Question

Write the base dissociation reaction for the conjugate base of acetic acid

A) find its Kb if the Ka of acetic acid is 1.8x10^-5

B) what is the pH of the solution if you start with 0.1 M acetic acid?

C) After the solution in part B reaches equilibrium, you add acetate to a concentration of 0.005M. What is the new pH of the solution? You can’t use assumptions here

Answer 1

Ans. A. Kb = 10^-14 / Ka

            Or, Kb = 10^-14 / (1.8 x 10^-5) = 5.56 x 10^-10

#B. Create an ICE table as shown in figure.

At equilibrium, Ka = [CH3COO^-] [H3O⁺] / [CH3COOH]

            Or, 1.8 x 10^-5 = (X) (X) / (0.10- X)

            Or, 0.000018 (0.10- X) = X²

            Or, X² + 0.000018 X - 0.0000018 = 0

Solving the quadratic equation, we get following two roots-

            X1 = 0.00133             ; X2 = - 0.00135

Since concentration can’t be negative, reject X2.

Therefore, X = 0.00133

So, [H₃O⁺] at equilibrium = X = 0.00133 M

Now,

            pH = -log [H₃O⁺] = - log 0.00133 = 2.88

Hence, pH = 2.88

#C. The initial concentrations at equilibrium in part B as follow-

            Initial [CH3COOH] = 0.10 – X = 0.10 – 0.00133 = 0.09867 M

            Initial [CH3COO^-] = X = 0.00133 M

            Since 0.005 M acetate ion is added to it, the total initial [CH3COO-] becomes equal to 0.00133 M + 0.005 M = 0.00633 M

            Therefore, total initial [CH3COO-] = 0.00633 M            

Initial [H³O⁺] = 0.00133 M

Following LeChatlier’s principle, the addition of acetate ion shifts the equilibrium to the left, i.e. favors formation of acetic acid. Create an ICE table in opposite direction of part B.   

Now, using ICE table (note- the reaction has reversed its direction)-

            1/ Ka = [CH3COOH] / [CH3COO-] [H3O+]

            Or, 1 / 0.000018 = (0.09867 + X) / [(0.00633-X) (0.00133- X)]

            Or, 55555.556 (0.0000084189 – 0.00766 X + X²) = (0.09867 + X)

            Or, 0.4677 – 425.556X + 55555.556 X² – 0.09867 – X = 0

            Or, 55555.556 X² - 424.556 X + 0.369 = 0

Solving the quadratic equation, we get following two roots-

            X1 = 0.00664            ; X2 = 0.001

Since X can’t be greater than any of the initial concentrations, reject X1.

Therefore, X = 0.001M

Therefore, [H₃O+] at equilibrium = 0.00133 – X = 0.00133 – 0.001 = 0.00033 M

Now, pH = [H₃O⁺] at equilibrium = - log 0.00033 = 3.481

Therefore, pH = 3.481