To 50.0 mL of 0.10 M aqueous ammonia solution, 25.0 mL of 0.10 M HCl is added. If \( K_b \) of ammonia is \( 1.77 \times 10^{-5} \) and \( \log 1.77 = 0.248 \), the pH of the resulting mixture at 298 K is:

Initial millimoles: \( \text{NH}_3 = 5.0 \), \( \text{HCl} = 2.5 \). After reaction, remaining \( \text{NH}_3 = 2.5\text{ mmol} \), formed \( \text{NH}_4^+ = 2.5\text{ mmol} \). Since \( [\text{base}] = [\text{salt}] \), \( pOH = pK_b + \log(1) = pK_b = 4.752 \). Thus, \( pH = 14 - 4.752 = 9.248 \approx 9.25 \).