀[H₂O] = 15.74. And the pK_eq, because it doesn't omit either [H₂O] in the standard general expression of K_eq, is 17.48 (14.00+2 log₁₀[H₂O]). Have I explained it understandably now? -User:Nightvid

[edit] Transforming

couldnt we add so that u can get the self ionization of any solution if u know its Ka or Kb? and also so that if we know its Ka we can figure out its Kb