[fredsmith3]

We  recently had an exam in my class  and I am not thoroughly convinced yet about the supposed correct answer to this question:

Which aqueous liquid is the least acidic?

.1 M  Al(NO3)2,  Ka=1E-5
.1 M Be(NO3)2,  Ka=4E-6
.1  M Pb(NO3)2, Ka= 3E-8
.1 M Ni(NO3)2, Ka= 1E-10
Pure Water


I personally  chose .1 M Ni(NO3)2 because it has  the smallest Ka.
My professor makes the case that  the answer  is pure water  because  the other choices (namely Ni(NO3)2) form hydrated metal ions in water.  However, as I was looking in my  book, I found that these Ka values actually  describe the Ka values for the hydrated metal ions themselves.  Who is correct here?

[UG]

Look at it from another perspective, what is the K_a of pure water?

[fredsmith3]

Look at it from another perspective, what is the K_a of pure water?

1E-7, that is still more than choice  d, though. 

[UG]

What makes you say the K_a of pure water is 1E-7? I can assure you it is not.

[fredsmith3]

I see now!  Ka = 1.8E-16
So it has nothing to do with the hydrated ion, but it has to do with the Ka of the water.

Thanks!

[Dan]

I see now!  Ka = 1.8E-16

Not for pure water it isn't... What is the source of this value?

[fredsmith3]

(1*10^-7)(1*10^-7) / 55.6 = 1.8*10^-16

[Dan]

[H₂O] is assumed constant and is already part of K_a:

http://www.chembuddy.com/?left=pH-calculation&right=water-ion-product

[fredsmith3]

I'm confused then, does that mean the Ka is 1E-7?

[Dan]

No, K_a(H₂O) = K_w = [H⁺][HO^-] = 10^-14

[Big-Daddy]

To back up Dan's logic is the fact that [H₂O] is actually present in every Ka expression, just that we remove it because it's taken to be constant. e.g. K_a=[H₃O⁺]*[A^-]/([H₂O]*[HA])≈[H₃O⁺]*[A^-]/[HA]. So we can also exclude H₂O from Kw and it will still be a fair comparison.