$50 mL$ of $0.2 M$ $NH X_{3}$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ) is mixed with $50 mL$ of $0.002 M$ $NaOH$ . Taking the ammonia ionisation in presence of the strong base into account, the pH of the mixture at 298 K is nearest to

Answer: C

After mixing,

[NH X_{3}] = 0.1 M

and

[OH X^{-}]_{NaOH} = 0.001 M

. Let

x

be the

OH X^{-}

from ammonia:

x (0.001 + x) = 1.8 \times 1 0^{- 5} (0.1 - x)

, giving

x = 9.26 e - 04 M

. Total

[OH X^{-}] = 1.93 e - 03 M

pOH = 2.72

pH = 11.28