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Home Q & A Board Science-Related Homework Help Chemistry (Moderator: Laser_3)
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julia.k18 julia.k18
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11 years ago
Here is my question:  If a 0.5 M solution of certain weak acid (HA) with a pk of 6.3 is in equilibrium and the concentration of the deprotonated form is equal to that of the protonated form (in other words [A-] = [HA]), what is the pH of the solution?
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#1
11 years ago
The equilibrium reaction is:

HA Leftwards Arrow-> H+ + A-

If 0.5M of HA ionizes to the point that [HA]=[A-], it must be the case that [HA]=[A-]=[H+]=0.25M. The pH can then be calculated from its definition:

pH = -log[H+] = -log(0.25) = 0.602
wrote...
#2
11 years ago
What Dan A says is just fine, but this question bothers me. Weak acids with a Ka = 5 x 10^-7 do not ionize 50%.

Without the [A-] = [HA] stipulation, the pH would be 3.3, a pH much more in line with a weak acid than a pH of 0.6.

Oh well.
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