Given that the \(K_{\mathrm{a}}\) value for acetic acid is \(1.8 \times 10^{-5}\) and the \(K_{\mathrm{a}}\) value for hypochlorous acid is \(3.5 \times 10^{-8},\) which is the stronger base, \(\mathrm{OCl}^{-}\) or $\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{O}_{2}^{-} ?$

In order to determine the stronger base, we first need to calculate the Kb values for OCl- and C2H3O2-. We can do this using the relationship between Ka, Kb, and Kw: \(K_\mathrm{a} \cdot K_\mathrm{b} = K_\mathrm{w}\) where Ka is the acid dissociation constant, Kb is the base dissociation constant, and Kw is the ion product of water, which is equal to \(1.0 \times 10^{-14}\) at 25°C. For acetic acid: \(K_\mathrm{a}(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2) = 1.8 \times 10^{-5}\) So, \(K_\mathrm{b}(\mathrm{C}_2 \mathrm{H}_3 \mathrm{O}_2^-) = \frac{K_\mathrm{w}}{K_\mathrm{a}(\mathrm{C}_2 \mathrm{H}_4 \mathrm{O}_2)} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} = 5.56 \times 10^{-10}\) For hypochlorous acid: \(K_\mathrm{a}(\mathrm{H}\mathrm{O}\mathrm{Cl}) = 3.5 \times 10^{-8}\) So, \(K_\mathrm{b}(\mathrm{O}\mathrm{Cl}^-) = \frac{K_\mathrm{w}}{K_\mathrm{a}(\mathrm{H}\mathrm{O}\mathrm{Cl})} = \frac{1.0 \times 10^{-14}}{3.5 \times 10^{-8}} = 2.86 \times 10^{-7}\)

Now that we have the Kb values for both OCl− and C2H3O2−, we can compare them to determine which is the stronger base: \(K_\mathrm{b}(\mathrm{O}\mathrm{Cl}^-) = 2.86 \times 10^{-7}\) \(K_\mathrm{b}(\mathrm{C}_2 \mathrm{H}_3 \mathrm{O}_2^-) = 5.56 \times 10^{-10}\) Since the Kb value of OCl− is larger than the Kb value of C2H3O2−, OCl− is the stronger base.