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 .

[CH₃NH₃⁺] = 0.00355 M

[CH₃NH₂] = 0.028 M

pH = 11.55

Explanation:

The reaction is the following:

CH₃NH₂  +  H₂O  ⇄  CH₃NH₃⁺ + OH⁻       (1)

0.0317 - x                      x               x

The equilibrium constant of the reaction (1) is:

    (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:

Hence, the pH of the solution is 11.55.  

I hope it helps you!  

c. 4.54 x 10–12 j happens to be correct