PH chemistry solutions

Question

What quantity in moles of CH₃NH₃Cl need to be added to 200.0 mL of a 0.500 M solution of CH₃NH₂ (Kb for CH₃NH₂ is 4.4 × 10⁻⁴) to make a buffer with a pH of 11.00?

Answer 1

The Henderson Hasselbalch equation for a basic buffer is...

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

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

pOH = 3.36 + log [CH₃NH₃Cl] / [CH₃NH₂]

pOH = 14 - pH = 14 - 11 = 3

3 = 3.36 + log [CH₃NH₃Cl] / [0.5]

log [CH₃NH₃Cl] / [0.5] = -0.36

[CH₃NH₃Cl] / [0.5] = 0.437

log [CH₃NH₃Cl] = 0.218 M

To obtain 0.218 M (mol/L) when volume is 200.0 mls we have...

0.218 mols / L x 0.200 L = 0.0436 mols CH₃NH₃Cl should be added (assuming no change in volume)