What are the major species present in \(0.250 \mathrm{M}\) solutions of each of the following acids? Calculate the \(\mathrm{pH}\) of each of these solutions. a. \(\mathrm{HNO}_{2}\) b. \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3} \mathrm{O}_{2}\right)\)

First, we need to identify the major species present in each of the 0.250 M solutions: a. For the \(\mathrm{HNO}_{2}\) solution, the major species are \(\mathrm{HNO}_{2}\), \(\mathrm{H}_{2}\mathrm{O}\), \(\mathrm{H}^{+}\), and \(\mathrm{NO}_{2}^{-}\). b. For the $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3}\mathrm{O}_{2}\right)\( solution, the major species are \)\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\(, \)\mathrm{H}_{2}\mathrm{O}\(, \)\mathrm{H}^{+}\(, and \)\mathrm{CH}_{3}\mathrm{CO}_{2}^{-}$.

Now, we should express the dissociation equilibrium constants for both acids: a. For \(\mathrm{HNO}_{2}\), the acid dissociation constant is \(K_{a1}\): \[K_{a1} = \frac{[\mathrm{H}^{+}][\mathrm{NO}_{2}^{-}]}{[\mathrm{HNO}_{2}]}\] b. For $\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}\left(\mathrm{HC}_{2} \mathrm{H}_{3}\mathrm{O}_{2}\right)\(, the acid dissociation constant is \)K_{a2}$: \[K_{a2}= \frac{[\mathrm{H}^{+}][\mathrm{CH}_{3}\mathrm{CO}_{2}^{-}]}{[\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}]}\]

We need to find the concentration of \(\mathrm{H}^{+}\) ions from the equilibrium constants before we can calculate the pH values. Typically, we should be given the values for \(K_{a1}\) and \(K_{a2}\) or refer to a table of dissociation constants. In this case, we will use hypothetical values, viz.: - For \(\mathrm{HNO}_{2}\), let \(K_{a1} = 7.1 \times 10^{-4}\). - For \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\), let \(K_{a2} = 1.8 \times 10^{-5}\). Now, we will solve for the concentration of \(\mathrm{H}^{+}\) for each solution: a. For \(\mathrm{HNO}_{2}\): \[7.1 \times 10^{-4} = \frac{[\mathrm{H}^{+}][\mathrm{NO}_{2}^{-}]}{[\mathrm{HNO}_{2}]} = \frac{[\mathrm{H}^{+}]^2}{0.250}\] \[ [\mathrm{H}^{+}] = \sqrt{7.1 \times 10^{-4} \times 0.250} = 0.0133 \mathrm{M}\] b. For \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\): \[1.8 \times 10^{-5} = \frac{[\mathrm{H}^{+}][\mathrm{CH}_{3}\mathrm{CO}_{2}^{-}]}{[\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}]} = \frac{[\mathrm{H}^{+}]^2}{0.250}\] \[[\mathrm{H}^{+}] = \sqrt{1.8 \times 10^{-5} \times 0.250} = 0.000948 \mathrm{M}\] Lastly, we can calculate the pH for each solution: a. For \(\mathrm{HNO}_{2}\): \[\mathrm{pH} = -\log([\mathrm{H}^{+}]) = -\log(0.0133) \approx 1.88\] b. For \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\): \[\mathrm{pH} = -\log([\mathrm{H}^{+}]) = -\log(0.000948) \approx 3.02\] So, the pH of the \(\mathrm{HNO}_{2}\) solution is approximately 1.88, and that of the \(\mathrm{HC}_{2}\mathrm{H}_{3}\mathrm{O}_{2}\) solution is approximately 3.02.