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I understand the Henderson-Hasselbalch equation for acids that do not completely dissociate but I am having trouble calculating the pH of a strong acid using the equation: for example a .25 m solution.

pH = p(Ka) + log([A-]/[HA])

which is 

pH = p([A-][H+]/[HA]) + log(.25/0)

which is approximately 

pH = p(.25^2/0) + infinity

and therefore 

pH = -infinity + infinity = 0

So why isn't the pH of any strong acid 0?

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    $\begingroup$ This is a question on math, not chemistry. In short, you just can't operate with infinities as if they were normal numbers, because they are not. $\endgroup$
    – Ivan Neretin
    Commented May 25, 2016 at 21:42
  • $\begingroup$ so we just arbitrarily define it to be the log of molar concentration of the strong acid? $\endgroup$
    – user2879934
    Commented May 25, 2016 at 21:47
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    $\begingroup$ We do not have to define anything special for this case, arbitrarily or not. The ordinary definition of pH still applies without problems. $\endgroup$
    – Ivan Neretin
    Commented May 25, 2016 at 21:58
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    $\begingroup$ People make the mistake that anything divided by 0 is infinity. In fact it is not defined. Besides that the Henderson-Hasselbalch equation comes from the equilibrium constant equation for the dissociation of a weak acid not a strong acid, so don't use it. $pH = -\log[H^+]$. So assuming you have a strong acid, you have complete dissociation of acidic Hs. So you now know the $[H^+]$ so simply get $-\log$ of it. Don't over complicate things :) $\endgroup$
    – Leeser
    Commented May 25, 2016 at 22:55
  • $\begingroup$ en.wikipedia.org/wiki/Henderson%E2%80%93Hasselbalch_equation especially "limitations" $\endgroup$
    – Mithoron
    Commented May 25, 2016 at 23:53

4 Answers 4

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The ${K_{a}}$ of a strong acid is not infinity. It is a defined number depending on solvent. Thus the assumptions you make do not hold and you don't have to worry about dealing with infinities.

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  • $\begingroup$ Indeed, that's true. $\endgroup$
    – Mithoron
    Commented May 25, 2016 at 22:56
  • $\begingroup$ Yeah that is true but then let's look at the the definition of ka = h a- / ha ....multiply both sides by ha and take the log we get log(ka) + log(ha) = log(h^2) since h and a- are equal... This still doesn't give the rift andwer $\endgroup$
    – user2879934
    Commented May 25, 2016 at 23:32
  • $\begingroup$ So it seems we just break from the normal rules $\endgroup$
    – user2879934
    Commented May 25, 2016 at 23:32
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    $\begingroup$ We very much do follow the normal rules. Broken, if anything, is your perception thereof. $\endgroup$
    – Ivan Neretin
    Commented May 26, 2016 at 5:35
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You are confusing infinitely strong acid with simply a strong acid.

For any real acid, there will be some undissociated acid (HA) present. While countless books say strong acids completely dissociate, this is just an approximation.

See Table of Acid and Base Strength for the Ka values of some common strong acids.

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One factor here is that of temperature. As the temperature rıses, ionızation ıncreases. pH ıs measured at room temperature. Heating the solutıon of a strong acıs wıll ıncrease ıonizatıon and drop the pH.

Melek

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One principle from mathematics is that you cannot say that two infinities are equal to one another. For example, while there are an infinite number of natural and real numbers, these two infinities are not the same. Here you can find a derivation of the Henderson–Hasselbalch equation. Since the Henderson–Hasselbalch equation bundles up the math in a way that is difficult to analyze, lets look at the original equilibrium equation, which you can derive from the Henderson–Hasselbalch equation simply by undoing all of the steps: $$k_a=\frac{[H^+][A^-]}{[HA]}$$ Since a strong acid dissociates completely, $k_a$ must be equal to $\frac{1}{0}$. In other words, the concentration of HA must be zero at equilibrium. Since we are doing no nuclear reactions here, this rule that my friends like to call "conservation of stuff" takes effect. In other-words, all of the strong acid that you started with must have gone somewhere (hint: it dissociated into $\ce{H+}$ and $\ce{A-}$). Now, to calculate the final pH of a strong acid solution, we use the number of protons that dissociated to find the concentration of $\ce{H+}$ and from there, the actual pH.

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  • $\begingroup$ The statement "the concentration of HA must be zero at equilibrium" is incorrect. Ka is 24 for nitric acid. For Ka of other strong acids see depts.washington.edu/eooptic/links/acidstrength.html The undissociated fraction depends upon Ka and concentration. $\endgroup$
    – DavePhD
    Commented Jun 13, 2016 at 15:14
  • $\begingroup$ I was using the assumption that a strong acid dissociates completely, which I know is not actually the case. If it dissociates completely, then the concentration of HA would be zero. $\endgroup$
    – Niels Kornerup
    Commented Jun 13, 2016 at 15:23
  • $\begingroup$ So you are basically repeating what the OP did to get infinities, rather than explaining that there are no infinities involved. $\endgroup$
    – DavePhD
    Commented Jun 13, 2016 at 15:51
  • $\begingroup$ I am explaining why the OP's infinity's would not imply that the pH is always zero. $\endgroup$
    – Niels Kornerup
    Commented Jun 13, 2016 at 15:57

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