Question

5. The ionization constant of acetic acid is 1.74x10*5. Calculate the degree of dissociation of acetic acid in its 0.05 M solution. Calculate the concentration of acetate ion in the solution and its pH.

Answer 1

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K = 1.74×10^-5
C = 5×10^-2 M

Let degree of association be alpha

K = alpha² C for alpha <0.05
1.74×10^-5 = alpha² 5×10^-2
alpha ≈ 0.018


Concentration of acetate ion = alpha×C
= 0.018×0.05
= 0.0009 M

Concentration of Hydrogen ion = alpha×C
= 9×10^-4 M

pH = -log 9×10^-4
= 4 - log 9
≈ 3.04

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Answer 2

Answer:

3.03 is the correct answer for acetate ion