Consider a \(0.67-M\) solution of $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right)$ a. Which of the following are major species in the solution? i. \(C_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\) ii. \(\mathrm{H}^{+}\) ii.. \(\mathrm{OH}^{-}\) iv. \(\mathrm{H}_{2} \mathrm{O}\) v. \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}+\) b. Calculate the \(\mathrm{pH}\) of this solution.

Ethylamine (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\)) acts as a weak base in water, as it can accept a proton (\(\mathrm{H}^{+}\)) from water. The balanced reaction is given by: \[\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_2(aq) + \mathrm{H}_2\mathrm{O}(l) \leftrightarrows \mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_3^{+}(aq) + \mathrm{OH}^{-}(aq)\]

Major species are those which have significant concentrations in the solution. Based on the reaction, we can identify following major species: i. \(\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_2\): As 0.67 M solution, it is the major species. ii. \(\mathrm{H}^{+}\): Not a major species, as it comes from the autoionization of water which is very small. iii. \(\mathrm{OH}^{-}\): A major species as ethylamine is a weak base and reacts with water to create hydroxide ions. iv. \(\mathrm{H}_2\mathrm{O}\): A major species, as it is a solvent. v. \(\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_3^+\): A major species, as it is formed when ethylamine reacts with water.

To calculate the concentration of hydroxide ions, we can set up an equilibrium expression using \(K_b\) and an ICE (Initial, Change, Equilibrium) table: \[K_b = \frac{[\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_3^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_2]}\] Initial concentrations: \([C_{2}H_5NH_2] = 0.67 M\), \([C_{2}H_5NH_3^+] = 0 M\), \([\mathrm{OH}^{-}] = 0 M\) Change in concentrations: \(+x\) moles of \(\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_2\) and \(-x\) moles of \(\mathrm{C}_{2}\mathrm{H}_5\mathrm{NH}_3^+\) and \(\mathrm{OH}^{-}\) will be in equilibrium. Equilibrium concentrations: \([C_{2}H_5NH_2] = 0.67 - x M\), \([C_{2}H_5NH_3^+] = x M\), \([\mathrm{OH}^{-}] = x M\) Plugging in concentrations into the \(K_b\) expression: \[5.6\times10^{-4} = \frac{x^2}{0.67-x}\] Since \(K_b\) is small, \(x\) will be small compared to 0.67. Thus, we can approximate: \[5.6\times10^{-4} \approx \frac{x^2}{0.67}\] Solving for \(x\): \[x = [\mathrm{OH}^{-}] = \sqrt{5.6\times10^{-4}\times0.67} \approx 0.0193\ M\]

Now that we have the concentration of hydroxide ions, we can calculate the concentration of hydrogen ions using the relationship: \[\mathrm{pK_w} = \mathrm{pH} + \mathrm{pOH} = 14\] First, calculate pOH: \[\mathrm{pOH} = -\log(0.0193) \approx 1.71\] Next, calculate pH: \[\mathrm{pH} = \mathrm{pK_w} - \mathrm{pOH} = 14 - 1.71 \approx 12.29\] So, the pH of the ethylamine solution is approximately 12.29.