$20$ mL of a weak monoacidic base($\text{BOH}$) requires $12$ mL of $0.3$ M $\text{HCl}$ solution for the equivalence point. During titration, the pH of the base solution was $10$ upon the addition of $4$ mL of $0.3$ M $\text{HCl}$ solution. What is the $\text{pKb}$ of the base($\text{BOH}$)?
My attempt:
The concentration of the base is given by,
$c * 20 = 12 * 0.3$ -- (equating the moles of the acid and base)
$c = 0.18$ M
So, initial moles of the base in the container $= 20 * 0.18 = 3.6$ mmol
Moles of acid that is added $ = 4 * 0.3 = 1.2$ mmol
Since the base is monoacidic, they will react in a 1:1-mole ratio. The acid is the limiting reagent, so it will be fully consumed. Therefore the moles of base left is,
$3.6 - 1.2 = 2.4$ mmol
The concentration of the base is,
$\frac{2.4}{24} = 0.1$ M
Since the pH of the solution is $10$, therefore the concentration of $\text{OH}^- = 10^{-4}$
Applying the approximated formula for calculating the $\text{k}_b$ of a weak base,
$$\text{K}_b = \frac{x^2}{c}$$
Where,
x = concentration of $\text{OH}^-$ ions
c = concetration of the base
In our case,
x = $10^{-4}$
c = $0.1$
Plugging in the values I got $pK_b = 7$. But the answer is given $4.3$.
Any help would be appreciated.