- #1

njh

- 19

- 4

- TL;DR Summary
- Where am I making a mistake in my understanding of the HH equation?

I understand the HH equation as:

##pH = pKa + log\frac{[A^-]}{[HA]}##

I think that somewhere above I am making a mistake, but I do not know what:

##pH = pKa + log\frac{[A^-]}{[HA]}##

- The Numerator of the fraction is the conjugate Base of the original acid.
- The denominator is the undissociated/ unprotonated Acid.
- I understand that when the pH equals the PkA, then the solution is 50% dissociated, so for example 1 Mol of conjugate Base and 1 Mol of undissociated Acid. Therefore, if pH = 2 and pKA = 2, then the solution is 50% dissociated, or ##log\frac{1}{1}####=0##.
- I know that if the pH fell from 2 to 1, and the pKa remains 2, then the lower pH means a higher level of protonation, and so more conjugate Base and less undissociated Acid. The ratio of ##\frac{[A^-]}{[HA]}## falls from ##\frac{1}{1}## to ##\frac{1}{10}##
- I know that if the pH of a solution falls, then it becomes a stronger protonator, thereby distributing more H ions.

I think that somewhere above I am making a mistake, but I do not know what:

- If a stronger acid protonates more, then this results in less undissociated Acid and more conjugate Base (the Acid distributes Hydrogen ions and converts to its conjugate Base). I would expect that to result in an INCREASED ratio of ##\frac{[A^-]}{[HA]}## i.e. MORE CONJUGATE BASE.
- In my example above, lowering the pH leads to the ratio of ##\frac{[A^-]}{[HA]}## falling from 1 to 0.1 (##\frac{1}{1}## to ##\frac{1}{10}##), which means the concentration of conjugate Base REDUCES (or the concentration of undissociated Acid INCREASES).