駿肥 發問時間: 科學化學 · 4 年前

請教化學大神有關酸鹼滴定的問題?

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  • 4 年前
    最佳解答

    1.

    It is known that Kw = [H⁺][OH⁻] = 1 × 10⁻¹⁴

    In pure water : [H⁺] = [OH⁻]

    Then [H⁺]² = 1 × 10⁻¹⁴

    [H⁺] = √(1 × 10⁻¹⁴) = 1 × 10⁻⁷ M

    pH = -log(1 × 10⁻⁷) = 7

    ====

    2.

    The volume of 0.008 mol of NaOH is negligible.

    [OH⁻] = 0.008 M >> (1 × 10⁻⁷ M)

    Thus, the self-dissociation of water can be neglected.

    pOH = -log(0.008) = 2.1

    pH = 14 - 2.1 = 11.9

    ====

    3.

    Denote the weak acid MOPS as HA, and its conjugate base as A⁻.

    HA(aq) ⇌ H⁺(aq) + A⁻(aq)

    pH = pKa - log([HA]/[A⁻]

    pH = 7.2 - log(61/39)

    pH = 7.0

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