Assume you dissolve 0.325 g of the weak acid benzoic acid, CsH5 CO2H, in enough

Question

Assume you dissolve 0.325 g of the weak acid benzoic acid, CsH5 CO2H, in enough water to make 1.00 x 102 mL of solution and then titrate the solution with 0.120 M NaOH. (Ka for benzoic acid 6.3 x 10-5) C6H,CO2H(aq) + OH_ (aq) ? C6H5 CO2-(aq) +H2O(1) a. What was the pH of the original benzoic acid solution? pH- b. What are the concentrations of all of the following ions at the equivalence point: Nat, Ho , OH-, and C6H, CO2_ ? Na - HO OH [C6H, C02-1 c. What is the pH of the solution at the equivalence point?

Explanation / Answer

inital [C6H5COOH] molarity = 0.325 g/122.12 g/mol x 0.1 L

                                             = 0.027 M

a. initial pH

C6H5COOH + H2O <==> C6H5COO- + H3O+

let x amount dissociated

Ka = [C6H5COO-][H3O+]/[C6H5COOH]

6.3 x 10^-5 = x^2/0.027

x = [H3O+] = 1.30 x 10^-3 M

pH = -log(1.30 x 10^-3) = 2.88

b. at equivalence point

volume NaOH added = 0.027 M x 100 ml/0.120 M = 22.5 ml

Total volume of solution = 122.5 ml

[Na+] = 0.120 M x 22.5 ml/122.5 ml = 0.022 M

[C6H5COO-] = 0.027 M x 100 ml/122.5 ml = 0.022 M

C6H5COO- + H2O <==> C6H5COOH + OH-

let x amount hydrolyzed

Kb = Kw/Ka = [C6H5COOH][OH-]/[C6H5COO-]

1 x 10^-14/6.3 x 10^-5 = x^2/0.022

x = [OH-] = 1.87 x 10^-6 M

[H3O+] = 1 x 10^-14/1.87 x 10^-6 = 5.35 x 10^-9 M

c. pH at equivalence point

pH = -log(5.35 x 10^-9)

      = 8.27

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