HardChemistryHigh School (11–12)Organic Compounds with Functional GroupsNumerical

Two solutions are made in water at 298 K, each with a concentration of 0.0500 mol/L…

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Two solutions are made in water at 298 K, each with a concentration of $0.0500\ \text{mol/L}$ . Chemists compare these two bases to show how a ring changes the behaviour of an amino group. Neglect the ionisation of water itself in both solutions, and keep the full precision in the middle steps. One holds aniline, with $\text{p}K_b = 9.38$ . The other holds methylamine, with $\text{p}K_b = 3.38$ . In aniline the lone pair on nitrogen is spread into the benzene ring, so it is much less ready to take a proton from water. In methylamine the methyl group pushes electrons onto the nitrogen. Solve the equilibrium for each solution without any shortcut, using $K_b = x^2/(c - x)$ . Find the pH of the aniline solution, the pH of the methylamine solution, the difference between them, and the percentage of aniline that is ionised.

Need a reminder? The key ideaShow

$K_b = \dfrac{[\text{RNH}_3^+][\text{OH}^-]}{[\text{RNH}_2]}$ , $\text{p}K_b = -\log K_b$ , and $\text{p}K_a + \text{p}K_b = 14$ at 298 K. For a solution of concentration $c$ , $K_b = \dfrac{x^2}{c - x}$ with $x = [\text{OH}^-]$ . Then $\text{pOH} = -\log x$ and $\text{pH} = 14 - \text{pOH}$ . A smaller $\text{p}K_b$ is a stronger base.

The worked solution

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