Consider a $0.050\;M$ solution of the weak base pyridine ($C_5H_5N$ , $K_b$ $=$ $1.7 \cdot 10^{-4}$)
$C_5H_5N$ $+$ $H_2O$ 
$HC_5H_5N^+$ $+$ $OH^-$
To what side will shift the reaction to attain its equilibrium?
a) to the left side b) to the right side c) it will not shift, because the equilibrium is already attained
Correct answer: b) The reaction will shift to the right side