Respuesta :
Answer: The pH of the solution is 9.33
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 HCl:
Molarity of HCl solution = 0.200 M
Volume of solution = 7.5 mL
Putting values in equation 1, we get:
[tex]0.200M=\frac{\text{Moles of HCl}\times 1000}{7.5}\\\\\text{Moles of HCl}=\frac{0.200\times 7.5}{1000}=0.0015mol[/tex]
- For trimethylamine:
Molarity of trimethylamine solution = 0.100 M
Volume of solution = 20 mL
Putting values in equation 1, we get:
[tex]0.100M=\frac{\text{Moles of trimethylamine}\times 1000}{20}\\\\\text{Moles of trimethylamine}=\frac{0.100\times 20}{1000}=0.002mol[/tex]
The chemical reaction for trimethylamine and HCl follows the equation:
[tex](CH_3)_3N+HCl\rightarrow (CH_3)_3NH^++Cl^-[/tex]
Initial: 0.002 0.0015
Final: 0.0005 - 0.0015
Total volume of the solution = [20 + 7.5] = 27.5 mL = 0.0275 L (Conversion factor: 1 L = 1000 mL)
To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})[/tex]
[tex]pOH=pK_b+\log(\frac{[(CH_3)_3NH^+]}{[(CH_3)_3N]})[/tex]
where,
[tex]pK_b[/tex] = negative logarithm of base dissociation constant of trimethylamine = 4.19
[tex][(CH_3)_3NH^+]=\frac{0.0015}{0.0275}[/tex]
[tex][(CH_3)_3NH]=\frac{0.0005}{0.0275}[/tex]
pOH = ?
Putting values in above equation, we get:
[tex]pOH=4.19+\log(\frac{(0.0015/0.0275)}{(0.0005/0.0275)})\\\\pOH=4.67[/tex]
To calculate pH of the solution, we use the equation:
[tex]pH+pOH=14\\pH=14-4.67=9.33[/tex]
Hence, the pH of the solution is 9.33
