A weak monoprotic acid, HA dissociates by 1.300 % and has a pH of 2.15. Calculate...

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Answer 1

Answer #1

assume weak acid = HA

HA <------------> H⁺ + A^-

pH = 2.15

[H⁺] = 10^-pH = 10^-2.15 = 0.0071 M

at equilibrium [H⁺] = 0.0071 M

% ionization = [H⁺]_equilibrium / [HA]_initial x 100

1.300 = (0.0071 / [HA]) x 100

0.0071 / [HA] = 0.01300

[HA] = 0.546 M

[HA]_equilibrium = 0.546 - 0.0071 = 0.5389

Ka = [H⁺][A^-] / [HA]

Ka = [0.0071][0.0071] / [0.5389]

Ka = 9.35 x 10^-5

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Answer 2

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