Write the reaction and the corresponding \(K_{\mathrm{b}}\) equilibrium expression for each of the following substances acting as bases in water. a. aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\) b. dimethylamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\)

Short Answer

Expert verified
a. Aniline, C6H5NH2: Reaction: \(C_{6}H_{5}NH_{2} + H_{2}O \rightarrow C_{6}H_{5}NH_{3}^{+} + OH^{-}\) Kb expression: \(K_{b} = \frac{[C_{6}H_{5}NH_{3}^{+}][OH^{-}]}{[C_{6}H_{5}NH_{2}]}\) b. Dimethylamine, (CH3)2NH: Reaction: \((CH_{3})_{2}NH + H_{2}O \rightarrow (CH_{3})_{2}NH_{2}^{+} + OH^{-}\) Kb expression: \(K_{b} = \frac{[(CH_{3})_{2}NH_{2}^{+}][OH^{-}]}{[(CH_{3})_{2}NH]}\)

Step by step solution

01

Write the reaction of aniline with water

When aniline (C6H5NH2) acts as a base in water, it accepts a proton (H+) from a water molecule, forming its conjugate acid (C6H5NH3+) and hydroxide ion (OH-): C6H5NH2 + H2O -> C6H5NH3+ + OH-
02

Write the Kb equilibrium expression for aniline

For the given reaction, the Kb equilibrium expression is defined as: Kb = \(\frac{[C_{6}H_{5}NH_{3}^{+}] [OH^{-}]}{[C_{6}H_{5}NH_{2}]}\) b. Dimethylamine, (CH3)2NH:
03

Write the reaction of dimethylamine with water

When dimethylamine ((CH3)2NH) acts as a base in water, it accepts a proton (H+) from a water molecule, forming its conjugate acid ((CH3)2NH2+) and hydroxide ion (OH-): (CH3)2NH + H2O -> (CH3)2NH2+ + OH-
04

Write the Kb equilibrium expression for dimethylamine

For the given reaction, the Kb equilibrium expression is defined as: Kb = \(\frac{[(CH_{3})_{2}NH_{2}^{+}] [OH^{-}]}{[(CH_{3})_{2}NH]}\)

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