Consider the titration of 25.00 mL of 0.150 M hydrazoic acid (Ka = 1.90 × 10^-5) with 0.200 M NaOH. What is the pH of the solution after the addition of 15.00 mL of NaOH solution?

• A. 7.00
• B. 6.25
• C. 5.10
• D. 4.75

Answer: (C)

Verified step by step guidance

1. Calculate the initial moles of hydrazoic acid (HN3) using the formula: \( \text{moles of HN3} = \text{volume (L)} \times \text{molarity (M)} \). Convert 25.00 mL to liters and multiply by 0.150 M.
2. Calculate the moles of NaOH added using the formula: \( \text{moles of NaOH} = \text{volume (L)} \times \text{molarity (M)} \). Convert 15.00 mL to liters and multiply by 0.200 M.
3. Determine the moles of HN3 remaining after reaction with NaOH. Since NaOH is a strong base, it will react completely with HN3. Subtract the moles of NaOH from the initial moles of HN3 to find the moles of HN3 remaining.
4. Calculate the concentration of the remaining HN3 in the solution. Add the volumes of HN3 and NaOH to find the total volume of the solution in liters. Use the formula: \( \text{concentration of HN3} = \frac{\text{moles of HN3 remaining}}{\text{total volume (L)}} \).
5. Use the Henderson-Hasselbalch equation to find the pH of the solution: \( \text{pH} = \text{pKa} + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \). Calculate \( \text{pKa} \) from \( \text{Ka} \) using \( \text{pKa} = -\log(\text{Ka}) \), and substitute the concentrations of the conjugate base \([\text{A}^-]\) and the weak acid \([\text{HA}]\) into the equation.