What is the $\mathrm{pH}$ of a $\pu{5M}$ solution of hydrochloric acid?
So in order to solve this, apparently all you need to do is plug it into the equation:
$$\mathrm{pH} = -\log[\ce{H3O+}]$$
where the concentration of $\ce{H3O+}$ is $\pu{5M}$. But why can we assume this? Don’t we need to create an ice table from $\ce{HCl + H2O-> H3O+ + Cl-}$ and use the $K_\text{a}$ value of $\ce{HCl}$ to solve this?
As explained in pH Paradoxes: Demonstrating That It Is Not True That pH ≡ -log[H+] Journal of Chemical Education Vol. 83 pages 752-757.
pH = -log (hydrogen ion activity).
and "pH of 7.6 M HCl is about -1.85 (not -0.88)"
So how far off is -log (5) = -0.69 from the real answer?
From this table of activity coefficients 5m HCl has an activity coefficient of 2.38, so activity is 11.9, which yields pH = -1.1.
Unfortunately the table is in terms of molality rather than molarity. 5M HCl is about 5.5m, and activity coefficient is dramaticly increasing with concentration (see table I of the pH paradox reference) so the true pH deviates somewhat further from -0.69, but pH = -1.1 is a reasonable estimate.
$$ \begin{array}{|c|c|c|c|c|}\hline &\text{Before}&&\text{After}&\\\hline &\ce{AH}&\ce{H2O}&\ce{A-}&\ce{H3O+}\\\hline \text{Initial}&n0&\text{excess}&0&0\\\hline \text{Equilbrium}&\epsilon&\text{excess}&n0&n0\\\hline \end{array} $$
Then we have $\mathrm{pH}=-\log\left[\ce{H3O+}\right]$ Now we have to verify if our result is true.
We consider that $\ce{H3O+}$ in the solution has exactly the same concentration as that of the acid. However they can react with $\ce{OH-}$ because of the water autoprotolysis! $K_\text{a}=10^{14}$ total reaction.
$$\ce{H3O^+ + HO^- -> 2H2O}$$
So the concentration of $\ce{H3O+}$ is approximatively constant if $\left[\ce{H3O+}\right]\gg\left[\ce{HO-}\right]$
For example if $\frac{1}{10}\left[\ce{H3O+}\right]>\left[\ce{HO-}\right]$ then we can find the limit for which our reasonning is false :
$$ \left[\ce{HO-}\right]\times \left[\ce{H3O+}\right] < \frac{1}{10} \times \left[\ce{H3O+}\right] \times \left[\ce{H3O+}\right] \iff K_\text{e}<\frac{1}{10}\left[\ce{H3O+}\right]^2 $$
Then we have $\mathrm{pH}<6.5$
So if with a strong monoacid you find $\mathrm{pH}=6.8$ with this formula the result is not correct and you have to make an other approximation or reasonning!
Here we have $\mathrm{pH}=-0.69<6.5$ so you find a reasonable answer. And in water $\mathrm{pH}$ is between $0$ and $14$ so your solution is at $\mathrm{pH} \approx 0$
There exists seven strong acids that completely disassociate with hydrogen ( $\ce{HCl}$, $\ce{HBr}$, $\ce{H2SO4}$, $\ce{HClO3}$, $\ce{HClO4}$, $\ce{HI}$, and $\ce{HNO3}$), so the only calculation you need to do is the negative log of the concentration (in molarity).
Therefore $\mathrm{pH} = -\log5.0=-0.69$. Yes, $\mathrm{pH} = −0.69$ even though the $\mathrm{pH}$ scale "normally" ends at $\ce{pH} = 0$.
(in case of $\ce{HCl}~\mathrm{pH}$ will be negative for any solution more then $\pu{2M}$)
Just to give a quick and logical answer, if you look at the following acids:
These acids above are Strong Acids, meaning that they completely dissociate, so for that purpose there is no need of a certain K value, since none of this is an equilibrium reaction.
Just a forward reaction. $\ce{->}$ not $\ce{<->}$
NOTE: This is usually for general purposes. Not to be super accurate.
