pH of an Extremely Dilute Acid

Problem

What is the pH of a 10-8 molar HCl solution?

Compare the numerical result of aqion with the analytical solution of the corresponding equations.

1. Numerical Solution with aqion

We begin with pure water (button H2O), then click on Reac and enter for “HCl” the value 1e-5 mmol/L, as shown in the right screenshot. (Please mind the concentration units: 10-8 mol/L = 10-5 mmol/L.)

Start the calculation by clicking on the Start button. The result appears immediately:

 pH = 6.98

The highly diluted HCl solution (in contrast to pure water) decreases the pH by a tiny amount from 7 to 6.98.

2. Analytical Solution (Theory)

Let us abbreviate the total HCl concentration by CT = 10-8 M. The math system involves three unknowns: [H+], [OH-], and [Cl-]. So we need three equations:

 (1.1) strong acid (completely dissociated): [Cl-] = CT (1.2) charge balance: [H+] = [OH-] + [Cl-] (1.3) self-ionization of water: Kw = [H+] [OH-] = 10-14

Inserting the first two equations into the third yields a quadratic equation in x = [H+]:

 (2) x2 – CT x – Kw = 0

The (non-negative) solution for x is

 (3) $$x \ = \ \dfrac{C_T}{2}\, \left\{ 1 + \sqrt{1+\dfrac{4K_w}{C_T^2}} \,\right\}$$

After inserting the numbers into the equation we get:

 (4) $$x \ = \ \dfrac{1.0\cdot 10^{-8} M}{2} \, \left( 1 + \sqrt{\strut 1+400} \ \right) \ = \ 10.51 \cdot 10^{-8} M$$

The negative decadic logarithm of x defines the pH1

 (5) pH  =  –lg [H+]  =  –lg x  =  6.978

This is in perfect agreement with the numerical result above.2

Pitfalls

One often makes the mistake of identifying the H+ concentration with that of HCl, that is, setting [H+] = 10-8 M into the last equation:

 (6) pH  =  –lg [H+]  =  –lg 10-8  =  8 ⇐  that’s wrong

This is definitely wrong, because an acid (no matter how much you dilute it) cannot have a pH value above 7. Where does the fallacy lie?

The answer is simple: We ignored the 10-7 mol/L H+ that comes from the self-ionization of water — see 1.3. To this “background concentration” of 1.0·10-7 M we should add the small amount of 0.1·10-7 M from HCl.