Question

Consider pyridine, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{~N},\) a pesticide and deer repellent. Its conjugate acid, \(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+},\) has \(K_{\mathrm{a}}=6.7 \times 10^{-6}\). (a) Write a balanced net ionic equation for the reaction that shows the basicity of aqueous solutions of pyridine. (b) Calculate \(K_{b}\) for the reaction in (a). (c) Find the \(\mathrm{pH}\) of a solution prepared by mixing \(2.74 \mathrm{~g}\) of pyridine in enough water to make \(685 \mathrm{~mL}\) of solution.

Short answer

(Given Ka for the conjugate acid of pyridine is 6.7 × 10^(-6)) Answer: The pH of the solution is 8.74.

Step-by-step solution

Balanced Net Ionic Equation

For pyridine (\(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}\)) to act as a base, it must accept a proton from water, forming its conjugate acid and a hydroxide ion. The balanced net ionic equation for the reaction is: $$\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}_{(aq)} + \mathrm{H}_{2} \mathrm{O}_{(l)} \rightleftharpoons \mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}_{(aq)} + \mathrm{OH}^{-}_{(aq)}$$ (b)

Calculate \(K_{b}\)

We have the \(K_a\) value for the conjugate acid of pyridine (\(\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}\)), which is \(6.7 \times 10^{-6}\). Using the relationship between \(K_a\), \(K_b\) and \(K_w\), we can find the \(K_b\) value for pyridine: $$K_a \times K_b = K_w$$ $$K_b = \frac{K_w}{K_a}$$ We know that \(K_w = 1.0 \times 10^{-14}\) at 25°C, so: $$K_b = \frac{1.0 \times 10^{-14}}{6.7 \times 10^{-6}} = 1.49 \times 10^{-9}$$ (c)

Find the pH of the Solution

First, calculate the initial concentration of pyridine in the solution: $$\mathrm{molarity} = \frac{\mathrm{moles}}{\mathrm{volume}}$$ $$\mathrm{moles} = \frac{\mathrm{grams}}{\mathrm{molar~mass}} = \frac{2.74 \mathrm{~g}}{79.10 \mathrm{~g/mol}} = 0.0346 \mathrm{~mol}$$ $$\mathrm{molarity} = \frac{0.0346 \mathrm{~mol}}{0.685 \mathrm{~L}} = 0.0505 \mathrm{~M}$$ Now, use an equilibrium table to analyze the reaction and solve for \(x\) which represents the hydroxide ion concentration: Initial: \([\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}] = 0.0505 \mathrm{M}\), \([\mathrm{OH}^{-}] = 0\) , \([\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}] = 0\) Change: \([\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}] = -x\), \([\mathrm{OH}^{-}] = +x\), \([\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}] = +x\) Equilibrium: \([\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}] = (0.0505 - x)\), \([\mathrm{OH}^{-}] = x\), \([\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}] = x\) Now substitute the equilibrium concentrations into the \(K_b\) expression and solve for \(x\): $$K_b = \frac{[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{NH}^{+}][\mathrm{OH}^{-}]}{[\mathrm{C}_{5} \mathrm{H}_{5} \mathrm{N}]}$$ $$1.49 \times 10^{-9} = \frac{x^2}{0.0505 - x}$$ Assuming \(x \ll 0.0505\), we can approximate: $$1.49 \times 10^{-9} \approx \frac{x^2}{0.0505}$$ $$x = \sqrt{1.49 \times 10^{-9} \times 0.0505} = 5.45 \times 10^{-6}\,\mathrm{M}$$ This is the hydroxide ion concentration, \([\mathrm{OH}^{-}]\). To find the pH, first find the pOH: $$\mathrm{pOH} = -\log{[\mathrm{OH}^{-}]} = -\log{(5.45 \times 10^{-6})} = 5.26$$ Now, use the relationship between pH and pOH: $$\mathrm{pH} = 14 - \mathrm{pOH} = 14 - 5.26 = 8.74$$ Thus, the pH of the solution is 8.74.