Question

Calculate the pH and concentrations of CH 3 NH 2 and CH 3 NH + 3 in a 0.0317 M methylamine ( CH 3 NH 2 ) solution. The K b of CH 3 NH 2 is 4.47 × 10 − 4 .

Answer 1

[CH₃NH₃⁺] = 0.00355 M

[CH₃NH₂] = 0.028 M

pH = 11.55

Step-by-step explanation:

The reaction is the following:

CH₃NH₂ + H₂O ⇄ CH₃NH₃⁺ + OH⁻ (1)

0.0317 - x x x

The equilibrium constant of the reaction (1) is:

`K_(b) = ([CH_(3)NH_(3)^(+)][OH^(-)])/([CH_(3)NH_(2)])`

`4.47\cdot 10^(-4) = (x \cdot x)/(0.0317 - x)`

`x^(2) - 4.47 \cdot 10^(-4)(0.0317 - x) = 0` (2)

By solving equation (2) for x, we have:

x₁ = -0.00399

x₂ = 0.00355

Taking the positive value, we have that:

x = [CH₃NH₃⁺] = [OH⁻] = 0.00355 M

[CH₃NH₂] = 0.0317 - x = (0.0317 - 0.00355)M = 0.028 M

Therefore, the concentrations of CH₃NH₂ and CH₃NH₃⁺ are 0.028 M and 0.00355 M, respectively.

The pH of the solution is:

`pOH = -log [OH^(-)] = -log (0.00355) = 2.45`

`pH = 14 - pOH = 14 - 2.45 = 11.55`

Hence, the pH of the solution is 11.55.

I hope it helps you!