Determination of pKb of a mono acidic base


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> $20$ 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($\text{BOH}$)?
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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.




I appreciate your effort you showed in your question, but you missed one important thing that when a weak base is reacted with a strong acid a *buffer solution* is formed and the $\mathrm{pH}$ is not only contributed by base but also by the acidic salt ($\ce{BCl}$ in this case).


So, simply applying the Henderson–Hasselbach equation (derivation can be found [here](https://chem.libretexts.org/Ancillary_Materials/Reference/Organic_Chemistry_Glossary/Henderson-Hasselbach_Equation))


$$\mathrm{pOH} = \mathrm{p}K\_\mathrm{b} + \log{\frac{[\ce{B-}]}{\ce{[BOH]}}}$$


we get


$$4 = \mathrm{p}K\_\mathrm{b} + \log{\frac{1.2}{2.4}},$$


which gives us


$$\mathrm{p}K\_\mathrm{b} = 4.3010$$


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