Anonymous
Anonymous asked in Science & MathematicsChemistry · 1 month ago

Ascorbic acid ( H2C6H6O2 ) has 𝐾a1=8.00×10−5 and 𝐾a2=1.60×10−12 . What's the pH of a 0.281 M ascorbic acid ( H2C6H6O2 ) solution?

2 Answers

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  • ?
    Lv 7
    1 month ago

    [H+] = sqrt( Ka *c)

    = sqrt(8.00×10−5*0.281 = 0.00474M

    pH = 2.32

  • 1 month ago

    In calculation of pH, we only consider Kₐ₁ because Kₐ₁ ≫ Kₐ₂.

               H₂C₆H₆O₂(aq) + H₂O(ℓ) ⇌ HC₆H₆O₂⁻(aq) + H₃O⁺(aq)     Kₐ₁ = 8.00 × 10⁻⁵

    Initial:    0.281 M                              0 M                 0 M

    Change:   -y M                               +y M               +y M

    Eqm:   (0.281 - y) M                         y M                 y M

               ≈ 0.281 M

    At equilibrium:

    Kₐ₁ = [HC₆H₆O₂⁻] [H₃O⁺] / [H₂C₆H₆O₂]

    8.00 × 10⁻⁵ = y² / 0.281

    y = √(0.281 × 8.00 × 10⁻⁵) = 4.74 × 10⁻³

    pH = -log[H₃O⁺] = -log(4.74 × 10⁻³) = 2.32

    ====

    OR:

    pH

    = -(1/2)[log(Kₐ₁) + log[H₂C₆H₆O₂]ₒ]

    = -(1/2)[log(8.00 × 10⁻⁵) + log(0.281)]

    = 2.32

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