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the ionization constant of acetic acid is 1 74 x 1...

Question:
The ionization constant of acetic acid is 1.74 x 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.

Answer:

1) CH₃COOH ↔ CH₃COO^- + H⁺ K_a = 1.74x10^-5

2) H₂O + H₂O ↔ H₃O⁺ + OH^- K_w = 1.0x10^-14

Since K_a > K_w

CH₃COOH + H₂O ↔ CH₃COO^- + H₃O⁺

Ci = 0.05 0 0

0.05- 0.05α 0.05α 0.05α

k_a = 0.05α x 0.05α / 0.05- 0.05α

= ~~0.05~~α x 0.05α / ~~0.05~~ (1-α)

= 0.05α² / (1-α)

⇒ 1.74x10^-5 = 0.05α² / (1-α)

⇒ 1.74x10^-5 - 1.74x10^-5 α = 0.05α²

⇒ 0.05α² + 1.74x10^-5 α - 1.74x10^-5 =0

D = b²- 4ac

= (1.74x10^-5)² - 4 (0.05) (1.74x10^-5)

= 3.02x10^-25 + 0.348x10^-5

Method 2:

Degree of Dissociation,

CH₃COOH ↔ CH₃COO^- + H⁺

Thus, concentration of CH₃COO^- = c.α

= 0.05x1.86x10^-2

= 0.93x10^-2

=.00093M

Since [oAc-] = [H⁺]

[H⁺] = .00093 = 0.093x10-2

pH = -log[H⁺]

= -log (0.93x10^-2)

∴ pH = 3.03

Hence, the concentration of acetate ion in the solution is 0.00093 M and its Ph is 3.03.

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Question:
The ionization constant of acetic acid is 1.74 x 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.

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Question: The ionization constant of acetic acid is 1.74 x 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.

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Question:
The ionization constant of acetic acid is 1.74 x 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.