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What is the pH at the equivalence point in the titration?

HEEEEELPPP ! What is the pH at the equivalence point for this titration .... ?

  • Answer:

    0.010 L x 0.350 M = 0.00350 mole CH3COOH 0.00350 mole / 0.150 M = 0.0233 L NaOH At the equivalence point the total volume = 0.0333 L (0.010 + 0.0233) The solution contains 0.00350 mole of CH3COO^-1 CH3COO^-1 + H2O <--> CH3COOH + OH^-1 Kb = [CH3COOH][OH-] / [CH3COO^-1] The Kb value for the acetate ion is calculated by- Ka x Kb = Kw = 1x10^-14 Ka for acetic acid = 1.75x10^-5 Kb = 5.71x10^-10 [CH3COOH] must equal [OH-] Let [OH-] = X Kb = 5.71x10^-10 = X^2 / [CH3COO^-1] At the equivalence point the concentrastion of CH3COO^-1 (before reaction with H2O to form some CH3COOH) = 0.00350 mole / 0.0333 L = 0.105 M 5.71x10^-10 = X^2 / (0.105 - X) Assume X << 0.105 0.105 x 5.71x10^-10 = X^2 X = 7.74x10^-6 pOH = 5.11 pH + pOH = 14 pH = 14 - 5.11 = 8.89

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0.010 L x 0.350 M = 0.00350 mole CH3COOH 0.00350 mole / 0.150 M = 0.0233 L NaOH At the equivalence point the total volume = 0.0333 L (0.010 + 0.0233) The solution contains 0.00350 mole of CH3COO^-1 CH3COO^-1 + H2O <--> CH3COOH + OH^-1 Kb = [CH3COOH][OH-] / [CH3COO^-1] The Kb value for the acetate ion is calculated by- Ka x Kb = Kw = 1x10^-14 Ka for acetic acid = 1.75x10^-5 Kb = 5.71x10^-10 [CH3COOH] must equal [OH-] Let [OH-] = X Kb = 5.71x10^-10 = X^2 / [CH3COO^-1] At the equivalence point the concentrastion of CH3COO^-1 (before reaction with H2O to form some CH3COOH) = 0.00350 mole / 0.0333 L = 0.105 M 5.71x10^-10 = X^2 / (0.105 - X) Assume X << 0.105 0.105 x 5.71x10^-10 = X^2 X = 7.74x10^-6 pOH = 5.11 pH + pOH = 14 pH = 14 - 5.11 = 8.89

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