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What is the equilibrium constant expression for the ionization of acetic acid (CH3COOH) in water, and what is the value of the equilibrium constant at 25°C?

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The ionization of acetic acid  CHCOOH  in water can be represented by the following chemical equation:CHCOOH  aq  + HO  l   CHCOO  aq  + HO  aq In this reaction, acetic acid  CHCOOH  donates a proton  H  to water  HO , forming the acetate ion  CHCOO  and the hydronium ion  HO .The equilibrium constant expression for this reaction, known as the acid dissociation constant  Ka , can be written as:Ka = [CHCOO][HO] / [CHCOOH]Here, [CHCOO] represents the equilibrium concentration of the acetate ion, [HO] represents the equilibrium concentration of the hydronium ion, and [CHCOOH] represents the equilibrium concentration of acetic acid.At 25C, the value of the equilibrium constant  Ka  for the ionization of acetic acid is approximately 1.8 x 10. This value indicates that the ionization of acetic acid in water is relatively weak, as the equilibrium lies towards the reactants  undissociated acetic acid  rather than the products  acetate and hydronium ions .

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