help would be greatly appreciated!

[kazuma]

this question is out to anyone, please help if you get the time.

it says 10 mols of a solution was added and the resulting pH was 5.7

what is the correct relationship between the Ka of the protonated form and the Kb of the deprotonated form.

the answer is Ka < Kb.

I'm having a little trouble with this, from my understanding the greated the Kb the stronger the base, meaning it will cause an abundance of OH- ions in the solution. Now if the Kb of the deprotonated form is great, than OH- molecules will be in solution in a greater ammount. but shouldn't that increase the pH?

[kazuma]

one other way i was looking at it, that might make some sense is this..

A- + H2O = HA + OH-

now since the Kb is great, the equation lies to the right. and also, since HA is the acid of a weak base, it will form a complex with OH- from the solution.

Much like NH4+ with OH-, making NH4+OH, and thus cause H+ to be left behind from the water. kind of a stretch, but plausible? suggestions ?

[MANDALAM]

Kazuma,

When you add 10(or whatever) moles of solute and the pH becomes slightly acidic (eg. 4<pH<7) you are dealing with a weak acid. Therefore, the acid has a Ka which is less than 1. However the reverse reaction is still less favourable than the forward reaction. The Kb here refers to the equilibrium constant of the reverse reaction and not much more. Think of it in terms of carbonic system in the blood.

Hope that helped.

[kazuma]

thx for the reply Mandalam, but if what you are saying is correct, why is the Kb of the anion greater than the Ka of the acid, doesnt that mean the reverse reaction is more favorable. Also, if it is a weak acid, in solution, doesnt it cause the solution to become more alkaline?

Kb of the anion means Kb = [OH-] [HA] / [A-], which would in turn make it more alkaline.

the answer says Ka of HA < Kb of A-

Im still kinda lost =/

[kazuma]

maybe im overthinking this, tell me if this is consistent with what you are saying;

the Ka is small, hence the HA disassociates very little (slightly less pH than 7). Since the Ka is so little, driving the forward reaction minutely, the reverse reaction has a bigger equilibrium value. if this is true, than I need to simplify things a bit more =[.


Mandalam, thx a bunch for the help, if you could, can you please answer the question I asked about weak acids, noone replied to it. its in this same gen chem section. thx a bunch!!