Question 18.p.8: Pyridine contains a nitrogen atom that accepts H^+ from wate......
18.p.8:Pyridine contains a nitrogen atom that accepts \text{H}^+ from water to form \text{OH}^- ions in aqueous solution. Write a balanced equation and equilibrium expression for the reaction, convert \text{p}K_\text{b}\text{ to }K_\text{b}, make simplifying assumptions (if valid) and solve for [\text{OH}^-]. Calculate \text{[H}_3\text{O}^+]\text{ using [H}_3\text{O}^+] [\text{OH}^-] = 1.0 \times 10^{-14} and convert to pH.
Assumptions:
1 ) Since K_\text{b} >> K_\text{w},\text{ disregard the [OH}^-] that results from the autoionization of water.
2) Since K_\text{b}\text{ is small, [C}_5\text{H}_5\text{N}] = 0. 10 – \text{x} ≈ 0.10. \\ K_\text{b}=\frac{[\text{C}_5\text{N}_5\text{NH}^+][\text{OH}^-]}{[\text{C}_5\text{N}_5\text{N}]} =\frac{\text{x}^2}{(0.10)} =1.69824\times 10^{-9}\text{ (unrounded)} \\ \text{x}=1.303165\times 10^{-5}=1.3\times 10^{-5} \ M=[\text{OH}^-]=[\text{C}_5\text{H}_5\text{NH}^+] \\ \text{Since}\frac{[\text{OH}^-]}{[\text{C}_5\text{H}_5\text{N}_5]} (100)=\frac{1.303265\times 10^{-5}}{0.10} (100)=0.01313 which < 5%, the assumption that the dissociation of C_5\text{H}_5\text{N}_5 is small is valid.
Check: Since pyridine is a weak base, a pH > 7 is expected.