100 mL of 0.30 M NH3 is mixed with 100 mL of 0.20 M CH3COOH. Given K_b(NH3) = 1.8 x 10^-5 (pK_b = 4.74), K_a(CH3COOH) = 1.8 x 10^-5 and log 2 = 0.301, the pH is closest to (acetate is a very weak base and does not alter the NH3/NH4+ ratio appreciably)

JEE Advanced levelSingle correct optionDifficulty 3 of 3ia104

$100 mL$ of $0.30 M$ $NH X_{3}$ is mixed with $100 mL$ of $0.20 M$ $CH X_{3} COOH$ . Given $K_{b} (NH X_{3}) = 1.8 \times 1 0^{- 5}$ ( $p K_{b} = 4.74$ ), $K_{a} (CH X_{3} COOH) = 1.8 \times 1 0^{- 5}$ and $lo g 2 = 0.301$ , the pH is closest to (acetate is a very weak base and does not alter the $NH X_{3} / NH X_{4} X^{+}$ ratio appreciably)

The acid reacts completely with

NH X_{3}

30 mmol

NH X_{3}

+

20 mmol

CH X_{3} COOH

gives

20 mmol

NH X_{4} X^{+}

and

10 mmol

NH X_{3}

remaining.

pOH = p K_{b} + lo g (20/10) = 4.74 + 0.301 = 5.04

, hence pH

= 14 - 5.04 = 8.96