### K and K(sub)a

Posted:

**Sun Feb 12, 2017 7:08 pm**What's the difference between K and K(sub)a in the context of concentration of solutions?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=139&t=19319

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Posted: **Sun Feb 12, 2017 7:08 pm**

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

Posted: **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.

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.

Posted: **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 K_{A} 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 K_{a} = ([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}).

Because we know that HF is a weak acid, we know that (1) it dissociates to H