Assume you dissolve 0.134 g of the weak acid benzoic acid, in enough water to ma

Question

Assume you dissolve 0.134 g of the weak acid benzoic acid, in enough water to make 100 mL of solution and then titrate the solution with 0.166 M NaOH( Ka for benzoic acid =6.3 x 10^-5Assume you dissolve 0.390 g of the weak acid benzoic acid, C6H5CO2H in enough water to make 100 mL of solution and then titrate the solution with 0.146 M NaOH( Ka for benzoic acid =6.3 x 10^-5 .)

C6H5CO2H (aq) + OH- (aq) ----> C6H5CO2- (aq) + H2O (l)

a)What was the pH of the original benzoic acid solution?

b)What are the concentrations of all of the following ions at the equivalence point:

Na+

H3O+

OH-

C6H5CO2-

c)What is the pH of the solution at the equivalence point?

Explanation / Answer

a) molar concentration of original benzoic acid solution = 0.134 g/122.12 g/mol x 0.1 L = 0.011 M

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

let x amount of acid has dissociated

6.3 x 10^-5 = x^2/0.011

x = [H+] = 8.32 x 10^-4 M

So pH of solution is, pH = -log[H+] = 3.08

b) at equivalence point, moles of acid present = moles of base added

moles of acid = 0.011 M x 100 ml = 1.1 mmol

Volume of NaOH added = 1.1 mmol/0.166 M = 6.62 ml

final concentration of acid in solution = 1.1 mmol/106.62 ml = 0.010 M

at equilibrium formed [C6H5COO-] = 0.010 M

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

1 x 10^-14/6.3 x 10^-5 = x^2/0.010

x = [C6H5COOH] = [OH-] = 1.26 x 10^-6 M

[C6H5COO-] = 9.999 x 10^-3 M or 0.01 M

[H3O+] = 1 x 10^-14/1.26 x 10^-6 = 7.94 x 10^-9 M

c) pH at equivalence point

pOH = -log[OH-] = 5.9

pH = 14 - pOH = 8.1

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