Question

Calculate the pH of a buffer that is prepared by mixing 25.0 mL of 0.300 M methylamine (CH3NH2) and 0.405 g of methylammonium chloride with water to make 500.0 mL of solution.

Answer 1

Answer: The pH of the solution is 10.74

Explanation:

• For methylammonium chloride:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of methylammonium chloride = 0.405 g

Molar mass of methylammonium chloride = 67.52 g/mol

Putting values in above equation, we get:

$\text{Moles of methylammonium chloride}=\frac{0.405g}{67.52g/mol}=0.00599mol$

• For methylamine:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of methylamine solution = 0.300 M

Volume of solution = 25 mL = 0.025 L

Putting values in above equation, we get:

$0.300M=\frac{\text{Moles of methylamine}}{0.025L}\\\\\text{Moles of methylamine}=(0.300mol/L\times 0.025L)=0.0075mol$

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:  

$pOH=pK_b+\log(\frac{[salt]}{[base]})$

$pOH=pK_b+\log(\frac{[CH_3NH_3^+Cl^-]}{[CH_3NH_2]})$

We are given:  

$pK_b$ = negative logarithm of base dissociation constant of methylamine = 3.36

$[CH_3NH_2]=\frac{0.0075}{0.525}$

$[CH_3NH_3^+Cl^-]=\frac{0.00599}{0.525}$

pOH = ?

Putting values in above equation, we get:

$pOH=3.36+\log (\frac{(0.00599/0.525)}{(0.0075/0.525)}\\\\pOH=3.26$

To calculate the pH of the solution, we use the equation:

$pH+pOH=14\\\\pH=14-3.26=10.74$

Hence, the pH of the solution is 10.74