A 0.02 M solution of pyridinium hydrochloride has pH=3.44 . Calculate the ionization constantof pyridine.

A 0.02 M solution of pyridinium hydrochloride has pH=3.44 . Calculate

1.	A 0.02 M solution of pyridinium hydrochloride has pH=3.44 . Calculate the ionization constantof pyridine.
Answer» SOLUTION :Pyridinium hydrochloride `C_6H_5N^(+) HCL^(-)` is conjugate ACID of pyridine `(C_6H_5N)`. pH=3.44 = -log `[H^+]` `therefore` log `[H^+] = -3.44 = bar4.56` `therefore [H^+] = 10^(-3.44) = 3.63xx10^(-4) = alpha` `{:(,C_6H_5N^(+)HCl^(-)+ aqhArr ,C_6H_5NCl_((aq))^(-) +, H_((aq))^(+)),("Molarity at equilibrium" , (0.012-alpha)M =0.02 M, 3.63xx10^(-4) , 3.63xx10^(-4)) :}` `K_a=([C_6H_5NCl^(-)][H^+])/([C_6H_5N^(+)HCl^-])` `=((3.63xx10^(-4))(3.63xx10^(-4)))/0.02=6.588xx10^(-6)` Pyridine is conjugate base of pyridinium hydrochloride , so its ionization constant = `K_b` `K_b=K_w/K_a=(1.0xx10^(-14))/(6.588xx10^(-6))=1.5179xx10^(-7)`