K and K(sub)a

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Carla Cruz Medina 1L
Posts: 12
Joined: Fri Sep 25, 2015 3:00 am

K and K(sub)a

Postby Carla Cruz Medina 1L » Sun Feb 12, 2017 7:08 pm

What's the difference between K and K(sub)a in the context of concentration of solutions?

Cinthia_Ramirez_3B
Posts: 10
Joined: Wed Sep 21, 2016 2:58 pm

Re: K and K(sub)a

Postby Cinthia_Ramirez_3B » Sun Feb 12, 2017 9:09 pm

From the solution, it appears that K is for 2 moles 2 HF (aq) ->2 H+(aq) + 2F-(aq) while k(sub)a is for 1 mole HF (aq) ->2 H+(aq) + 2F-(aq)

K(sub)a = ([H+][F-])/[HF].

K = ([H+]^2*[F-]^2])/[HF]^2

That is why in the end you take the square root of K to get K(sub)a since K has the products and reactants squared.

Celina N 2H
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Re: K and K(sub)a

Postby Celina N 2H » Mon Feb 13, 2017 10:44 am

There is a good example of this in the Winter 2014 Midterm, Question 8, where you are asked to find the value of KA for HF using given standard cell potentials.

Because we know that HF is a weak acid, we know that (1) it dissociates to H+ (aq) and F- (aq), and (2) it has an acid dissociation constant of Ka = ([H+][F-])/([HF]). K, the equilibrium constant, is constant for the complete and final balanced redox reaction: the final reaction is 2HF (aq) --> 2H+ (aq) + 2F- (aq), which gives K = ([H+]2[F-]2) / ([HF]2).


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