What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M NH3 with 20.00 mL of 0.10 M NH4Cl? Assume that the volume of the solutions are additive and that Kb = 1.8 × 10-5 for NH3.

To solve this problem, we need to use the Henderson-Hasselbalch equation, which is pH = pKa + log([A-]/[HA]). In this case, NH3 is the weak base (B) and NH4+ is its conjugate acid (HB+).

Step 1: Calculate the initial moles of NH3 and NH4Cl Moles of NH3 = Molarity * Volume = 0.10 M * 50.00 mL = 5.00 mmol Moles of NH4Cl = Molarity * Volume = 0.10 M * 20.00 mL = 2.00 mmol

Step 2: Determine the moles of NH3 and NH4+ in the final solution Since NH4Cl is a strong electrolyte, it will completely dissociate into NH4+ and Cl-. Therefore, the moles of NH4+ will be equal to the initial moles of NH4Cl. Moles of NH3 = Initial moles of NH3 - moles of NH4+ = 5.00 mmol - 2.00 mmol = 3.00 mmol Moles of NH4+ = 2.00 mmol

Step 3: Calculate the concentrations of NH3 and NH4+ in the final solution The total volume of the solution is 50.00 mL + 20.00 mL = 70.00 mL Concentration of NH3 = Moles / Volume = 3.00 mmol / 70.00 mL = 0.043 M Concentration of NH4+ = Moles / Volume = 2.00 mmol / 70.00 mL = 0.029 M

Step 4: Calculate the pKa from the given Kb We know that Ka * Kb = Kw, where Kw is the ion product of water, which is 1.0 x 10^-14 at 25°C. Ka = Kw / Kb = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.56 x 10^-10 pKa = -log(Ka) = -log(5.56 x 10^-10) = 9.25

Step 5: Substitute the values into the Henderson-Hasselbalch equation pH = pKa + log([NH3]/[NH4+]) = 9.25 + log(0.043/0.029) = 9.25 + 0.19 = 9.44

Therefore, the pH of the solution prepared by mixing 50.00 mL of 0.10 M NH3 with 20.00 mL of 0.10 M NH4Cl is approximately 9.44.