What ratio of CH₃NH₂ to CH₃NH₃⁺ is needed to prepare a pH 10.00 buffer? (Kb for CH₃NH₂ is 4.4 × 10⁻⁴)

Answer 1

This types of problems can generally be answered using the Henderson Hasselbalch equation:

pOH = pKb + log [conj.acid] / [base]

pOH = 14 - pH = 14 = 10 = 4.00

pKb = -log Kb = -log 4.4x10^-4 = 3.36

[CH₃NH₃⁺] / [CH₃NH₂] = ?

Plug in values and solve for [CH₃NH₃⁺] / [CH₃NH₂]

4.00 = 3.36 + log [CH₃NH₃⁺] / [CH₃NH₂]

0.64 = log [CH₃NH₃⁺] / [CH₃NH₂]

[CH₃NH₃⁺] / [CH₃NH₂] = 4.37

[CH₃NH₂] / [CH₃NH₃⁺] = 1 / 4.37 = 0.229

If you prefer to use the more conventional Henderson Hasselbalch equation, we can arrive at the same:

pH = pKa + log [conj.base] / [acid]

pH = 10

pKa = 14 - pKb = 14 - 3.36 = 10.64

Plug in values and solve for [CH₃NH₂] / [CH₃NH₃⁺]

10.00 = 10.64 + log [CH₃NH₂] / [CH₃NH₃⁺]

log [CH₃NH₂] / [CH₃NH₃⁺] = -0.64

[CH₃NH₂] / [CH₃NH₃⁺] = 0.229