In an experiment to determine the molecular weight and the ionization constant f

Question

In an experiment to determine the molecular weight and the ionization constant for ascorbic acid, a student dissolved 1.3717 grams of the acid in water to make 50.00 milliliters of solution. The entire solution was titrated with a 0.2211 M NaOH solution. The pH was monitored throughout the titration. The equivalence point was reached when 35.32 mL of the base jad been added. Under the conditions of this experiment, ascorbic acid acts as a monoprotic acid that can be represented as HA.

a) Calculate the molecular weight of ascorbic acid.

b)When 20.00 milliliters of NaOH had been added during the titrations, the pH of the solution was 4.23. Caculate the ionization constant for ascorbic acids.

c) Caculate the Equilibrium constant for the reaction of the ascorbate ions, A-, with water.

Explanation / Answer

(a) (0.2211)(0.03523) = 7.789 x 10 -3 mol

1.3717 / 7.789x10-3 = 176.1 g/ mol

(b) at pH = 4.23, [H+] = 8.0 x 10 -8M

[A-] = (0.02 x 0.2211)/0.07 = 0.06317M

[HA] = (0.01523 x 0.2211)/0.7 = 0.04810 M

K = [H+] [A-] / [HA] = 5.9x 10-9 x 0.06317 / 0.04810 = 7.7 x 10 -5

(c) equilibrium constant K = Kw/Ka = 10 -14 / 7.7x10-5 = 1.3 x 10-10

(c) A- + H2O <=> HA + OH-

(d) at equiv. pt.

[OH-]2 = (1.3x10-10)(9.14x10-2) = 1.2x10-11

[OH-] = 3.4x10-6M

pOH =-log(3.4x10-6) =5.47; pH = (14-5.47)= 8.53

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