# Acid solution — lower pH when diluted

Today I came across a "thought experiment" regarding $$\mathrm{pH}$$ of a diluted acid solution. This idea looks somewhat logical but obviously contradicts the reality. Can you show me where is the error?

Let's suppose we have $$\pu{100 g}$$ of concentrated 95% (mass) sulfuric acid. That means we have a mixture of $$\pu{95 g}$$ of H2SO4 and $$\pu{5 g}$$ of $$\ce{H2O}.$$ Let's calculate the amount of both compounds.

$$n = \frac{m}{M}$$

$$n(\ce{H2SO4}) = \frac{\pu{95 g}}{\pu{98.08 g mol-1}} = \pu{0.9686 mol}$$

$$n(\ce{H2O}) = \frac{\pu{5 g}}{\pu{18.02 g mol-1}} = \pu{0.2778 mol}$$

Since the amount of the solvent molecules is much much lower than the amount of the $$\ce{H2SO4}$$ molecules, it is not possible for the acid to be completely dissolved, which leads to the relatively small amount of $$\ce{H3O+}$$ cations, and $$\mathrm{pH}$$ of the solution is relatively high $$(\mathrm{pH} = -\log [\ce{H3O+}]).$$

$$\ce{H2SO4 + H2O -> HSO4^- + H3O+}$$

If we increase the amount of the $$\ce{H2O}$$ molecules (dilute solution), more $$\ce{H3O+}$$ cations can be formed, and the $$\mathrm{pH}$$ will decrease.

Conclusion: dilution of acid-water solutions decreases $$\mathrm{pH}.$$

In reality it is completely opposite. So what is the catch?

• 95% sulfuric acid isn't an "aqueous" solution. // At higher concentrations, certainly anything beyond 1 molar, one needs to use activities of the ions not their concentrations.
– MaxW
Jul 9, 2019 at 0:10
• Also note, that activity coefficients will be much higher than 1, and that sulphuric acid will dissociate just to hydrogensulphate. $$\ce{H2SO4 + H2O -> HSO4- + H3O+}$$ Jul 9, 2019 at 4:11
• Jul 10, 2019 at 18:08 