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QJEEAdv20Chc0b4: JEE Advanced 2020 Chemistry - Class 11 JEEAdv20Chc0b4 Solution: A solution of 0.1 M weak base (B) is titrated with 0.1 M of a…

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 neutralisation reaction is given by

B+HA⟶BH++A−

JEE Advanced 2020 Paper 2 Offline Chemistry - Ionic Equilibrium Question 7 English

Step-by-Step Verified Solution

From the given diagram, 63 mL volume of HA used till equivalence point. 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] (∵ pH = 11)

Therefore, pOH = pKb = 14 − 11 = 3

∵ pKb = 3.00

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