What is the pH at the equivalence point for the titration of 25.0 mL of 0.414 M HNO2 (Ka = 5.6 × 10^-4) with 0.414 M NaOH?

• A. 7.00
• B. 8.08
• C. 9.25
• D. 5.32

Answer: (B)

Verified step by step guidance

1. Identify the type of titration: This is a weak acid (HNO2) being titrated with a strong base (NaOH). At the equivalence point, the weak acid is completely neutralized, forming its conjugate base (NO2^-).
2. Calculate the moles of HNO2 initially present: Use the formula \( \text{moles} = \text{concentration} \times \text{volume} \). Convert the volume from mL to L by dividing by 1000.
3. Determine the moles of NaOH required to reach the equivalence point: Since the concentrations of HNO2 and NaOH are equal, the moles of NaOH needed will be the same as the moles of HNO2.
4. At the equivalence point, all HNO2 is converted to NO2^-. Calculate the concentration of NO2^- in the solution: The total volume of the solution is the sum of the volumes of HNO2 and NaOH. Use this total volume to find the concentration of NO2^-.
5. Calculate the pH at the equivalence point: Use the hydrolysis of the conjugate base NO2^- to find the pH. Set up the equilibrium expression for the reaction of NO2^- with water, and use the Kb value (derived from Kw and Ka) to find the concentration of OH^-. Finally, calculate the pH from the pOH.