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JEE Advanced levelSingle correct optionDifficulty 3 of 3g2044

The pKa of the anilinium ion is 4.6 at 298 K ( $K_{w} = 1 0^{- 14}$ ). Neglecting the autoionisation of water and the small extent of protonation, the pH of a 0.10 M aqueous solution of aniline is closest to

A4.6
B8.8
C5.2
D9.2

Answer: B

p K_{b} = 14 - 4.6 = 9.4

, so

K_{b} = 1 0^{- 9.4} = 3.98 e - 10

.

[OH X^{-}] = K_{b} c = 3.98 e - 10 \times 0.10 = 6.31 e - 06 M

, so pOH = 5.2 and pH = 8.8. The value 5.2 is the pOH and 4.6 the pKa.

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