At what pKa value do we consider an acid strong?
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 &quot;Strong&quot; 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:
$$\ce{HA (aq) + H2O (aq) &lt;=&gt; A- (aq) + H3O+ (aq)}.$$
The equilibrium constants for this reaction is
\begin{align}
K  &amp;=       \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} &amp;=
    \frac{c(\ce{A-})\,c(\ce{H3O+})}{c(\ce{HA})}\\
\mathrm{p}K_\mathrm{a} &amp;= -\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) &amp;= \infty&amp; 
&amp;\therefore&amp;
\lim_{c(\ce{HA})\to0}\left(\mathrm{p}K_\mathrm{a}\right) &amp;= -\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 $\mathrm{p}K_\mathrm{a}(\text{strong acid}) &lt; -0.7$ (in water). Taking this literal, however, might lead to many misconceptions, and using a relative approach of &ldquo;this acid is stronger than this acid&rdquo; should  be preferred in any case.