# The ionization constant of acetic acid is 1.74 × 10–5. 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.

Given:

Ionization constant of acetic acid = 1.74 × 10–5

0.05 M solution

Ionization of CH3COOH

CH3COOH CH3COO + H+

By applying the formula,

Ka = Ka = At equilibrium [CH3COO ]= [H+]

Hence,

Ka= As Ka = 1.74 × 10–5 (given)

[CH3COOH] = 0.05 M

1.74 × 10–5 = [H+]2 = 1.74 × 10–5 × 5 × 10-2M

[H+] = [H+] = 9.33 × 10-4 M

As, [CH3COO] = [H+]

Thus, the concentration of acetate ion is 9.33 × 10-4M

To calculate pH, we apply the formula,

pH = -log[H+]

pH = -log(9.33 × 10-4)

pH = -log 9.33 – log 10-4

pH = - log 9.33 – (-4) log 10

pH = 4 - log 9.33

pH = 4 – 0.9699

pH = 3.03

Thus, its pH is 3.03.

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