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First, I know that strong acids are assumed to dissociate completely in water, so that the concentration of $\ce{H+}$ which they provide equals their concentration.

What I am asking for is not the concentration of $\ce{H+}$ which they provide, but $\ce{[H+]}$ for the solution, since the $\ce{H+}$ ions that they provide will not stay aloof, but rather will react with water and with $\ce{OH-}$ ions that are produced in the self-ionization of water.

Won't the reaction of $\ce{H+}$ ions with water shift the self-ionization of water $$\ce{H2O <=> H+ + OH-}$$ to the left and won't their reaction with $\ce{OH-}$ shift it to the right according to Le Châtelier’s principle?

So, what will $\ce{[H+]}$ be for the solution?

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    $\begingroup$ Autoionization of water (though it certainly does happen) is quite negligible here. $\endgroup$ – Ivan Neretin Sep 26 '18 at 10:33
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    $\begingroup$ @VicLineal Thanks for polishing this post. For future reference, we prefer to not use MathJax in the title field, see here for details. So while you're editing a question, please remove it. $\endgroup$ – Martin - マーチン Sep 26 '18 at 12:04
  • $\begingroup$ Is this a case of common ion effect? $\endgroup$ – rv7 Sep 28 '18 at 14:43
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The statement $[\ce{H^+}] = [\ce{A^-}]$ is an approximation for not too diluted acids. However you can pretty fast tell that this won't hold if you add a strong acid to water so that the acid has a concentration of let's say $\pu{10^-8 mol//l}$. Via the formula $\mathrm{pH} = \log\left([\ce{H^+}]\right)$ this would result in $\mathrm{pH} = 8$ so if that was true, you just made the solution less acidic by adding an acid.

In fact the total concentration $[\ce{H^+}]$ is given by $[\ce{H^+}] = [\ce{A^-}] + \pu{10^-7 mol//l}$ where the last term stems from the autoionization of water. However in most use-cases $[\ce{A^-}]$ is far greater than $10^{-7}$. In those cases the autoionization of water can be neglected and thus the approximation $[\ce{H^+}] = [\ce{A^-}]$ gives sufficient results.
However, if you are working with really diluted solutions of acids this approximation is no longer applicable (as demonstrated above) which forces you to actually respect the autoionization of water.

If you want to be on the safe side you can always respect the autoionization. This will always give you the correct pH (even for not so diluted solutions) but you may neglect it if the acid-concentration is high enough (this of course depends on the accuracy of the results you need).

EDIT:
As @MaxW pointed out in the comment a general expression would be $[\ce{H+}] = [\ce{A-}] + [\ce{OH-}]$ which for a neutral solution at 25°C (thanks to @Martin for pointing this out) becomes the formula proved above.

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    $\begingroup$ Please note, that the value for the auto-ionisation is only true at about room temperature, where $\mathrm{p}K_\mathrm{w}^{\pu{25 ^\circ C}} = 14$ $\endgroup$ – Martin - マーチン Sep 26 '18 at 12:16
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    $\begingroup$ RE: "In fact the total concentration $[\ce{H^+}]$ is given by $[\ce{H^+}] = [\ce{A^-}] + \pu{10^-7 mol//l}$" --- This is not the correct expression. It should be $\ce{[H+] = [A-] + [OH-]}$ $\endgroup$ – MaxW Sep 26 '18 at 14:27
  • $\begingroup$ @MaxW Well I was assuming a neutral solution... Though in a general case you are of course correct. I'll add your comment to my answer. $\endgroup$ – Raven Sep 26 '18 at 16:46
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    $\begingroup$ The solution can't be neutral and acidic at the same time. Yes the solution can start out neutral, but after adding the acid $\ce{[A-] > [H+]}$ because some of the $\ce{H+}$ has reacted with the $\ce{OH-}$ from the autoionization of the water. $\endgroup$ – MaxW Sep 26 '18 at 22:50
  • $\begingroup$ What about the reaction of some hydrogen ions with some hydroxide ions? Won't this reduce the amount of the hydrogen ions? $\endgroup$ – Mason Sep 27 '18 at 5:43
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For "strong" acids the usual assumption is that the acid is completely ionized. So rather than being an equilibrium

$$\ce{HA -> H+ + A-}$$

Also this means that the conjugated base does not protonate.

$$\text{NO! }\quad\ce{A- + H+ -> HA}$$

The autoionization of is given by the reaction $$\ce{H2O <=> H+ + OH-}$$

The addition of $\ce{H+}$ ions from the acid react with $\ce{OH-}$ ions from the water to shift the reaction to the left according to Le Châtelier’s principle. Most pKa's are only good for two significant figures, so in most cases (eg when pH > 5) you can assume $\ce{[A-] = [H+]}$ for strong acids. But since the charges in the solution must balance the exact expression is

$$\ce{[H+] = [A-] + [OH-]}$$

Nominally the equilibrium for the autoionization of water would be given by

$$K_{eq} = \frac{\ce{[H+][OH-]}}{\ce{[H2O]}}$$

But the assumption is that for aqueous solutions $\ce{[H2O]}$ is always $55.5 \text{ molar}$ so instead the expression used is:

$$K_w = 1.0\times10^{-14} = \ce{[H+][OH-]}$$

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