Question

20mL of a weak monoacidic base BOH requires 12 mL 0.3 M HCL for the equivalence point. During titration pH of the base solution was 10 upon addition of 4 mL 0.3 M HCL solution . What is the pKb of the base BOH ?

Answer 1

$BOH+H_{2}O\rightleftharpoons B^{+}+O{H}^{-}$ ...(i)
(c-x) (x) (x)
$P{K}_b=\frac{\left[B^{+}\right]\left[O{H}^{-}\right]}{\left[BOH\right]}$
20 ml of base is neutralized by 12 ml of 0.3 M HCl.
$\Rightarrow O{H}^{-}$ ions is base is balanced by $H^{+}$ from the acid.
$V_1\times C_1=V_2\times C_2$
$20\times C=12\times 0.3$
$\Rightarrow C=\frac{\text{/}{12}^3\times 0.3}{\text{/}{20}^5}$
[BOH]=0.18 M
Given $\left[H^{+}\right]$ in terms of $p^H$ as 10.
$\Rightarrow$ POH = 4
$\Rightarrow \left[O{H}^{-}\right]={10}^{-4}\left(from\right(i\left)\right[B^{+}]={10}^{-4}]$
$P{K}_b=\frac{{10}^{-4}\times {10}^{-4}}{0.18}=5.5\times {10}^{-10}P{K}_b==5.5\times {10}^{-10}$
NOTE: $\left[BOH\right]=0.18-{10}^{-4}\simeq 0.18$ (low dissociation).