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
The reaction and \(K_{\mathrm{b}}\) equilibrium expression for aniline as a base in water are: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+H_{2}O \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-}\) \[K_{\mathrm{b}} =\frac{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right]}\] The reaction and \(K_{\mathrm{b}}\) equilibrium expression for dimethylamine as a base in water are: \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}+H_{2}O \rightarrow \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}^{+}+\mathrm{OH}^{-}\) \[K_{\mathrm{b}} =\frac{\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\right]}\]

Step by step solution

01

Write the chemical reaction for aniline as a base in water

Aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\), can act as a base in water because it has a lone pair of electrons on the nitrogen atom. When it acts as a base, it accepts a proton (H+) from water, forming a conjugate acid and hydroxide ion species. The chemical reaction for aniline acting as a base in water is: \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}+H_{2}O \rightarrow \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}+\mathrm{OH}^{-}\)
02

Write the \(K_{\mathrm{b}}\) equilibrium expression for the aniline reaction

To find the \(K_{\mathrm{b}}\) (base dissociation constant) expression, we need to write the equilibrium expression considering the molar concentrations of all species involved in the reaction at equilibrium. The \(K_{\mathrm{b}}\) expression for aniline in water is: \[K_{\mathrm{b}} =\frac{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\right]}\]
03

Write the chemical reaction for dimethylamine as a base in water

Dimethylamine, \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\), can also act as a base in water because it has a lone pair of electrons on the nitrogen atom. When it acts as a base, it accepts a proton (H+) from water, forming a conjugate acid and hydroxide ion species. The chemical reaction for dimethylamine acting as a base in water is: \(\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}+H_{2}O \rightarrow \left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}^{+}+\mathrm{OH}^{-}\)
04

Write the \(K_{\mathrm{b}}\) equilibrium expression for the dimethylamine reaction

To find the \(K_{\mathrm{b}}\) (base dissociation constant) expression, we need to write the equilibrium expression considering the molar concentrations of all species involved in the reaction at equilibrium. The \(K_{\mathrm{b}}\) expression for dimethylamine in water is: \[K_{\mathrm{b}} =\frac{\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}_{2}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\left(\mathrm{CH}_{3}\right)_{2} \mathrm{NH}\right]}\]

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