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 p𝐾b of the base? The neutralization reaction is given by B+HA → BH+ + A− .

B+HA⟶BH++A−0.1 M, V ml0.1 V m mol 0.1V m mol 0.1 V0.1 VBH+=0.1V2V=0.05MpH at eq. pt = 6 to 6.28pH=7−12pKb+log⁡0.05So pKb = 2.30 – 2.80PossibleSolution - 2at V = 6 ml                   rxn is completeSo V = 3 ml                  is half of eq. ptat which                        pH = 11pOH = (14 – 11) = pKb + log1pKb = 3