How to prepare buffer?

Question

What ratio of CH₃NH₂ to CH₃NH₃⁺ is needed to prepare a pH 10.30 buffer? (Kb for CH₃NH₂ is 4.4 × 10⁻⁴)

I got 0.23 but it said I was off by a factor of 2

Answer 1

Using the Henderson Hasselbalch equation for a basic buffer:

pOH = pKb + log [salt] / [base]

pOH = 14 - pH = 14 - 10.30 = 3.70

pKb = -log 4.4x10^-4 = 3.36

log [salt] / [base] = ?

3.70 = 3.36 + log [salt] / [base]

log [salt] / [base] = 0.34

[salt] / [base] = 2.18

[base] / [salt] = 1/2.18 = 0.46

This answer is 2 times what you reported. I hope you can find where and how you made the error. Good luck.