Question

A solution of a weak acid HA has initial concentration c and acid ionization constant Ka. To what concentration should the acid be diluted to make [H3O+] half of what it was? Answer in terms of c and Ka.

Answer 1

First, assume the acid:

HF

to be HA, for simplicity, so it will ionize as follows:

HA <-> H+ + A-

where, H+ is the proton and A- the conjugate base, HA is molecular acid

Ka = [H+][A-]/[HA]; by definition

initially

[H+] = 0

[A-] = 0

[HA] = c;

the change

initially

[H+] = + x

[A-] = + x

[HA] = - x

in equilbrirum

[H+] = 0 + x

[A-] = 0 + x

[HA] = c - x

substitute in Ka

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

x^2 + Kax - c*Ka = 0

"find" c given [H+] = 1/2c

(1/2c)^2 + Ka*1/2*c - c*Ka = 0

(C^2)/2 + Ka(c/2 - c) = 0

(C^2)/2 + -Ka*c/2 = 0

c^2 -cKa = 0

c-Ka = 0

c = ka

concentration must be equal to Ka