Question

How would you calculate the pH of 0.268 M CCl3COOH?

pKa = 0.52

Answer 1

With difficulty, inasmuch you neglected to include the acid dissociation constant, `K_a`, of `Cl_3C(=O)OH`. I got `pH` `=` `0.804`

Explanation:

From this site I learn that `K_a`, `Cl_3C(=O)OH` `=` `2.2xx10^-1`.

This is fairly large as an organic acid, and likely the method of successive approximations will be inadequate:

`Cl_3C(=O)OH + H_2O rightleftharpoons Cl_3C(=O)O^(-) + H_3O^+`

So `K_a` `=` `2.2xx10^-1` `=` `([H_3O^+][Cl_3C(=O)O^(-)])/[[Cl_3C(=O)OH]]`

If I put `[H_3O^+]=x`, then `K_a` `=` `2.2xx10^-1` `=` `x^2/(0.268-x)`

Solving this quadratic equation, I get two solutions: `x_1=0.157`; `x_2=-0.377`. I had to use the quadratic equation to get this answer.

Clearly, `x_1=0.157` is the required answer.

Thus `[H_3O^+]` `=` `0.157` `mol*L^-1` `=` `[Cl_3C(=O)O^(-)]` AND `[Cl_3C(=O)OH]` `=` `(0.268-0.157)*mol*L^-1` `=` `0.111*mol*L^-1`.

Thus `pH` `=` `-log_10{0.157}` `=` `0.804`

Please check the validity of this calculation. I am still using a web based calculator because my calculator (which has this programmed) has gone walkabout. Note that the `pH` of this solution is low because trichloroacetic acid is strongish, certainly stronger than acetic acid. As far as I know this is an entropy rather than a enthalpy phenomenon. Trichloroacetate is a much less polarizing ion than acetate, and is entropically favoured.