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








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[pH of aqueous solution of HCl of low concentration [duplicate]](/questions/66843/ph-of-aqueous-solution-of-hcl-of-low-concentration)

 (2 answers)
 


[Calculating the pH of a highly dilute solution of HCl](/questions/35559/calculating-the-ph-of-a-highly-dilute-solution-of-hcl)

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Closed 5 months ago.



We had a chemistry test and one of the questions was this:
  

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.




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).


You start with 3 types of equations


* 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:


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


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