Respuesta :
Answer: The pH of the solution is 12.73
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 cyanic acid:
Molarity of cyanic acid = 0.8600 M (Assuming)
Volume of solution = 179.5 mL (Assuming)
Putting values in equation 1, we get:
[tex]0.8600M=\frac{\text{Moles of cyanic acid}\times 1000}{179.5mL}\\\\\text{Moles of cyanic acid}=\frac{0.8600\times 179.5}{1000}=0.15437mol[/tex]
- For NaOH:
Molarity of NaOH = 0.6500 M (Assuming)
Volume of solution = 274.6 mL (Assuming)
Putting values in equation 1, we get:
[tex]0.6500M=\frac{\text{Moles of NaOH}\times 1000}{274.6mL}\\\\\text{Moles of NaOH}=\frac{0.6500\times 274.6}{1000}=0.17849mol[/tex]
Calculating the remaining moles of NaOH, we get:
[tex]\text{Remaining moles of NaOH}=\text{Moles of NaOH - Moles of cyanic acid}\\\\\text{Remaining moles of NaOH}=0.17849-0.15437=0.02412mol[/tex]
Calculating the concentration of NaOH by using equation 1, we get:
Moles of NaOH = 0.02412 moles
Volume of solution = 179.5 + 274.6 = 454.1 mL
Putting values in above equation, we get:
[tex]\text{Molarity of NaOH}=\frac{0.02412\times 1000}{454.1}\\\\\text{Molarity of NaOH}=0.0531M[/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^-]=0.0531M[/tex]
Putting values in above equation, we get:
[tex]pOH=-\log(0.0531)\\\\pOH=1.27[/tex]
To calculate pH of the solution, we use the equation:
[tex]pH+pOH=14\\pH=14-1.27=12.73[/tex]
Hence, the pH of the solution is 12.73
