<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:
![1.200M=\frac{\text{Moles of hydrazoic acid}\times 1000}{242.5mL}\\\\\text{Moles of hydrazoic acid}=0.291mol](https://tex.z-dn.net/?f=1.200M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20hydrazoic%20acid%7D%5Ctimes%201000%7D%7B242.5mL%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20hydrazoic%20acid%7D%3D0.291mol)
Molarity of NaOH solution = 0.3400 M
Volume of solution = 1006 mL
Putting values in equation 1, we get:
![0.3400M=\frac{\text{Moles of NaOH}\times 1000}{1006mL}\\\\\text{Moles of NaOH}=0.342mol](https://tex.z-dn.net/?f=0.3400M%3D%5Cfrac%7B%5Ctext%7BMoles%20of%20NaOH%7D%5Ctimes%201000%7D%7B1006mL%7D%5C%5C%5C%5C%5Ctext%7BMoles%20of%20NaOH%7D%3D0.342mol)
The chemical reaction for hydrazoic acid and NaOH follows the equation:
![HN_3+NaOH\rightarrow NaN_3+H_2O](https://tex.z-dn.net/?f=HN_3%2BNaOH%5Crightarrow%20NaN_3%2BH_2O)
<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:
![\text{Molarity of NaOH}=\frac{0.051mol}{1.2485L}=0.0408M](https://tex.z-dn.net/?f=%5Ctext%7BMolarity%20of%20NaOH%7D%3D%5Cfrac%7B0.051mol%7D%7B1.2485L%7D%3D0.0408M)
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:
![pOH=-\log(0.0408)\\\\pOH=1.39](https://tex.z-dn.net/?f=pOH%3D-%5Clog%280.0408%29%5C%5C%5C%5CpOH%3D1.39)
To calculate pH of the solution, we use the equation:
![pH+pOH=14\\\\pH=14-1.39=12.61](https://tex.z-dn.net/?f=pH%2BpOH%3D14%5C%5C%5C%5CpH%3D14-1.39%3D12.61)
Hence, the pH of the solution is 12.61