(30 pts) Sample Calculations (9 pts) o Concentration of H3O* from the pH of 0.1

Question

(30 pts) Sample Calculations (9 pts) o Concentration of H3O* from the pH of 0.1 M HCI. o Concentration of H3O* from the pH of 0.1 M acetic acid. o Equilibrium concentration of acetate ion in the 0.1 M acetic acid solution. o Equilibrium concentration acetic acid in the 0.1 M acetic acid solution. o Acid dissociation constant of acetic acid as calculated from the 0.1 M acetic acid solution o Concentration of acetate ion in the 0.5 g sample. o Equilibrium concentration of H30* in the 0.5 g sample of acetate and 0.1 M acetic acid. o Equilibrium concentration of acetate ion in the 0.5 g sample of acetate and 0.1 M acetic acid. Equilibrium concentration of acetic acid in the 0.5 g sample of acetate and 0.1 M acetic acid o a ^ AAAN :) CO

Explanation / Answer

I am solving the first four sub-parts of the problem, please post multiple question to get the remaining answer

a) Since HCl is a very strong acid, hence it will dissociate completely

HCl ----> H+ + Cl-

Therefore, [H+] = [H3O+] = 0.1M

pH = -log[H+] = -log[0.1] = 1

b) Since acetic acid is a weak acid, hence it will not ionize completely

CH3COOH ====== CH3COO- + H+

Initial 0.1 0 0

Final (0.1-x) x x

Ka value of acetic acid = 1.8 * 10^(-5)

Ka = [Products]/[Reactants] = [CH3COO-][H+]/[CH3COOH]

1.8 * 10^(-5) = x^2/0.1-x

x^2 = 1.8 * 10^(-6) [assuming x is very small in comparison of 0.1]

x = 1.34 * 10^(-3)

pH = 3 - log(1.34) = 2.872

c)

equilibrium cocentration of acetate ion = equilibrium concentration of H+ ion = 1.34 * 10^(-3)

d)

Equilibrium concentration of acetic acid = 0.1 - 1.34 * 10^(-3) = 0.09866

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