Question

A $0.25M$ solution of pyridinium chloride $C_5{H}_5{NH}^{+}{CI}^{-}$ was found to have a pH of $2.75$. What is $K_b$ for pyridine, $C_5{H}_{5}N$?

Answer 1

$\left[{py}^{-1}\right]+[H,O]----->\left[p_{y}OH\right]+\left[H^{+}\right]$

$C[1-a]$ $ca$ $ca$

$\frac{\left[p_{y}OH\right]\left[H^{+}\right]}{\left[{py}^{-1}\right]\left[H_{2}O\right]}=K_h$

$\Rightarrow K_h=\frac{kw}{k_b}$

$\left[H^{+}\right]=ca=\sqrt{k_{h}c}$

to we get $p^h=\frac{1}{2}[\log c+\log kw-logcb]$

${PK}_b=-\log c+PKW-2\left[ph\right]$

$=-\log 0.25+14-2\times 2.75$

$=-0.6+14-5.4$

$=8$

$K_b=1\times {10}^{-8}$