Question

0.25 M solution of pyridine chloride $C_5{H}_6{N}^{+}C{l}^{-}$ was found to have a pH of 2.699. What is $K_b$ for pyridine, $C_5{H}_{5}N$?(log2=0.3010)

Answer 1

For a salt of weal base and strong acid,

$pH=7-\frac{1}{2}p{K}_b-\frac{1}{2}\log c$

Where $C$ is the concentration of the salt.

$C_5{H}_6{N}^{+}C{e}^{-}$ is the salt given

$C_5{H}_6{N}^{+}C{e}^{-}+H_{2}O\rightleftharpoons C_5{H}_{6}NOH+HCe$

$C_5{H}_{6}NOH\rightleftharpoons C_5{H}_{5}N+H_{2}O$

Given,

$pH=2.699$

$C=0.25M$

$2.699=7-\frac{1}{2}p{K}_b-\frac{1}{2}\log \left(0.25\right)$

$\frac{1}{2}p{K}_b=4.602$

$p{K}_b=9.204$

$-\log \left[K_b\right]=9.204$

$K_b=6.25\times {10}^{-10}$