A chemist studies the anion share of two phenols at two pH values…

Step 1. Phenol at pH 13.0. $\dfrac{100}{1 + 10^{-3.0}} = 99.9\ \%$.

Step 2. Phenol at pH 7.40. $\dfrac{100}{1 + 10^{2.6}} = \dfrac{100}{1 + 398} = 0.25\ \%$.

Step 3. p-Nitrophenol at pH 7.40. $\dfrac{100}{1 + 10^{-0.2}} = \dfrac{100}{1 + 0.631} = 61.3\ \%$.