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Sphinxa [80]
3 years ago
12

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 o

f 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.
Chemistry
1 answer:
Arlecino [84]3 years ago
6 0

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

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