Question

$0.25M$ solution of pyridinium chloride $C_5{H}_{5}NHCl$ was found to have a $pH$ of $2.699$. What is $K_b$ for $C_5{H}_{5}N$ ?

Answer 1

$C_5{H}_{5}N{H}^{+}+H_{2}O\leftrightharpoons C_5{H}_{5}N+H_3{O}^{+}$

$K_a=\frac{\left[C_5{H}_{5}N\right]\left[H_3{O}^{+}\right]}{\left[C_5{H}_{5}N{H}^{+}\right]}$

$pH=2.669$; $\left[H_3{O}^{+}\right]={10}^{-2.699}\approx 0.002M$

$\left[C_5{H}_{5}N\right]=\left[H_3{O}^{+}\right]=0.002M$

$\left[C_5{H}_{5}N{H}^{+}\right]=0.25-0.002=0.248M$

$K_b=\frac{\left[C_5{H}_{5}N{H}^{+}\right]\left[{OH}^{-}\right]\left[H^{+}\right]}{\left[C_5{H}_{5}N\right]\left[H^{+}\right]}$

$K_b=\frac{\left[C_5{H}_{5}N{H}^{+}\right]\times (1.0\times {10}^{-14})}{({{10}^{-2.699})}^2}$

$K_b=6.2\times {10}^{-10}$