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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ie055

For $0.05 M$ $NH X_{3}$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ , 298 K; $lo g 9.487 = 0.977$ ) with $α ≪ 1$ , which statements are correct?

A $[OH X^{-}] \approx 9.5 \times 1 0^{- 4} M$ .
BThe pH is about $10.98$ .
CThe degree of ionisation is about $3.8%$ .
DAdding solid $NH X_{4} Cl$ raises the pH.

Answer: A, B

[OH X^{-}] = K_{b} c = 9 \times 1 0^{- 7} = 9.5 \times 1 0^{- 4} M

;

pOH = - lo g (9.5 \times 1 0^{- 4}) = 3.023

, so

pH = 10.98

.

α = 9.487 e - 04/0.05 = 1.9%

(not

3.8%

). Added

NH X_{4} X^{+}

suppresses ionisation, lowering

[OH X^{-}]

and pH.

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