A 20.00-mL sample of 0.150 M NH3 is being titrated with 0.200 M HCl. What is the pH after 25.00 mL of HCl has been added? Kb of NH3 = 1.8 × 10⁻⁵

• A. 5.25
• B. 3.50
• C. 9.26
• D. 4.74

Answer: (D)

Verified step by step guidance

1. Calculate the initial moles of NH3 in the solution using the formula: \( \text{moles} = \text{volume} \times \text{molarity} \). For NH3, \( \text{moles} = 20.00 \text{ mL} \times 0.150 \text{ M} \). Convert the volume from mL to L by dividing by 1000.
2. Calculate the moles of HCl added using the same formula: \( \text{moles} = \text{volume} \times \text{molarity} \). For HCl, \( \text{moles} = 25.00 \text{ mL} \times 0.200 \text{ M} \). Again, convert the volume from mL to L.
3. Determine the limiting reactant by comparing the moles of NH3 and HCl. Since HCl is a strong acid and NH3 is a weak base, they will react in a 1:1 ratio to form NH4+. Subtract the moles of HCl from the moles of NH3 to find the remaining moles of NH3 or the moles of NH4+ formed.
4. Calculate the concentration of NH4+ in the solution. The total volume of the solution after the addition of HCl is the sum of the initial NH3 solution volume and the added HCl volume. Use this total volume to find the concentration: \( \text{concentration} = \frac{\text{moles of NH4+}}{\text{total volume in L}} \).
5. Use the Henderson-Hasselbalch equation to find the pH of the solution: \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \). First, calculate \( \text{pKa} \) from \( \text{pKb} \) using the relation \( \text{pKa} + \text{pKb} = 14 \). Then, substitute the concentrations of NH3 and NH4+ into the equation to find the pH.