
What does the ${{K}_{b}}$ value for amine stands for?
Answer
562.2k+ views
Hint: ${{K}_{a}}$ and ${{K}_{b}}$ are the equilibrium constants for the hydrolysis of acids and bases in water. Equilibrium constant of an acid is defined as a term used to determine the strength of an acid whereas the equilibrium constant of a base is defined as a term used to determine the strength of a base.
Complete step by step answer:
Amines are aliphatic and aromatic derivatives of ammonia. Amines are also basic in nature due to the unshared pair of electrons. When amines are dissolved in water they furnish hydroxide ions in the solution. This can be shown as follows:
$R\overset{..}{\mathop{N}}\,{{H}_{2}}+{{H}_{2}}O\to RN{{H}_{3}}^{+}O{{H}^{-}}\rightleftharpoons RN{{H}_{3}}^{+}+O{{H}^{-}}$
So, due to the presence of hydroxide ions the aqueous solutions of amines are basic in nature.
The basic strength of amine is expressed in terms of dissociation constant, ${{K}_{b}}$ . For the reaction:
$R\overset{..}{\mathop{N}}\,{{H}_{2}}+{{H}_{2}}O\rightleftharpoons RN{{H}_{3}}^{+}+O{{H}^{-}}$
${{K}_{eq}}=\dfrac{[RN{{H}_{3}}][O{{H}^{-}}]}{[RN{{H}_{2}}][{{H}_{2}}O]}$
Since $[{{H}_{2}}O]$ is constant it is convenient to incorporate it into equilibrium constant as;
${{K}_{eq}}[{{H}_{2}}O]=\dfrac{[RN{{H}_{3}}^{+}][O{{H}^{-}}]}{[RN{{H}_{2}}]}$
Thus dissociation constant
${{K}_{b}}=\dfrac{\left[ RN{{H}_{3}}^{+} \right]\left[ O{{H}^{-}} \right]}{\left[ RN{{H}_{2}} \right]}$
So, greater the value of ${{K}_{b}}$ value, stronger is the base.
Thus the value of ${{K}_{b}}$ for an amine is a measure of basic strength.
Note: With the increase in the number of alkyl groups the electron density on the nitrogen atom increases and hence donation tendency increases. Therefore ethylamine is more basic than ammonia and hence has the larger value of ${{K}_{b}}$ than ammonia. The amines are basic because of the presence of lone pairs on the nitrogen atom.
The strong base will have a higher value of equilibrium constant of a base and it will completely dissociate in aqueous solution.
Complete step by step answer:
Amines are aliphatic and aromatic derivatives of ammonia. Amines are also basic in nature due to the unshared pair of electrons. When amines are dissolved in water they furnish hydroxide ions in the solution. This can be shown as follows:
$R\overset{..}{\mathop{N}}\,{{H}_{2}}+{{H}_{2}}O\to RN{{H}_{3}}^{+}O{{H}^{-}}\rightleftharpoons RN{{H}_{3}}^{+}+O{{H}^{-}}$
So, due to the presence of hydroxide ions the aqueous solutions of amines are basic in nature.
The basic strength of amine is expressed in terms of dissociation constant, ${{K}_{b}}$ . For the reaction:
$R\overset{..}{\mathop{N}}\,{{H}_{2}}+{{H}_{2}}O\rightleftharpoons RN{{H}_{3}}^{+}+O{{H}^{-}}$
${{K}_{eq}}=\dfrac{[RN{{H}_{3}}][O{{H}^{-}}]}{[RN{{H}_{2}}][{{H}_{2}}O]}$
Since $[{{H}_{2}}O]$ is constant it is convenient to incorporate it into equilibrium constant as;
${{K}_{eq}}[{{H}_{2}}O]=\dfrac{[RN{{H}_{3}}^{+}][O{{H}^{-}}]}{[RN{{H}_{2}}]}$
Thus dissociation constant
${{K}_{b}}=\dfrac{\left[ RN{{H}_{3}}^{+} \right]\left[ O{{H}^{-}} \right]}{\left[ RN{{H}_{2}} \right]}$
So, greater the value of ${{K}_{b}}$ value, stronger is the base.
Thus the value of ${{K}_{b}}$ for an amine is a measure of basic strength.
Note: With the increase in the number of alkyl groups the electron density on the nitrogen atom increases and hence donation tendency increases. Therefore ethylamine is more basic than ammonia and hence has the larger value of ${{K}_{b}}$ than ammonia. The amines are basic because of the presence of lone pairs on the nitrogen atom.
The strong base will have a higher value of equilibrium constant of a base and it will completely dissociate in aqueous solution.
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