What is the pH of HCl solution at $10^{-8}$M? [duplicate]








**This question already has an answer here**:
 
 




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

 (1 answer)
 

Closed 5 years ago.



After using the negative logarithmic value of $\ce{H^+}$ ion concentration, I get a value of pH that results in base. Can you please help me?


Thank you in advance




Well this is not pretty difficult, let's go ! 




---


Because chloride ions don't react with water we have,
$$c\_0=\ce{[HCl]=[Cl^-]}\tag1$$


Now for hydrogen ions, $$\ce{[H^+]=[HO^-]}\ce{+[Cl^-]}\tag2$$


We all, **must**, know that $$K\_e=\ce{[H^+][HO^-]}\tag3$$


Then we have $$\ce{[H^+]}^2-c\_0\ce{[H^+]}-K\_e=0\tag4$$


You'll find your pH solving $(4)$.




---


Now between theory and practice you may not find a big difference... 


 3

When you want to find $\mathrm{pH}$ of acid whose concentration is less than $10^{-6}$ you have to consider water's dissociation as $\ce{H+}$ and $\ce{OH-}$ too. Which is $10^{-7}$ at $\pu{25 ^\circ C}$. So concentration of $\ce{H+}$ into this case is actually, $10^{-6} + 10^{-7}$, giving $1.1 \times 10^{-6}$. Which will give you a $\mathrm{pH}$ of about $6.9$. We usually ignore water's dissociation because it is negligible but in cases like this it becomes necessary to consider it too.


 1