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?