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What is the pH of a 203 mL sample of 3.601 M acetic acid (CH3COOH) (Ka...

What is the pH of a 203 mL sample of 3.601 M acetic acid (CH3COOH) (Ka = 1.8 x 10-5)?

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Expert Solution

CH3COOH -----------------------------> CH3COO- + H+

3.601                                                    0                 0 ----------------------> initial

3.601-x                                                 x                  x-------------------------> after dissociation

Ka = [CH3COOH][H+]/[CH3COOH]

Ka = x^2 / 3.601-x

1.8 x 10^-5 = x^2 / 3.601-x

x^2 + 1.8 x 10^-5 x -6.48 x 10^-5 = 0

x = 8.04 x 10^-3

[H+] = 8.04 x 10^-3   M

pH = -log[H+]

pH = -log(8.04 x 10^-3 )

pH = 2.09

queen_honey_blossom answered 2 weeks ago
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