Three acids are each prepared at 0.100 mol/L…

Step 1. Perchloric acid. $[\text{H}^+] = 0.100\ \text{mol/L}$, so $\text{pH} = 1.00$.

Step 2. Chlorous acid. $x^2 + 0.011x - 0.0011 = 0$, so $x = \dfrac{-0.011 + \sqrt{0.011^2 + 4 \times 0.0011}}{2} = 0.0281\ \text{mol/L}$. The pH is $-\log_{10}(0.0281) = 1.55$.

Step 3. Hypochlorous acid. $x = \sqrt{2.9 \times 10^{-8} \times 0.100} = 5.39 \times 10^{-5}\ \text{mol/L}$. The pH is $4.27$.