For relatively high concentrations of $\ce{HCl}$, I usually just assume that $[\ce{H+}] = [\ce{HCl}]$, because $\ce{HCl}$ is a strong acid and is completely ionized in solution. By taking the negative logarithm of the concentration of $\ce{[H+]}$, the pH comes out more or less correctly.

However, at very low concentrations, this method doesn't work out so well, because water has an amount of auto-ionization, which is described by $K_\mathrm{w} = 10^{-14}$. My old method was just to say that

$$[\ce{H+}] = [\ce{HCl}] + \pu{10^-7 mol dm-3},$$

to include the auto-ionization of water in my calculations. I can't really describe why, but I believe this method to be wrong, because the number $\ce{10^{-7}}$ comes from water just by itself, not with $\ce{HCl}$ added to it.

Then I came across this website, which implies that the above method is an OK method, but mentions that "However, that more complex technique will not be discussed here.". What is that 'more complex technique', and how could I use it?


1 Answer 1


At first you have to think of where the protons come from. That is as you mentioned it, hydrochloric acid (for every chloride ion in solution there has to be a proton) and water (for every hydroxide ion from the autoprotolysis there also has to be a proton). $$\ce{[H+] = [OH-] + [Cl-]}$$

The concentration for the hydroxide ions can be derived from the autoprotolysis of water, as I've said before. $$k_\mathrm w = \ce{[H+]~[OH-]} \Rightarrow \ce{[OH-]}=\frac{k_\mathrm w}{\ce{[H+]}}$$

Using some funny math, one can find out, that the concentration of the chloride ions in solution can be described the following eqation. Recognizing shortly after that the hard way was not necessary yields to the simplification that hydrochloric acid is completely dissociated. $$\ce{[Cl-]} = \frac{\ce{[HCl]0}~k_\mathrm a}{\ce{[H+]} + k_\mathrm a} \xrightarrow{k_\mathrm a \gg \ce{[H+]~~}} \frac{\ce{[HCl]0}~k_\mathrm a}{k_\mathrm a} = \ce{[HCl]0}~(=c_0)$$

This changes the first equation to be like $$\ce{[H+]} = \frac{k_\mathrm w}{\ce{[H+]}} + c_0$$

Rearranging everything gives you a second order equation in $\ce{[H+]}~(=x)$ $$x^2-c_0~x-k_\mathrm w = 0$$ Wherefrom you get two solutions (from which only the positive one makes sense): $$x = \frac{c_0}{2} + \sqrt{\frac{c_0^2}{4}+k_\mathrm w}$$

Let's compare this solution to your mentioned version. The following graph shows you the difference between both assumptions as $$\operatorname{abs}(f_1(c)-f_2(c))$$

$\hskip3.5cm$enter image description here

A maximum difference of 0.1 pH between the easier and the "more complex" version is acceptable.

  • 3
    $\begingroup$ The gist is that assuming that any HCl added will completely dissociate in less than 1 molar solution is absolutely reasonable. If $x$ moles of HCl are added the the autoionization equilibrium is shifted such that $$x = \ce{[H^+] -[OH^-]}$$ $\endgroup$
    – MaxW
    Jan 22, 2017 at 21:17
  • 1
    $\begingroup$ Would anyone be able to show where the expression for $$\ce{[Cl-]=\frac{[HCl]_0K_a}{[H+] + K_a}}$$ comes from? $\endgroup$
    – H.Linkhorn
    Apr 11, 2019 at 19:56
  • $\begingroup$ @pH13 why is it that k_a >>$ [H^+]$? I mean how we are sure regarding [H+] to be coming out to be very less than ka $\endgroup$ Apr 26, 2022 at 22:48
  • $\begingroup$ @H.Linkhorn Solve $k_a [HCl] - [H^+][Cl^-] = 0$ and $[HCl] + [Cl^-] = [HCl]_0$ together for $[Cl^-]$ $\endgroup$ Jan 31, 2023 at 14:52
  • $\begingroup$ @ProblemDestroyer to replace activities with concentrations is only ok for diluted solutions and when you look up the k_a for HCl, you will find it is much higher than the acceptable concentrations within this approximation. $\endgroup$ Jan 31, 2023 at 14:55

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