[vande060]

My book asks me to show calculations proving that pKa of H3O+ is -1.74

the way i tried to go about this

H30+ + H20 --> H3O+ + H20

[H3O+] [H2O]
-------------
[H3O+]

since H3O+ is equal to H2O in the numerator, the the Ka of water equals the Ka of H2O(55.5)

thus

-Log(55.5 Ka) = -1.74 = pKa

is my logic correct here?

[Smrt guy]

My book asks me to show calculations proving that pKa of H3O+ is -1.74


since H3O+ is equal to H2O in the numerator, the the Ka of water equals the Ka of H2O(55.5)

This makes no sense.  If you use the Henderson-Hasselbach equation:

pH = pK_a + log([H₂O]/[H₃O⁺])

Since pH = -log[H₃O⁺] = pK_a + log[H₂O] - log[H₃O⁺],

pKa = -log[H₂O].

However, I'm not sure [H2O] should be considered as the normal (55.5 M) under these conditions.  Clearly that is how you get the pK_a of -1.74 though.