What volume of 0.10 M HCl (in mL, nearest integer) must be added to 100 mL of 0.15 M NH3 to obtain a solution of pH 9.00? pK_b(NH3) = 4.74, K_w = 1.0 x 10^-14. Treat the product as an NH3/NH4+ buffer and take 10^-0.26 = 0.55.

JEE Advanced levelNumerical answerDifficulty 3 of 3ia108

What volume of $0.10 M$ $HCl$ (in $mL$ , nearest integer) must be added to $100 mL$ of $0.15 M$ $NH X_{3}$ to obtain a solution of pH $9.00$ ? $p K_{b} (NH X_{3}) = 4.74$ , $K_{w} = 1.0 \times 1 0^{- 14}$ . Treat the product as an $NH X_{3} / NH X_{4} X^{+}$ buffer and take $1 0^{- 0.26} = 0.55$ .

Numerical answer type, answer in mL, tolerance 0.5.

pH = 14 - p K_{b} + lo g ([NH X_{3}] / [NH X_{4} X^{+}])

, so

9.00 = 9.26 + lo g (ratio)

and

[NH X_{3}] / [NH X_{4} X^{+}] = 1 0^{- 0.26} = 0.5495

. With

15.0 mmol

NH X_{3}

initially and

x

mmol

HCl

(15 - x) / x = 0.5495

, so

x = 15/ (1 + 0.5495) = 9.68 mmol

. Volume

= 9.68/0.10 = 96.8 mL

, about 97 mL.