pH of an Extremely Diluted Acid (10-8 M HCl)
What is the pH of a 10-8 molar HCl solution? Compare the numerical result of aqion with the exact solution of the corresponding equations.
1. Numerical Solution with aqion
We begin with pure water (click on button New), then click on Reac and enter for “HCl” the value 1e-5 mmol/L – see the right screenshot.
(Please mind the concentration units: 10-8 mol/L = 10-5 mmol/L.)
Run the calculation with Start. The result will be instantly displayed:
pH = 6.98
The highly diluted HCl solution has pH 6.98.
2. Exact Solution (Theory)
Let’s abbreviate the initial HCl concentration by c0 = 10-8 M. The aqueous system consists of 3 components: [H+], [OH-], and [Cl-]. Thus, we need 3 equations:
|(1) || strong acid (completely dissociated): || [Cl-] = c0 |
|(2) || charge balance: || [H+] = [OH-] + [Cl-] |
|(3) || self-ionization of water: || KW = [H+] [OH-] = 10-14 |
Inserting the first two equations into the third yields a quadratic equation in x = [H+]:
The (non-negative) solution for x is
|(5) || |
After entering the above values one gets:
|(6) || |
The negative decadic logarithm of x defines the pH
|(7) || pH = -log [H+] = -log x = 6.978 |
This is in perfect agreement with the numerical result above.
We fall into a deep error if we identify the H+ concentration with that of HCl, that is, if we set [H+] = 10-8 M and plug it into Eq.(7):
|(8) || pH = -log [H+] = -log 10-8 = 8 || (that’s wrong!) |
That’s evidently wrong because an acid cannot have pH > 7. Where we made the mistake?
The answer is simple: We ignored the 10-7 mol/L H+ that comes from the self-ionization of water — see Eq.(3). Instead, to this background concentration of 10-7 M we should add the small amount of 0.1 10-7 M from HCl.
[last modified: 2013-08-05]