# Is the common way to calculate the pH of a solution with low concentration of acid wrong? [duplicate]

What is the pH of $$1,4^{-8} M$$ $$HCl$$ in water?

This is such a small amount of acid that you need to take into concideration the self ionization of water. So the way our chemistry teacher did it on the test was something similar to the first answer on this site: https://www.quora.com/What-is-the-pH-of-10-6-M-HCl-aq

They find the total $$[H_3O^+]$$ in the solution by adding the $$1,0*10^{-7}M$$ $$H_3O$$ from self ionization of water and get:
$$1,14*10^{-7}M$$.
Then they just find the pH:
$$-\log(1,14*10^{-7}M)$$
This is $$6,943$$ which becomes a pH of $$6,9$$

I did this and got right on the test, but something didn't sit right with me:
If $$[H_3O^+]$$ was $$1,14*10^{-7}M$$ and $$[OH^-]$$ was $$1,0*10^{-7}$$ (from self ionization of water), then $$1,0*10^{-14}M^2≠[H_3O]*[OH^-]$$ and the solution is not in equilibrium. So I came up with another way to calculate the pH:

I also first find $$[H_3O^+]$$ to be $$1,14*10^{-7}M$$ but then I think that $$x$$ amounts of $$H_3O^+$$ and $$OH^-$$ has to react so the solution is in equilibrium, so I set up an equation:
$$1,0*10^{-14}M^2=(1,14*10^{-7}M-x)*(1,0*10^{-7}M-x)$$
When I solve for $$x$$ I find that $$6,755*10^{-9}$$ mole has reacted per liter. Then I take away that from the original $$[H_3O^+]$$ and get that the current $$[H_3O^+]$$ is $$1,07245*10^{-7}M$$. Then I just find the pH of the solution:
$$-\log(1,07245*10^{-7}M)$$
This is $$6,9696$$ which becomes a pH of $$7,0$$

A pH of $$7,0$$! Not $$6,9$$! And the teacher deducted points for $$7,0$$! I got $$6,9$$ on the test and full marks so I'm not that mad anyway, but I'm curious what is the most right answer. I also discussed it with my teacher and I think he somewhat understood my problem, but not my solution.

Edit:
I see that my question was a little unclear. I would like to know which answer is most right and if my thought process is correct.

• The simple calculation for strong acids pH = - log c neglects water autodissociation and is justified only for c >> 10^-7 mol/L. Commented Apr 16, 2023 at 16:42
• I never understood why they keep teaching people 'approximate' methods for pH calculation, which only seems to generate confusion and a lot of playing with sketchy assumptions. Nowadays one can solve exact systems numerically using open-source software, so there is really no need to approximate anything (except the activity vs concentration story). And for a solution of a single strong acid in water, even the exact system is analytical, so... Maybe they should teach students how to build the exact system first, and then show how it can be simplified in special cases. Commented Apr 16, 2023 at 16:52
• See for instance chemistry.stackexchange.com/a/97840/41751 . The equation that the user derived is valid for any concentration of HCl, you only solve it as a 2nd degree one, without making any assumptions or simplifications. And remember to replace $1$ with the actual $[HCl]_0$. Commented Apr 16, 2023 at 16:55
• I believe using approximations and "doing stuff in your head" is the only way to get a proper grasp on the more complex topics and learning about chemistry. There are already too many people using computers without actually knowing what they do. But this is actually not the place to discuss this. However, this question does show that critical thinking is not lost when using approximations. Commented Apr 16, 2023 at 17:12
• Knowing what approximations you can use in the given situation is actually a sign you have a good qualitative understanding of the system (what is important, what is not). This is one reason such calculations are still taught as exercize
– Greg
Commented Apr 16, 2023 at 17:39

If no usable equation is ready, you can always use in acido-basic calculations the scenario below. It is less straight-forward than some ready to use (often simplified) equations, but it is generally valid ( putting aside the Pandora's box of activities).

• The law of charge conservation - total charge has to be zero
$$[\ce{H+}] = [\ce{OH-}] + [\ce{A-}] \label{1}\tag{1}$$

• The law of mass conservation - in a specific form of conservation of total molar amount/concentration of particular forms of given substance:
$$c(\ce{HA}) = [\ce{HA}] + [\ce{A-}] \label{2}\tag{2}$$

• The equations for all present acido-basic equilibrii
$$K_\mathrm{w} = [\ce{H+}][\ce{OH-}]=\pu{e-14} \label{3}\tag{3}$$
$$K_\mathrm{a} = \frac{[\ce{H+}][\ce{A^-}]}{[\ce{HA}]} \label{4}\tag{4}$$

Generally, we get a set of N equations for N variable and one often has to apply justified simplifications.

But in our case, things go smoothy to the simple solution:

We have welcome simplification, as we can assume there is not HA present for strong acids as $$K_\mathrm{a}$$ in \eqref{4} is effectively infinite and therefore
$$c(\ce{HA}) = [\ce{A-}]\label{2a}\tag{2a}$$

After substitution of \eqref{2a} and \eqref{3} to \eqref{1}, we get:

$$[\ce{H+}] = \frac{K_\mathrm{w}}{[\ce{H+}]} + c(\ce{HA}) \label{5}\tag{5}$$

$$[\ce{H+}]^2 - c(\ce{HA}){[\ce{H+}]} - K_\mathrm{w} = 0 \label{6}\tag{6}$$

$$[\ce{H+}] = \frac{c(\ce{HA}) \pm \sqrt{(c(\ce{HA}))^2 + 4K_\mathrm{w}}}{2} \label{7}\tag{7}$$

We accept only the plus sign variant, the other root is negative.

$$[\ce{H+}] = \frac{c(\ce{HA}) + \sqrt{(c(\ce{HA}))^2 + 4K_\mathrm{w}}}{2} \label{8}\tag{8}$$

$$\mathrm{pH} = -\log{[\ce{H+}]}\label{9}\tag{9}$$

If it neglect water auto-dissociation and pronounce $$K_\ce{w} = 0$$ then the equation \eqref{8} reduces itself to simple:

$$[\ce{H+}] = c(\ce{HA})\label{10}\tag{10}$$