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JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ie106

For $0.1 M$ ammonia solution ( $K_{b} = 1.8 \times 1 0^{- 5}$ at 298 K; $lo g 1.34 = 0.127$ , use $[OH X^{-}] = K_{b} c$ ), select the correct statement(s).

A $[NH X_{4} X^{+}] \approx 1.34 \times 1 0^{- 2} M$ .
BThe pH is about 11.13.
CA $0.1 M$ $NaOH$ solution has a higher pH (13).
DThe degree of ionisation is about $1.34%$ .

Answer: B, C, D

[OH X^{-}] = [NH X_{4} X^{+}] = 1.8 \times 1 0^{- 5} \times 0.1 = 1.34 \times 1 0^{- 3} M

(not

1 0^{- 2}

).

α = 1.34 \times 1 0^{- 3} /0.1 = 1.34%

.

pOH = 3 - 0.127 = 2.87

and

pH = 11.13

.

0.1 M

NaOH

gives pH 13.

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