Answer
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Hint: The law of dilution gives us a relation between dissociation constant and the degree of dissociation. This relation can be explained as: the degree of dissociation of a weak electrolyte (\[{{\alpha }}\]) is directly proportional to dilution constant \[{{\text{K}}_b}\], and it is inversely proportional to the concentration \[{{\text{C}}_{\text{0}}}\].
Complete step by step solution:
For the reaction,
Dilution constant can be written as:
\[{K_b} = \dfrac{{\left[ {{A^ + }} \right]\left[ {{B^ - }} \right]}}{{\left[ {AB} \right]}} = \dfrac{{\left( {\alpha {C_0}} \right)\left( {\alpha {C_0}} \right)}}{{\left( {1 - \alpha } \right){C_0}}} = \dfrac{{{\alpha ^2}}}{{1 - \alpha }} \times {C_0}\]
Where \[{{\alpha }}\]is the degree of dissociation of a weak electrolyte. And \[{{\text{C}}_{\text{0}}}\] is the concentration.
For weak electrolyte, \[\alpha < < 0\] so \[\left( {1 - \alpha } \right)\] can be neglected and the resulting equation is:
So, \[\alpha = \sqrt {\dfrac{{{K_b}}}{{{C_0}}}} \]
So, for pyridine on dilution with water results in pyridinium ion. In question, we are given that molarity of pyridine solution is 0.10M and \[{{\text{K}}_{\text{b}}}\] for \[{C_5}{H_5}N\]\[ = 1.7 \times {10^{ - 9}}\].
\[\alpha = \sqrt {\dfrac{{{K_b}}}{{{C_0}}}} = \sqrt {\dfrac{{1.7 \times {{10}^{ - 9}}}}{{0.10}} = } 1.30 \times {10^{ - 4}}\]
So the degree of dissociation of pyridinium ion \[{{\alpha = }}\]\[1.30 \times {10^{ - 4}}\].
Therefore, percentage of pyridine that forms pyridinium ion is \[{{1}}{{.30 \times 1}}{{\text{0}}^{{\text{ - 4}}}}{{ \times 100 = 0}}{\text{.013% }}\].
Hence the correct option is (D).
Note: The Degree of dissociation of any solute within a solvent is basically the ratio of molar conductivity at C concentration and limiting molar conductivity at zero concentration or infinite dilution. This can be mathematically represented as \[\alpha = \dfrac{{{\Lambda _C}}}{{{\Lambda _0}}}\].
Complete step by step solution:
For the reaction,
Dilution constant can be written as:
\[{K_b} = \dfrac{{\left[ {{A^ + }} \right]\left[ {{B^ - }} \right]}}{{\left[ {AB} \right]}} = \dfrac{{\left( {\alpha {C_0}} \right)\left( {\alpha {C_0}} \right)}}{{\left( {1 - \alpha } \right){C_0}}} = \dfrac{{{\alpha ^2}}}{{1 - \alpha }} \times {C_0}\]
Where \[{{\alpha }}\]is the degree of dissociation of a weak electrolyte. And \[{{\text{C}}_{\text{0}}}\] is the concentration.
For weak electrolyte, \[\alpha < < 0\] so \[\left( {1 - \alpha } \right)\] can be neglected and the resulting equation is:
So, \[\alpha = \sqrt {\dfrac{{{K_b}}}{{{C_0}}}} \]
So, for pyridine on dilution with water results in pyridinium ion. In question, we are given that molarity of pyridine solution is 0.10M and \[{{\text{K}}_{\text{b}}}\] for \[{C_5}{H_5}N\]\[ = 1.7 \times {10^{ - 9}}\].
\[\alpha = \sqrt {\dfrac{{{K_b}}}{{{C_0}}}} = \sqrt {\dfrac{{1.7 \times {{10}^{ - 9}}}}{{0.10}} = } 1.30 \times {10^{ - 4}}\]
So the degree of dissociation of pyridinium ion \[{{\alpha = }}\]\[1.30 \times {10^{ - 4}}\].
Therefore, percentage of pyridine that forms pyridinium ion is \[{{1}}{{.30 \times 1}}{{\text{0}}^{{\text{ - 4}}}}{{ \times 100 = 0}}{\text{.013% }}\].
Hence the correct option is (D).
Note: The Degree of dissociation of any solute within a solvent is basically the ratio of molar conductivity at C concentration and limiting molar conductivity at zero concentration or infinite dilution. This can be mathematically represented as \[\alpha = \dfrac{{{\Lambda _C}}}{{{\Lambda _0}}}\].
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