Respuesta :
Answer: The pH of the solution is 10.74
Explanation:
- For methylammonium chloride:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
Given mass of methylammonium chloride = 0.405 g
Molar mass of methylammonium chloride = 67.52 g/mol
Putting values in above equation, we get:
[tex]\text{Moles of methylammonium chloride}=\frac{0.405g}{67.52g/mol}=0.00599mol[/tex]
- For methylamine:
To calculate the number of moles for given molarity, we use the equation:
[tex]\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}[/tex]
Molarity of methylamine solution = 0.300 M
Volume of solution = 25 mL = 0.025 L
Putting values in above equation, we get:
[tex]0.300M=\frac{\text{Moles of methylamine}}{0.025L}\\\\\text{Moles of methylamine}=(0.300mol/L\times 0.025L)=0.0075mol[/tex]
Total volume of the solution = [25 + 500] = 525 mL = 0.525 L (Conversion factor: 1 L = 1000 mL)
To calculate the pOH of basic buffer, we use the equation given by Henderson Hasselbalch:
[tex]pOH=pK_b+\log(\frac{[salt]}{[base]})[/tex]
[tex]pOH=pK_b+\log(\frac{[CH_3NH_3^+Cl^-]}{[CH_3NH_2]})[/tex]
We are given:
[tex]pK_b[/tex] = negative logarithm of base dissociation constant of methylamine = 3.36
[tex][CH_3NH_2]=\frac{0.0075}{0.525}[/tex]
[tex][CH_3NH_3^+Cl^-]=\frac{0.00599}{0.525}[/tex]
pOH = ?
Putting values in above equation, we get:
[tex]pOH=3.36+\log (\frac{(0.00599/0.525)}{(0.0075/0.525)}\\\\pOH=3.26[/tex]
To calculate the pH of the solution, we use the equation:
[tex]pH+pOH=14\\\\pH=14-3.26=10.74[/tex]
Hence, the pH of the solution is 10.74
