<u>Answer:</u> The pH of the solution is 12.61
<u>Explanation:</u>
To calculate the number of moles for given molarity, we use the equation:
......(1)
- <u>For hydrazoic acid:</u>
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:

<u>Initial:</u> 0.291 0.342
<u>Final:</u> 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:
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
We are given:
![[OH^-]=0.0408M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D0.0408M)
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