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pH of a weak base solution before and after strong acid is added

(to contrast with the buffer solution calculation on 12/3/2007)

Recall, the pH of 1.0 L of a buffer that was 0.10 M NH₃ and 0.10 M NH₄⁺ was ~9.25

After adding 0.020 mol HCl, the pH was ~9.08 for a change in pH of only 0.17 pH units.

pH of 1.0 L of 0.10 M NH₃, K_b = 1.8´10^-5

Equilibrium: NH₃ + H₂O ↔ NH₄⁺ + OH^-

K_b = 1.8´10^-5 ≈ [x][x] / [0.10] so [OH^-] ≈ (1.8´10^-5 ^. 0.10)^1/2 = 0.0013 M

[H⁺] ≈ (1.0´10^-14 / 0.0013) = 7.4 ´10^-12 M & pH = 11.13

pH after 0.020 mol HCl is added

Reaction: NH₃ + H⁺ → NH₄⁺ (assume complete)

–find initial #moles of each species before the reaction

#mol NH₃ = (1.0 L)(0.10 mol/L) = 0.10 mol NH₃

#mol H⁺ = 0.020 mol;

#mol NH₃ = assumed small (only source is NH₃ equilibrium and we’ll do that shortly)

–find new “initial” molarities of each species before equilibrium is reached (watch for volume changes)

M NH₃ = (0.080 mol NH₃) / (1.0 L total) = 0.080 M NH₃

M NH₄⁺ = (0.020 mol NH₄⁺) / (1.0 L total) = 0.020 M NH₃

M OH^- ≈ 0.00 M OH^- (assumes OH^–from H₂O equilibrium is negligible)

Equilibrium: NH₃ + H₂O ↔ NH₄⁺ + OH^-

K_b = 1.8´10^-5 ≈ [x][0.020] / [0.080] so [OH^-] ≈ (1.8´10^-5 ^. 0.080) / (0.020) = 7.2´10^-5 M

[H⁺] ≈ (1.0´10^-14 / 7.2´10^-5) = 1.4 ´10^-10 M & pH = 9.86

NOTE: the change in pH is larger (~1.27 pH units)

And: recall in pure water the pH change was rather humongous (~5.30 pH units)

Good luck and try not to stress out!