Question

Part A: Calculate the pH of a 0.316 M aqueous solution of hydrofluoric acid (HF, K_a = 7.2×10^-4) and the equilibrium concentrations of the weak acid and its conjugate base.

pH=? [HF]equilibrium= ? [F^-]equilibrium= ?

Part B:Calculate the pH of a 0.0193 M aqueous solution of chloroacetic acid (CH₂ClCOOH, K_a = 1.4×10^-3) and the equilibrium concentrations of the weak acid and its conjugate base.

pH=? [CH₂ClCOOH]_equilibrium = ? [CH₂ClCOO^- ]_equilibrium = ?

Answer 1

Part:a

By using ICE table-

                       HF (aq) + H2O (l)   <---------> H₃O⁺ (l) + F^- (aq)

      I(M)          0.316           -            0 0

C    - x +x +x

   Eq (0.316-x) x    x

Therefore -

   K_a = (x)² / (0.316-x) = 7.2*10^-4

By solving - x = 0.015 M = [H₃O⁺] = [F^-]

Thus - pH = -log [H₃O⁺] = -log (0.015) = 1.82

[HF] = 0.316 - 0.015 = 0.301 M

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Part:2

By using ICE table-

                       ClCH₂COOH (aq) + H₂O (l)   <---------> H₃O⁺ (l) + F^- (aq)

      I(M)          0.0193           -            0 0

C    - x     +x +x

   Eq    (0.0193-x) x    x

Therefore -

   K_a = (x)² / (0.0193-x) = 1.4*10^-3

By solving - x = 0.00519 M = [H₃O⁺] = [ClCH2COO^--]

Thus - pH = -log [H₃O⁺] = -log (0.00519) = 2.3

[HF] = 0.0193 - 0.00519 = 0.01411 M