A solution of 0.1 M weak base (B) is titrated with 0.1 M of a strong acid (HA). The variation of pH of the solution with the volume of HA added is shown in the figure below. What is the p𝐾_b of the base? The neutralization reaction is given by B+HA → BH⁺ + A⁻

 

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$\mathrm{B}+\mathrm{HA}⟶{\mathrm{BH}}^{+}+{\mathrm{A}}^{-}$

0.1 M, V ml
0.1 V m mol 0.1V m mol 0.1 V0.1 V

$\left[{\mathrm{BH}}^{+}\right]=\frac{0.1\mathrm{V}}{2\mathrm{V}}=0.05\mathrm{M}$

pH at eq. pt = 6 to 6.28

$\mathrm{pH}=7-\frac{1}{2}\left[{\mathrm{pK}}_{\mathrm{b}}+\log 0.05\right]$

So pK_b = 2.30 – 2.80

Possible

Solution - 2

at V = 6 ml                   rxn is complete
So V = 3 ml                  is half of eq. pt
at which                        pH = 11
pOH = (14 – 11) = pK_b + log1
pK_b = 3