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.

The concentration c is equal to Ka

Explanation:

The acid will ionize as observed in the following reaction:

HA = H+ + A-

H+ is the proton of the acid and A- is the conjugate base . The equation to calculate the Ka is as follows:

Ka = ([H+]*[A -])/[HA]

Initially we have to:

[H+] = 0

[A-] = 0

[HA] = c

During the change we have:

[H+] = +x

[A-] = +x

[HA] = -x

During balance we have:

[H+] = 0 + x

[A-] = 0 + x

[HA] = c - x

Substituting the Ka equation we have:

Ka = ([H+]*[A-])/[HA]

Ka = (x * x)/(c-x)

x^2 + Kax - (c * Ka) = 0

We must find c, having as [H+] = 1/2c. Replacing we have:

(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 -(c*Ka) = 0

c-Ka = 0

Ka = c

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