1
$\begingroup$

In between the traditional strong mineral acid that dissociate completely in aqueous solution, and obvious weak organic acid like acetic acid, things get murky.

For example, oxalic acid has a $\mathrm{p}K_\mathrm{a} = 1.46$ which is about 1000 times stronger than acetic acid, but no one seems to qualify it as a "strong" acid. How about trichloroacetic acid, $\mathrm{p}K_\mathrm{a}\approx 0.5$? Well, some people start calling it "strong" acid.

So, what $\mathrm{p}K_\mathrm{a}$ value would qualify an acid as a strong acid?

I would like to know a concrete numbers, not merely “smaller $\mathrm{p}K_\mathrm{a}$ means stronger acid” as found in other questions. Just like a specific body mass of $90\ \mathrm{kg}$ would qualify a boxer to be a heavyweight boxer.

$\endgroup$
5
$\begingroup$

Your question is not as easy answered as you might want it to be. The reasons for that are mainly given in the answer by Zen to What is a "Strong" Acid? The main problem is, that there hardly can be an absolute acid strength and you will always have to take the reference system into account. Another problem arises, that there is no consistent (or official by IUPAC) definition of the acid constants. Usually, when you are talking about $\mathrm{p}K_\mathrm{a}$ values, then you are most likely already implying aqueous solution.

It is a quite accepted classification of strong acids (in water), to state that they completely dissociate. Hence in the equilibrium is entirely shifted to the right side: $$\ce{HA (aq) + H2O (aq) <=> A- (aq) + H3O+ (aq)}.$$ The equilibrium constants for this reaction is \begin{align} K &= \frac{a(\ce{A-})\,a(\ce{H3O+})}{a(\ce{HA})\,a(\ce{H2O})} \approx \frac{c(\ce{A-})\,c(\ce{H3O+})}{c(\ce{HA})\,c(\ce{H2O})} \approx \frac{K_\mathrm{a}}{c(\ce{H2O})}\\ K_\mathrm{a} &= \frac{c(\ce{A-})\,c(\ce{H3O+})}{c(\ce{HA})}\\ \mathrm{p}K_\mathrm{a} &= -\lg K_\mathrm{a}. \end{align}

For a strong acid we can see that the concentration of undissociated acid is (nearly) zero. Therefore we write \begin{align} \lim_{c(\ce{HA})\to0} \left(K_\mathrm{a}\right) &= \infty& &\therefore& \lim_{c(\ce{HA})\to0}\left(\mathrm{p}K_\mathrm{a}\right) &= -\infty. \end{align}

Strictly speaking, in aqueous solutions the acidity constants cannot be measured, since the hydronium ion is the strongest acid in this medium.

So in priniciple, the value that separates strong from weak acids would be the $\mathrm{p}K_\mathrm{a}$ of the hydronium ion. With that we arrive at a completely new problem, which is quite efficiently discussed here: What is the pKa of the hydronium, or oxonium, ion (H3O+)?

From this question we can take home the suggestion the value of $\mathrm{p}K_\mathrm{a}(\ce{H2O}) = -0.7$ for the purpose of comparing it to other acids. In other words $\mathrm{p}K_\mathrm{a}(\text{strong acid}) < -0.7$ (in water). Taking this literal, however, might lead to many misconceptions, and using a relative approach of “this acid is stronger than this acid” should be preferred in any case.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.