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.

The pH of the solution is 12.61

Explanation:

To calculate the number of moles for given molarity, we use the equation:

      ......(1)

Molarity of hydrazoic acid solution = 1.200 M

Volume of solution = 242.5 mL

Putting values in equation 1, we get:

Molarity of NaOH solution = 0.3400 M

Volume of solution = 1006 mL

Putting values in equation 1, we get:

The chemical reaction for hydrazoic acid and NaOH follows the equation:

Initial:           0.291        0.342

Final:                  0          0.051                 0.291      0.291

Volume of solution = 242.5 + 1006 = 1248.5 mL = 1.2485 L    (Conversion factor:  1 L = 1000 mL)

Left moles of NaOH = 0.051 moles

Volume of the solution = 1.2485 L

Putting values in equation 1, we get:

1 mole of NaOH produces 1 mole of sodium ions and 1 mole of hydroxide ions

To calculate pOH of the solution, we use the equation:

We are given:

Putting values in above equation, we get:

To calculate pH of the solution, we use the equation:

Hence, the pH of the solution is 12.61

it takes on an electronical energy