I'm quite a newbie in chemistry and I'm trying to learn how logarithmic diagrams work. I have been told that you can calculate pKa and pH graphically:
- pKa, when [HA] = [A-]
- pH, when [H+] = [A-]
Why does this happen?
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3$\begingroup$ $\mathrm{p}K_{\mathrm{a}} = \mathrm{pH}$ when $\ce{[HA]} = \ce{[A-]}$. This is easy to see from the form of the Henderson-Hasselbalch equation. So, we the concentration of hydronium is equal to concentration of conjugate base, you can read off the $\mathrm{p}K_{\mathrm{a}}$ from whatever the $\mathrm{pH}$ reads. It's not clear what you're second statement means... $\endgroup$ – Zhe May 16 '17 at 21:24
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You should state the charge balance or the proton balance.
[H3O+] = [OH-] + [CH3COO-]
We take the logarithm of this and get: log [H3O+] = log([OH-] + [CH3COO-]).
Now, you can plot this in your nice diagram.
log [H3O+] will follow the straight line in your diagram, i.e. log [H+].
The line for log([OH-] + [CH3COO-]) will at very high pH follow the line for log([OH-], but quite soon it will bend off and only follow the line for log [CH3COO-]. This is because [OH+] << [CH3COO-] at lower pH.
When [OH+] << [CH3COO-], you get log [H3O+] = log [CH3COO-]). When the line for log [H3O+] crosses the line for log [CH3COO-], you can calculate the pH of the solution from the diagram. The lines are crossing each other at right angel: pH = (1.00 + 4.65)/2 = 2.82(5). Very good, I am encouraging you to use logarithmic diagrams for solving acid-base equilibria.
