Calculate the pH and concentrations of CH3NH2 and CH3NH3+

Question

Calculate the pH and concentrations of CH₃NH₂ and CH₃NH₃⁺ in a 0.0447 M methylamine (CH₃NH₂)  solution. The 𝐾b of CH₃NH₂ is 4.47×10⁻⁴.

pH=

[CH₃NH₂]= M

[CH₃NH₃⁺]=

Answer 1

CH₃NH₂ + H2O ==> CH₃NH₃⁺ + OH^-

Kb = 4.47x10^-4 = [CH₃NH₃⁺][OH-] / [CH₃NH₂]

4.47x10^-4 = (x)(x) / 0.0447 - x

Use the quadratic equation and solve for x. This will be the [OH-] and [CH₃NH₃⁺]

To get [CH₃NH₂] just subtract the value of x from 0.0447.

To obtain pH, take negative log of the value you got for x (which will give you pOH), and subtract it from 14 to get pH.