200 mL of 0.50 M $NH X_{3}$ ( $p K_{b} = 4.74$ , $p K_{w} = 14$ ) is titrated by adding V mL of 0.50 M HCl until the pH of the solution is 9.00. Find V in mL, to the nearest integer.

Numerical answer type, answer in mL, tolerance 1.

Answer: 129 mL

NH3 = 100 mmol. With x mmol HCl,

pOH = 5.00 = 4.74 + lo g (x / (100 - x))

, so

x / (100 - x) = 1 0^{0.26} = 1.820

, x = 64.54 mmol. V = 64.54/0.50 = 129.1 mL, about 129 mL.