Questions & Answers

Question

Answers

Answer
Verified

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 $5.55\times {{10}^{-12}}$.