Question

A 0.25 M solution of pyridinium chloride CsH5NH CH was found to have a pH of 2.75. What is Ko for
pyridine, CsH5N ?
​

Answer 1

Answer:

$C_{5} H_{5} NH^{+} (aq)+H_{2} O(I)$⇔$C_{5} H_{5} N(aq)+H_{3} O^{+} (aq)$

pH=2.699

$[H_{3} O^{+} ]=10^{-2.699} M=0.002M$

Then

$[C_{5} H_{5} N]=[H_{3} O^{+} ]=0.002M$

and $[C_{5} H_{5} NH^{+} ]=(0.25-0.002)M=0.248M$

Kb=$[C_{5}H_{5} NH^{+}  ][OH^{-} ]/[C_{5} H_{5} N]$

Kb=$[H_{3} O^{+} ][OH^{-} ][C_{5} H_{5} NH^{+} /[C_{5} H_{5} N][H_{3} O^{+} ]$

Kb=Kw*${[C_{5} H_{5} NH^{+} ]/[C_{5} H_{5} N][H_{3} O^{+} ]}$

Kb=$(1*10^{-14} )*(0.25-10^{-2.699} )/(10^{-2.699} )^{2}$

Kb=$6.2*10^{-10}$