40.0 mL of 0.10 M aqueous $NH X_{3}$ ( $K_{b} = 1.8 \times 1 0^{- 5}$ ) is titrated with 0.10 M HCl. Using $lo g 3 = 0.477$ , calculate the pH after adding 30.0 mL of HCl. Give two decimal places.

Numerical answer type, tolerance 0.02.

Answer: 8.78

NH X_{3} = 4.0

mmol, HCl

= 3.0

mmol, giving 3.0 mmol

NH X_{4} X^{+}

and 1.0 mmol

NH X_{3}

(a buffer). pOH

= p K_{b} + lo g ([NH X_{4} X^{+}] / [NH X_{3}]) = 4.745 + 0.477 = 5.222

, so pH

= 8.78