2. A 10.0 mL portion of 0.10 M acetic acid HCaH,O2 (K 1.8 x 10) was diluted with

Question

2. A 10.0 mL portion of 0.10 M acetic acid HCaH,O2 (K 1.8 x 10) was diluted with 90.0 mL of deionized water. A 10.0 mL portion of the resulting diluted solution was diluted further by the addition of 90.0 mL of water a. Write the chemical equation for the dissociation of acetic acid. b. Write the equation for the dissociation constant of acetic acid [Equation (21-7) can be used as a guide in writing the equation]. c. Use the equation to calculate the concentration of Ht and C2H:02 in the 0.10 M acetic acid solution (you might want to review your lecture notes and text for details of the calculation). d. Use the [H'] from 2c to calculate the pH of the solution by using Equation (21-1). e. Similarly calculate the pH of the solution prepared after the first dilution and the pH of the solution prepared after the second dilution.

Explanation / Answer

Ans -

(a) The chemical equation -

Acetic acid - HC2H3O2

Deionised water - H2O

So the chemical reaction is -

HC2H3O2(aq) + H2O(l) ------> CH3COO- (aq)+ H3O+(aq)

(b) Dssociation constant can be written as -

Ka = Product / Reactant

Ka = [CH3COO-] [H3O+] / [HC2H3O2]

H2O is not included as it is liquid and constant through out the reaction.

(c)

HC2H3O2(aq) + H2O(l) ------> CH3COO- (aq)+ H3O+(aq)

initial conc. 0.10 M 0 0 0

Final conc. 0.10-x x x

Ka can be written as for final conc. as -

Ka = x.x / 0.10-x

Ka = 1.8 x 10-5

1.8 x 10-5 = x2/ 0.10-x

x is versy small then dinominator can be written as .10

1.8 x10-5 = x2 / .10

x2 = 1.8 x10-6

x = 1.36 x 10-3

conc. of H+ and CH3COO- is 1.36 x10-3 M

(d) we know that -

pH = -log [H+]

H+ = 1.36x10-3

pH = -log [1.36x10-3]

pH = - (-2.87)

pH = 2.87

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