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JEE Advanced 2020Paper 2

Exam: JEE Advanced
Year: 2020
Paper: Paper 2
Subject: Chemistry
Chapter: Ionic Equilibrium

JEE Advanced 2020 · Paper 2 · Chemistry · Ionic Equilibrium

A solution of 0 . 1M weak base ( B ) is titrated with 0 . 1M of a...

A solution of

0 . 1M

weak base

( B )

is titrated with

0 . 1M

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 \to {BH}^{+} + {A}^{-}

.

A solution of 0 . 1M weak base ( B ) is titrated with 0 . 1M of a strong acid ( HA ) . The variation of pH of the

Correct answer

3

Step-by-step solution

At $3mlpH$ of solution is $11$ .

$B + HA \to {BH}^{+} + {A}^{-}$

at $3$ best buffer action is shown

So, ${P}_{OH} = {P}_{{K}_{b}} = 3$ (As basic buffer is formed)

So, ${P}_{{K}_{b}} = 3$

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