Question

What is the pH of a solution prepared by mixing 50.00 mL of 0.10 M NH3 with 5.00 mL of 0.10 M NH4Cl? Assume that the volume of the solutions are additive and that Kb = 1.8 × 10^-5 for NH3.

Answer 1

To find the pH of the solution prepared by mixing NH3 and NH4Cl, calculate the concentrations of OH- and H+ ions using the given volumes and concentrations. The pH is calculated from the pOH value, which is derived from the concentration of OH-. The resulting pH is 10.26.

Step-by-step explanation:

To find the pH of the solution prepared by mixing NH3 and NH4Cl, we need to calculate the concentrations of OH- and H+ ions.

Step 1: Calculate the moles of NH3 and NH4Cl using the given volume and concentration of each solution.

• moles of NH3 = (50.00 mL)(0.10 M) = 5.00 x 10^-3 moles
• moles of NH4Cl = (5.00 mL)(0.10 M) = 5.00 x 10^-4 moles

Step 2: Use the moles of NH3 and NH4Cl to find the concentrations of NH3 and NH4+ ions.

• concentration of NH3 = moles of NH3 / total volume of solution = 5.00 x 10^-3 moles / (50.00 mL + 5.00 mL) = 0.10 M
• concentration of NH4+ = moles of NH4Cl / total volume of solution = 5.00 x 10^-4 moles / (50.00 mL + 5.00 mL) = 0.01 M

Step 3: Use the concentration of NH3 and NH4+ ions to calculate the concentration of OH- and H+ ions using the Kb value of NH3.

• Kb = [NH4+][OH-] / [NH3]
• [OH-] = Kb * [NH3] / [NH4+] = (1.8 x 10^-5)(0.10) / 0.01 = 1.8 x 10^-4 M
• pOH = -log10([OH-]) = -log10(1.8 x 10^-4) = 3.74
• pH = 14 - pOH = 14 - 3.74 = 10.26

Therefore, the pH of the solution prepared by mixing 50.00 mL of 0.10 M NH3 with 5.00 mL of 0.10 M NH4Cl is 10.26.