B+HA→BH++A– Volume of HA used till equivalence point = 6 mL (from given diagram) At half of equivalence point, solution will be basic buffer with B and BH‌+ ∵ pOH=pKb+log‌
[BH+]
[B]
At half equivalence point : [BH+]=[B]. pOH=pKb=14−11=3 ∴ [pKb=3]