Question

A solution of weak base BOH was titrated with $1MHCI$. The pH of the solution was found to be $10.04$ and $9.14$ after the addition of $5ml$ and $20ml$ of the acid respectively. Find the dissociation constant of the base.

Answer 1

Let $N_B$ be the normality of base $BOH$

Substituting the given data,

$14-10.04=p{K}_b+\log \left(\frac{5.0\times 1}{40\times N_B-5.0\times 1}\right)⟶\left(1\right)$

$\log \left[\right(\frac{20}{40N_B-20})\times (\frac{40N_B-5}{5}\left)\right]=0.9$

$\left(\frac{20}{40N_B-20}\right)\times \left(\frac{40N_B-5}{5}\right)=7.943$

Solving for $N_B,$ we get, $N_B=0.875$

$Eqn\left(1\right)⟶$

$3.96=p{K}_b+\log \left(\frac{5}{40\times 0.815-5}\right)=p{K}_b+\log \left(\frac{5}{30}\right)$

$p{K}_b=3.96-\log \left(\frac{1}{6}\right)=3.96+0.778$

$p{K}_b=4.738K_b=1.83\times {10}^{-5}$