Chemistry Titration

Question

Calculate the pH when 82.5 mL of 0.200 M HBr is mixed with 30.0 mL of 0.400 M CH₃NH₂ (Kb = 4.4 × 10⁻⁴).

Answer 1

HBr + CH₃NH₂ ==> CH₃NH₃⁺ + Br^- .. balanced equation

mols HBr = 82.5 ml x 1 L / 1000 ml x 0.200 mol / L = 0.0165 mols

mols CH3NH2 = 30.0 ml x 1 L / 1000 ml x 0.400 mol / L = 0.0120 mols

Set up an ICE table:

HBr + CH₃NH₂ ==> CH₃NH₃⁺ + Br^-

0.0165...0.0120............0............0..........Initial

-0.012....-0.012.......+0.012.....+0.012....Change

4.5x10^-3....0...........0.012.........0.012.....Equilibrium

Final volume = 82.5 ml + 30 ml = 112 mls = 0.112 L

[HBr] = 4.5x10^-3 mol / 0.112 L = 0.0402 M

[H⁺] = 0.0402 M

Contribution to H⁺ from the CH₃NH₃⁺ will be negligible and won't contribute significantly to the pH(see below)

pH = -log [H⁺] = -log 0.0402

pH = 1.40

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Contribution to [H⁺] from CH₃NH₃⁺

[CH₃NH₃⁺] = 0.012 mol / 0.112 L = 0.107 M

Ka = 1x10^-14 / 4.4x10^-4 = 2.27x10^-11

2.27x10^-11 = (x)(x) / 0.012 - x (assume x is small relative to 0.012 M and ignore it)

2.27x10^-11 = x² / 0.012

x² = 2.72x10^-13

x = 5.22x10^-7 M = [H⁺] insignificant compared to 0.0402 M from the HBr