Calculate the pH when 45.0 mL of 0.200 M HBr is mixed with 30.0 mL of 0.400 M CH₃NH₂ (Kb = 4.4 × 10⁻⁴).

Question

what would you do with the volume and concentrations of the acid and base?

Answer 1

In cases like this, where no other additions are being made, you can use moles and ignore the volume, or you can convert to moles / liter (M) concentration. They both will provide the same answer (see below).

HBr + CH₃NH₂ ==> CH₃NH₃⁺ + H2O

mols HBr = 45.0 ml x 1 L / 1000 ml x 0.200 mol / L = 0.009 mols

mols CH₃NH₂ = 30.0 ml x 1 L / 1000 ml x 0.400 mol / L = 0.012 mols

Set up ICE table

HBr + CH₃NH₂ ==> CH₃NH₃⁺ + H2O

0.009....0.012.................0................Initial

-0.009...-0.009..............+0.009........Change

0............0.003..............0.009...........Equilibrium

Henderson Hasselbalch equation

pOH = pKb + [salt]/[base]

pOH = 3.36 + log [0.009] / [0.003]

pOH = 3.36 + log 3

pOH = 3.36 + 0.48

pOH = 3.84

pH = 14 - 3.94

pH = 10.16

If you use molarity (M), then you have the following solution:

Final volume = 45 ml + 30 ml = 75 ml = 0.075 L

Final [CH₃NH₃⁺] = 0.009 mol / 0.075 L = 0.12 M

Final [CH₃NH₂] = 0.003 mol / 0.075 L = 0.04 M

Henderson Hasselbalch

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

pOH = 3.36 + log (0.12 / 0.04)

pOH = 3.36 + log 3

Same solution as above and pH = 10.16