A solution contains 0.20 M NH3 (K_b = 1.8x10^-5) and 0.05 M NH4Cl (fully ionised); the ionisation of NH3 is small. Its pH at 298 K is (log 7.2 = 0.857)

JEE Advanced levelSingle correct optionDifficulty 3 of 3ie050

A solution contains $0.20 M$ $NH X_{3}$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ) and $0.05 M$ $NH X_{4} Cl$ (fully ionised); the ionisation of $NH X_{3}$ is small. Its pH at 298 K is ( $lo g 7.2 = 0.857$ )

With common ion

NH X_{4} X^{+}

from the salt,

[OH X^{-}] = K_{b} [NH X_{3}] / [NH X_{4} X^{+}] = 1.8 \times 1 0^{- 5} \times 0.20/0.05 = 7.2 \times 1 0^{- 5} M

pOH = 5 - 0.857 = 4.143

, so

pH = 9.86

. Ignoring the salt gives

11.28