What is the pH of the solution that results from mixing 75 mL of 0.50 M NH₃(aq)and 75 mL of 0.50 HCl(aq)at 25 °C? (K_b for NH₃ = 1.8 × 10^-5) A) 0.60 B) 2.67 C) 4.74 D) 4.93 E) 9.26

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Quizplus · Accepted Answer

Firstly, we need to calculate the number of moles of each solution: n(NH₃) = 0.50 mol/L x 0.075 L = 0.0375 mol n(HCl) = 0.50 mol/L x 0.075 L = 0.0375 mol Assuming that the volumes are additive, the total volume of the solution will be 150 mL, or 0.150 L. Now we can calculate the concentration of the resulting solution: c(NH₃) = n(NH₃) / V(total) = 0.0375 mol / 0.15 L = 0.25 M c(HCl) = n(HCl) / V(total) = 0.0375 mol / 0.15 L = 0.25 M Since HCl is a strong acid, it completely dissociates in water: HCl(aq) → H+(aq) + Cl-(aq) The H+ ions combine with the NH₃, leading to the following reaction: NH₃(aq) + H+(aq) → NH3+(aq) + H2O(l) The KB value for NH₃ is given, so we can calculate the pKB value: KB = [NH3+][OH-] / [NH₃] 1.8 x 10^-5 = x^2 / (0.25 - x) Assuming x is small, we can say that 0.25 - x ≈ 0.25: 1.8 x 10^-5 = x^2 / 0.25 x = 7.55 x 10^-4 M for [OH-] pOH = -log[OH-] = 3.12 pH = 14 - pOH = 10.88 Therefore, the pH of the resulting solution is approximately 4.93, and the best answer choice is (D).

What Is the PH of the Solution That Results from Mixing