For example, if I have a solution of 50mM acetlycholine (pKa = 4.5), how would I go about calculating the pH of the solution?



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  • $\begingroup$ You wouldn't. In situations like this, you would typically have some other solutes at much greater concentrations, and it is them that will ultimately define the resulting pH. Otherwise, go for the Ostwald’s dilution law. $\endgroup$ – Ivan Neretin Jan 9 at 17:04
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    $\begingroup$ A pKa of 4.5 would be typical for a carboxylic acid group, but I don't see any functional groups that look like it (there is a quaternary amine and an ester group, neither one have acid/base chemistry in water). What else is in the solution, and what is the solvent? $\endgroup$ – Karsten Theis Jan 9 at 17:09

I am assuming in your example that there is only one solute. If you have more than one then ionisation is mutually suppressed and things are a bit more complicated.

A useful expression that you can derive from an ICE table for a weak acid is:

$\qquad \qquad\qquad\qquad\qquad\rm pH=\dfrac{1}{2}[pK_a-\log{c}]$

Where c is the concentration of the acid which we assume to be just about equal to the equilibrium concentration.

This is reasonable for $\rm pK_a$ values between 4 and 10.

Putting in the numbers:

$\rm pH=\dfrac{1}{2}[4.5-(-1.3)]$

$\rm \underline{pH=2.9}$

Acetylcholine has a quaternary nitrogen so I can only assume that, given that the $\rm pK_a$ value is correct, the deprotonisation occurs at a different site on the molecule.

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    $\begingroup$ How did you get log(0.05) = -0.7? $\endgroup$ – Karsten Theis Jan 9 at 18:42
  • $\begingroup$ Oops ! Sorry about that. Should be -1.3. Have corrected answer. $\endgroup$ – Michael F. Jan 9 at 20:01

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