Question

An analytical chemist is titrating 242.5mL of a 1.200M solution of hydrazoic acid HN3 with a 0.3400M solution of NaOH. The pKa of hydrazoic acid is 4.72.
Calculate the pH of the acid solution after the chemist has added 1006.mL of the NaOH solution to it.
Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places.

Answer 1

Answer:

The pH of the solution is 12.61

Explanation:

Step 1: Data given

Concentration of hydrazoic acid solution = 1.200 M

Volume of the solution = 242.5 mL = 0.2425 L

Concentration of NaOH solution = 0.3400 M

The pKa of hydrazoic acid is 4.72

Step 2: The balanced equation

NaOH + HN3  → NaN3 + H2O

Step 3: Calculate moles hydrazoic acid

Moles hydrazoic acid (HN3) = concentration * volume

Moles hydrazoic acid = 1.200 M * 0.2425 L

Moles hydrazoic acid = 0.291 moles

Step 4: Calculate moles NaOH

Moles NaOH = 0.3400 M * 1.006 L

Moles NaOH = 0.342 moles

Step 5: Determine the limiting reactant

HN3 is the limiting reactant. It will completely be consumed (0.291 moles)

NaOH is in excess. There reacts 0.291 moles. There will remain 0.342 - 0.291 = 0.051 moles NaOH

Step 6: Calculate total volume

Total volume = 242.5 mL + 1006 mL = 1248.5 mL = 1.2485 L

Step 7: Calculate concentration of NaOH

concentration NaOH = 0.051 moles / 1.2485 L

concentration NaOH = 0.0408 M

Step 8: Calculate pOH

pOH = -log [OH-] = -log(0.0408)

pOH = 1.39

Step 9: Calculate pH

pH + pOH = 14

pH = 14 - pOH

pH = 14 - 1.39

pH = 12.61

The pH of the solution is 12.61