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A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine.
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Hint: This question deals with the hydrolysis of salt of weak base with strong acid. The strong acid here is $\text{HCl}$, the weak base is pyridine and the salt of the weak base is pyridinium hydrochloride. The reaction of hydrolysis is ${{\text{C}}_{5}}{{\text{H}}_{6}}{{\text{N}}^{+}}\text{C}{{\text{l}}^{-}}+{{\text{H}}_{2}}\text{O}\rightleftharpoons \left[ {{\text{C}}_{5}}{{\text{H}}_{6}}{{\text{N}}^{+}} \right]\left[ \text{O}{{\text{H}}^{-}} \right]+H \text{Cl}$ .
Complete answer:
Let us solve this question step by step like derivation of hydrolysis of salt of weak base and strong acid.
Step (1)- Write the reaction and introduce degree of hydrolysis and hydrolysis constant:
${{\text{C}}_{5}}{{\text{H}}_{6}}{{\text{N}}^{+}}\text{C}{{\text{l}}^{-}}+{{\text{H}}_{2}}\text{O}\rightleftharpoons \left[ {{\text{C}}_{5}}{{\text{H}}_{6}}{{\text{N}}^{+}} \right]\left[ \text{O}{{\text{H}}^{-}} \right]+H \text{Cl}$ .
h is the degree of hydrolysis which represents the extent of hydrolysis.
${{\text{K}}_{\text{h}}}$ is like other dissociation or equilibrium constants which defines equilibrium of hydrolysis. It is expressed as \[{{\text{K}}_{\text{h}}}=\dfrac{\left[ \left( {{\text{C}}_{5}}{{\text{H}}_{6}}{{\text{N}}^{+}} \right)\left( \text{O}{{\text{H}}^{-}} \right) \right]\left[ {{\text{H}}^{+}} \right]\left[ \text{C}{{\text{l}}^{-}} \right]}{\left[ {{\text{C}}_{5}}{{\text{H}}_{6}}{{\text{N}}^{+}} \right]\left[ \text{C}{{\text{l}}^{-}} \right]}\] .
Step (2)- Calculate the concentration of ${{\text{H}}^{+}}$ ion from the pH given in the question using the formula $\left[ {{\text{H}}^{+}} \right]={{10}^{-\text{pH}}}$ and pH is given as 3.44. So, the $\left[ {{\text{H}}^{+}} \right]$ will be 0.00036 M.
Step (3)- Calculate the hydrolysis constant or ${{\text{K}}_{\text{h}}}$ using the formula ${{\text{K}}_{\text{h}}}=\dfrac{\left[ {{\text{H}}^{+}} \right]}{\text{c}}$ where $\left[ {{\text{H}}^{+}} \right]$ is the concentration of hydrogen ion which is 0.00036 M and c is the concentration of the salt or pyridinium chloride. The value of c is given as 0.2 M. The ${{\text{K}}_{\text{h}}}$ will be $\dfrac{0.00036}{0.2}$ or ${{\text{K}}_{\text{h}}}=0.0018$.
Step (4)- In this question, we have to find the ionization constant of pyridine or the weak base. The formula to be used here is ${{\text{K}}_{\text{b}}}=\dfrac{{{\text{K}}_{\text{w}}}}{{{\text{K}}_{\text{h}}}}$ , where ${{\text{K}}_{\text{h}}}$is the hydrolysis constant and ${{\text{K}}_{\text{w}}}$ is the ionic product of water. The value of ${{\text{K}}_{\text{w}}}$ is fixed at standard temperature of ${{25}^{\text{o}}}\text{C}$. This constant value is ${{10}^{-14}}$. So, the dissociation constant of base will be $\dfrac{{{10}^{-14}}}{0.0018}$ or $5.55\times {{10}^{-12}}$.
The dissociation constant of pyridine is $5.55\times {{10}^{-12}}$.
Note: These formulae are to be used when the hydrolysis of salt is considered to be negligible. Here, pyridine is a weak base because it is an aromatic base. So, we can consider the hydrolysis of its salt to be very less. We can consider hydrolysis negligible till the time it is not mentioned to consider it and compound is basic.
Complete answer:
Let us solve this question step by step like derivation of hydrolysis of salt of weak base and strong acid.
Step (1)- Write the reaction and introduce degree of hydrolysis and hydrolysis constant:
${{\text{C}}_{5}}{{\text{H}}_{6}}{{\text{N}}^{+}}\text{C}{{\text{l}}^{-}}+{{\text{H}}_{2}}\text{O}\rightleftharpoons \left[ {{\text{C}}_{5}}{{\text{H}}_{6}}{{\text{N}}^{+}} \right]\left[ \text{O}{{\text{H}}^{-}} \right]+H \text{Cl}$ .
h is the degree of hydrolysis which represents the extent of hydrolysis.
${{\text{K}}_{\text{h}}}$ is like other dissociation or equilibrium constants which defines equilibrium of hydrolysis. It is expressed as \[{{\text{K}}_{\text{h}}}=\dfrac{\left[ \left( {{\text{C}}_{5}}{{\text{H}}_{6}}{{\text{N}}^{+}} \right)\left( \text{O}{{\text{H}}^{-}} \right) \right]\left[ {{\text{H}}^{+}} \right]\left[ \text{C}{{\text{l}}^{-}} \right]}{\left[ {{\text{C}}_{5}}{{\text{H}}_{6}}{{\text{N}}^{+}} \right]\left[ \text{C}{{\text{l}}^{-}} \right]}\] .
Step (2)- Calculate the concentration of ${{\text{H}}^{+}}$ ion from the pH given in the question using the formula $\left[ {{\text{H}}^{+}} \right]={{10}^{-\text{pH}}}$ and pH is given as 3.44. So, the $\left[ {{\text{H}}^{+}} \right]$ will be 0.00036 M.
Step (3)- Calculate the hydrolysis constant or ${{\text{K}}_{\text{h}}}$ using the formula ${{\text{K}}_{\text{h}}}=\dfrac{\left[ {{\text{H}}^{+}} \right]}{\text{c}}$ where $\left[ {{\text{H}}^{+}} \right]$ is the concentration of hydrogen ion which is 0.00036 M and c is the concentration of the salt or pyridinium chloride. The value of c is given as 0.2 M. The ${{\text{K}}_{\text{h}}}$ will be $\dfrac{0.00036}{0.2}$ or ${{\text{K}}_{\text{h}}}=0.0018$.
Step (4)- In this question, we have to find the ionization constant of pyridine or the weak base. The formula to be used here is ${{\text{K}}_{\text{b}}}=\dfrac{{{\text{K}}_{\text{w}}}}{{{\text{K}}_{\text{h}}}}$ , where ${{\text{K}}_{\text{h}}}$is the hydrolysis constant and ${{\text{K}}_{\text{w}}}$ is the ionic product of water. The value of ${{\text{K}}_{\text{w}}}$ is fixed at standard temperature of ${{25}^{\text{o}}}\text{C}$. This constant value is ${{10}^{-14}}$. So, the dissociation constant of base will be $\dfrac{{{10}^{-14}}}{0.0018}$ or $5.55\times {{10}^{-12}}$.
The dissociation constant of pyridine is $5.55\times {{10}^{-12}}$.
Note: These formulae are to be used when the hydrolysis of salt is considered to be negligible. Here, pyridine is a weak base because it is an aromatic base. So, we can consider the hydrolysis of its salt to be very less. We can consider hydrolysis negligible till the time it is not mentioned to consider it and compound is basic.
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