Bob asked in Science & MathematicsChemistry · 2 years ago

The Ka of acetic acid (HC2H3O2) is 1.8 × 10-5 What is the pH at 25.0 °C of an aqueous solution that is 0.100M in acetic acid.?

1 Answer

Relevance
  • 2 years ago
    Favorite Answer

    Acetic acid is a weak acid that dissociates to a small extent: :

    CH3COOH ↔ CH3COO- + H+

    The Ka equation is:

    Ka = [CH3COO-] [H+] / [CH3COOH]

    Note that [CH3COO-] and [H+] are equal

    The dissociation is small so [CH3COOH] can be taken as the starting concentration

    The Ka expression can be rewritten:

    Ka = [H+]² / [CH3COOH]

    Substitute:

    1.8*10^-5 = [H+]² / 0.1

    [H+]² = 0.1( 1.8*10^-5)

    [H+]² = 1.8*10^-6

    [H+] = 1.34*10^-3M

    pH = -log [H+]

    pH = -log ( 1.34*10^-3)

    pH = 2.87

Still have questions? Get your answers by asking now.