Calculate the molar concentration of \(\mathrm{OH}^{-}\) in a \(0.050 \mathrm{M}\) solution of ethylamine $\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} ; K_{b}=6.4 \times 10^{-4}\right) .$ Calculate the \(\mathrm{pH}\) of this solution.

The ethylamine (\(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\)) reacts with water, as it is a weak base: \[ \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2} \left(\mathrm{aq}\right) + \mathrm{H}_{2}\mathrm{O} \left(\mathrm{l}\right) \rightleftarrows\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+} \left(\mathrm{aq}\right) + \mathrm{OH}^{-} \left(\mathrm{aq}\right). \]

Using given \(K_{b}=6.4 \times 10^{-4}\) and initial concentrations, we can create a table to determine the equilibrium concentrations. Let \(x\) be the concentration of \(\mathrm{OH}^{-}\) at equilibrium. Initial concentrations (M): Ethylamine: \(0.050\ \mathrm{M}\) Ethylammonium ion: \(0\ \mathrm{M}\) Hydroxide ion: \(0\ \mathrm{M}\) Changes (M): Ethylamine: \(-x\) Ethylammonium ion: \(+x\) Hydroxide ion: \(+x\) Equilibrium concentrations (M): Ethylamine: \((0.050-x)\ \mathrm{M}\) Ethylammonium ion: \(x\ \mathrm{M}\) Hydroxide ion: \(x\ \mathrm{M}\) Thus, the equilibrium expression for this reaction is: \[ K_b = \frac{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right] \left[\mathrm{OH}^{-}\right]}{\left[\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\right]}. \] Substituting the equilibrium concentrations: \[ 6.4 \times 10^{-4} = \frac{x \cdot x}{0.050 - x}. \]

Assuming \(x << 0.050\), we can approximate the denominator to simply \(0.050\). Solving for x: \[ 6.4 \times 10^{-4} = \frac{x^2}{0.050} \Rightarrow x^2 = 6.4 \times 10^{-4} \times 0.050 \Rightarrow x = \sqrt{6.4\times 10^{-4}\times 0.050} = 1.8\times 10^{-2}. \] Thus, the concentration of \(\mathrm{OH}^{-}\) ions at equilibrium is \(1.8\times 10^{-2}\ \mathrm{M}\).

Now that we have the concentration of \(\mathrm{OH}^{-}\) ions, we can calculate the \(\mathrm{pOH}\), and subsequently, the \(\mathrm{pH}\) of the solution. Formula for \(\mathrm{pOH}\) and \(\mathrm{pH}\): \[ \mathrm{pOH}=-\log_{10}\left[\mathrm{OH}^{-}\right], \] \[ \mathrm{pH} = 14 - \mathrm{pOH}. \] So, the \(\mathrm{pOH}\) is: \[ \mathrm{pOH} = -\log_{10}\left(1.8 \times 10^{-2}\right) = 1.74. \] And the \(\mathrm{pH}\) is: \[ \mathrm{pH} = 14 - 1.74 = 12.26. \] Thus, the \(\mathrm{pH}\) of the solution is \(12.26\).