How does molarity affect Ka?

↳Redirected from "Suppose that I don't have a formula for #g(x)# but I know that #g(1) = 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
1 Answer
Dr. Hayek
Oct 19, 2015

No effect.

Explanation:

#K_a# is the equilibrium constant of the dissociation of an acid.

In general, the acid dissociation is written as:

#HA(aq) rightleftharpoons H^+(aq) + A^(-)(aq)#

The expression of #K_a# is: #K_a = ([H^+][A^-])/([HA])#

The concentrations in the expression of #K_a# are the equilibrium concentration, they change however, their ratio remains constant, and therefore, the name of constant #K_a#.

The ONLY factor that affect #K_a# is temperature, and therefore there is no influence of concentration change on the value of #K_a#.

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