Question

$ 0.25{\text{ M}} $ solution of pyridinium chloride $ {C_5}{H_5}NHCl $ was found to have pH of $ 2.699 $ . What is $ {K_b} $ for $ {C_5}{H_5}N $ .

Answer 1

Hint: Pyridinium chloride is a salt of weak base $ {C_5}{H_5}N $ and a strong acid $ HCl $ . Since $ HCl $ is a strong acid, that’s why pH of salt is given $ 2.699 $ . We have to find the $ {K_b} $ for $ {C_5}{H_5}N $ . We will use the relation between $ pH $ , $ p{K_b} $ and concentration of the salt for finding the value of $ {K_b} $ .
 $ pH{\text{ = 7 - }}\dfrac{1}{2}p{K_b}{\text{ - }}\dfrac{1}{2}\log C $
Here C is the concentration of the salt.

Complete answer:
Pyridinium Chloride $ {C_5}{H_5}NHCl $ is a salt which is formed when a weak base $ {C_5}{H_5}N $ is made to react with a strong acid $ HCl $ . Since it is an acid-base reaction so formation of salt is must. Since we have used a strong acid then the amount of $ {H^ + } $ ions will be more in the solution. Hence the salt will be acidic in nature. For finding $ {K_b} $ we know that,
 $ pH{\text{ = 7 - }}\dfrac{1}{2}p{K_b}{\text{ - }}\dfrac{1}{2}\log C $
We are given the molarity of the salt which is $ 0.25{\text{ M}} $ . Therefore the value of C is $ 0.25{\text{ M}} $ . We can also derive the base according to following reactions,
 $ \dfrac{1}{2}p{K_b}{\text{ = 4}}{\text{.602}} $
Thus when it is added with water, it produces $ HCl $ again.
 $ {C_5}{H_6}NOH{\text{ }} \rightleftharpoons {\text{ }}{C_5}{H_5}N{\text{ + }}{{\text{H}}_2}O $
And then further it is in equilibrium with the weak base of pyridine.
Now substituting the values in the above formula,
 $ pH{\text{ = 7 - }}\dfrac{1}{2}p{K_b}{\text{ - }}\dfrac{1}{2}\log C $
 $ {\text{2}}{\text{.699 = 7 - }}\dfrac{1}{2}p{K_b}{\text{ - }}\dfrac{1}{2}\log \left( {0.25} \right) $
On solving the equation we get,
 $ \dfrac{1}{2}p{K_b}{\text{ = 4}}{\text{.602}} $
 $ p{K_b}{\text{ = 9}}{\text{.204}} $
 $ {\text{ - log}}\left( {{K_b}} \right){\text{ = 9}}{\text{.204}} $
Now taking antilog both sides we get the result as,
 $ {{\text{K}}_b}{\text{ = 6}}{\text{.25 }} \times {\text{ 1}}{{\text{0}}^{ - 10}} $
The we get the value of $ {{\text{K}}_b}{\text{ = 6}}{\text{.25 }} \times {\text{ 1}}{{\text{0}}^{ - 10}} $ .

Note:
This formula is valid only when weak base and strong acid reacts to make salt. For finding the value of log and antilog, refer to log and antilog tables. We can check the acidic part of the salt by adding water to the given salt. More the value of $ {K_b} $ , the more will be its basicity.