Calculating the pH of a highly dilute solution of HCl

Question

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?

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 $$\operatorname{abs}(f_1(c)-f_2(c))$$

$\hskip3.5cm$

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