An analytical chemist is titrating 242.5 mL of a 1.200 M solution of hydrazoic acid HN3 with a 0.3400 M 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.

The pH of the solution is 12.61

Explanation:

Step 1: Data given

Molarity of hydrazoic acid solution = 1.200 M

Volume of solution = 242.5 mL

Molarity of NaOH solution = 0.3400 M

pKa = 4.72

Step 2: The balanced equation

HN3 + NaOH → NaN3 + H2O

Step 3: Calculate moles hydrazoic acid

Moles hydrazoic acid = molarity * volume

Moles hydra

zoic 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: Calculate the limiting reactant

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

NaOH is in excess. There reacts 0.29 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 molarity of NaOH

Molarity NaOH = 0.051 moles / 1.2485 L

Molarity NaOH = 0.0408 M

Step 8: Calculate pOH

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

pOH = 1.39

Step 9: Calculate pH

pH = 14 - pOH

pH = 14 - 1.39

pH = 12.61

The pH of the solution is 12.61

thats how its supposed to be

a pan containing water is placed in a positio n where it is exposed to a 10 mile per hour wind. the pan contains a uniform depth of 1.27 cm of water, is quite wide, and has a length of 2.44 m in the wind direction. the water in the pan is at a constant temperature of 15.5 °c.