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M asked in Science & MathematicsChemistry · 1 month ago

chem titrations buffer?

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  • 1 month ago
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    A.

    C₆H₅NH₂(aq) + H₂O(l) ⇌ C₆H₅NH₃⁺(aq) + OH⁻(aq)

    Henderson-Hasselbach equation:

    pOH = pKb + log([C₆H₅NH₃⁺]/[ C₆H₅NH₂])

    pKw - pH = pKb + log([C₆H₅NH₃⁺]/[ C₆H₅NH₂])

    14.00 - 5.40 = 9.13 + log([C₆H₅NH₃⁺]/0.335)

    log([C₆H₅NH₃⁺]/0.335) = -0.53

    [C₆H₅NH₃⁺]/0.335 = 10⁻⁰˙⁵³

    [C₆H₅NH₃⁺] = 0.335 × 10⁻⁰˙⁵³

    [C₆H₅NH₃⁺] = 0.0989 M

    ====

    B.

    Molar mass of NaOH = (23.0 + 16.0 + 1.0) g/mol = 40.0 g/mol

    Moles of NaOH added = (0.357 g) / (40.0 g/mol) = 0.00893 mol

    On addition of NaOH, part of C₆H₅NH₃⁺ ion is neutralize to C₆H₅NH₂.

    After addition:

    Moles of C₆H₅NH₃⁺ = (0.0989 mol/L) × (1.75 L) - (0.00893 mol) = 0.164 mol

    Moles of C₆H₅NH₂ = (0.335 mol/L) × (1.75 L) + (0.00893 mol) = 0.595 mol

    Henderson-Hasselbach equation again:

    pKw - pH = pKb + log([C₆H₅NH₃⁺]/[ C₆H₅NH₂])

    14.00 - pH = 9.13 + log(0.164/0.595)

    pH = 14.00 - 9.13 - log(0.164/0.595)

    pH = 5.43

    ΔpH = 5.43 - 5.40 = 0.03

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