# How would you calculate the pH of 0.268 M CCl3COOH?

## pKa = 0.52

Apr 7, 2016

With difficulty, inasmuch you neglected to include the acid dissociation constant, ${K}_{a}$, of $C {l}_{3} C \left(= O\right) O H$. I got $p H$ $=$ $0.804$

#### Explanation:

From this site I learn that ${K}_{a}$, $C {l}_{3} C \left(= O\right) O H$ $=$ $2.2 \times {10}^{-} 1$.

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

$C {l}_{3} C \left(= O\right) O H + {H}_{2} O r i g h t \le f t h a r p \infty n s C {l}_{3} C \left(= O\right) {O}^{-} + {H}_{3} {O}^{+}$

So ${K}_{a}$ $=$ $2.2 \times {10}^{-} 1$ $=$ $\frac{\left[{H}_{3} {O}^{+}\right] \left[C {l}_{3} C \left(= O\right) {O}^{-}\right]}{\left[C {l}_{3} C \left(= O\right) O H\right]}$

If I put $\left[{H}_{3} {O}^{+}\right] = x$, then ${K}_{a}$ $=$ $2.2 \times {10}^{-} 1$ $=$ ${x}^{2} / \left(0.268 - x\right)$

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 $\left[{H}_{3} {O}^{+}\right]$ $=$ $0.157$ $m o l \cdot {L}^{-} 1$ $=$ $\left[C {l}_{3} C \left(= O\right) {O}^{-}\right]$ AND $\left[C {l}_{3} C \left(= O\right) O H\right]$ $=$ $\left(0.268 - 0.157\right) \cdot m o l \cdot {L}^{-} 1$ $=$ $0.111 \cdot m o l \cdot {L}^{-} 1$.

Thus $p H$ $=$ $- {\log}_{10} \left\{0.157\right\}$ $=$ $0.804$

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 $p H$ 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.