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

Answer 1

You are mixing a strong acid (HBr) with a weak base (CH₃NH₂), and up until the equivalence point, this will form a buffer.

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

moles HBr = 30.0 ml x 1 L / 1000 ml x 0.200 mol/L = 0.006 mol HBr

CH₃NH₂ + HBr <==> CH₃NH₃⁺ + Br^-

0.012........0.006..............0...............0............Initial

-0.006....-0.006............+0.006.......................Change

0.006...........0................0.006.......................Equilibrium

When the [CH₃NH₂] equals [CH₃NH₃⁺], the pH = pKa or pOH = pKb. This is according to the Henderson Hasselbalch equation which is as follows:

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

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

In either case you can see that when the conjugate and the acid or base are equal, the log of 1 = 0 and

pH = pKa or pOH = pKb

Thus, pOH = pKb = -log 4.4x10^-4

pOH = 3.357

pH = 14 - pOH

pH = 10.643

(be sure to check all of the math)