A 20.00-mL sample of 0.150 M NH3 is being titrated with 0.200 M HCl. What is the pH after 15.00 mL of HCl has been added? Kb of NH3 = 1.8 × 10⁻⁵

• A. 5.24
• B. 4.98
• C. 11.02
• D. 9.26

Answer: (A)

Verified step by step guidance

1. Calculate the initial moles of NH3 in the solution using the formula: \( \text{moles of NH3} = M \times V \), where \( M \) is the molarity and \( V \) is the volume in liters.
2. Calculate the moles of HCl added using the formula: \( \text{moles of HCl} = M \times V \), where \( M \) is the molarity and \( V \) is the volume in liters.
3. Determine the moles of NH3 remaining and the moles of NH4+ formed after the reaction between NH3 and HCl. The reaction is: \( \text{NH3} + \text{HCl} \rightarrow \text{NH4+} + \text{Cl-} \).
4. Calculate the concentration of NH4+ and NH3 in the solution after the reaction. Use the total volume of the solution (initial volume of NH3 plus the volume of HCl added) to find these concentrations.
5. Use the Henderson-Hasselbalch equation to find the pH of the solution: \( \text{pH} = \text{pK}_a + \log \left( \frac{[\text{base}]}{[\text{acid}]} \right) \), where \( \text{pK}_a = 14 - \text{pK}_b \) and \( [\text{base}] \) and \( [\text{acid}] \) are the concentrations of NH3 and NH4+ respectively.