Respuesta :
Answer: The pH of the solution is 12.61
Explanation:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}\times 1000}{\text{Volume of solution (in mL)}}[/tex] ......(1)
- For hydrazoic acid:
Molarity of hydrazoic acid solution = 1.200 M
Volume of solution = 242.5 mL
Putting values in equation 1, we get:
[tex]1.200M=\frac{\text{Moles of hydrazoic acid}\times 1000}{242.5mL}\\\\\text{Moles of hydrazoic acid}=0.291mol[/tex]
- For NaOH:
Molarity of NaOH solution = 0.3400 M
Volume of solution = 1006 mL
Putting values in equation 1, we get:
[tex]0.3400M=\frac{\text{Moles of NaOH}\times 1000}{1006mL}\\\\\text{Moles of NaOH}=0.342mol[/tex]
The chemical reaction for hydrazoic acid and NaOH follows the equation:
[tex]HN_3+NaOH\rightarrow NaN_3+H_2O[/tex]
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)
- For NaOH left:
Left moles of NaOH = 0.051 moles
Volume of the solution = 1.2485 L
Putting values in equation 1, we get:
[tex]\text{Molarity of NaOH}=\frac{0.051mol}{1.2485L}=0.0408M[/tex]
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:
[tex]pOH=-\log[OH^-][/tex]
We are given:
[tex][OH^-]=0.0408M[/tex]
Putting values in above equation, we get:
[tex]pOH=-\log(0.0408)\\\\pOH=1.39[/tex]
To calculate pH of the solution, we use the equation:
[tex]pH+pOH=14\\\\pH=14-1.39=12.61[/tex]
Hence, the pH of the solution is 12.61
