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JEE Main levelSingle correct optionDifficulty 2 of 3ie029

A $0.25 M$ solution of a weak base has pH $11.0$ at 298 K. Its $K_{b}$ (neglect ionised fraction relative to $c$ ) is

A $4 \times 1 0^{- 6}$
B $1 \times 1 0^{- 5}$
C $1 \times 1 0^{- 6}$
D $2.5 \times 1 0^{- 4}$

Answer: A

pOH

= 3

, so

[OH X^{-}] = 1 0^{- 3} M

and

K_{b} = [OH X^{-}]^{2} / c = 1 0^{- 6} /0.25 = 4 \times 1 0^{- 6}

.

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