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13 July, 04:43

A 25.0 mL solution of quinine was titrated with 1.00 M hydrochloric acid, HCl. It was found that the solution contained 0.125 moles of quinine. What was the pH of the solution after 50.00 mL of the HCl solution were added

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  1. 13 July, 05:12
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    pH = 9.08

    Explanation:

    Quinine, C₂₀H₂₄O₂N₂, Q, is a weak base that, in water, has as equilibrium:

    Q + H₂O ⇄ QH⁺ + OH⁻

    Where pKb is 5.10

    Using H-H equation for weak bases:

    pOH = pKb + log₁₀ [QH⁺] / [Q]

    The reaction of quinine with HCl is:

    Q + HCl → QH⁺ + Cl⁻

    Initial moles of quinine are 0.125 moles and moles added of HCl are:

    0.05000L * (1.00mol / L) = 0.05000moles.

    That means after the addition of 50.00mL of the HCl solution, moles of Q and QH⁺ are:

    Q = 0.125mol - 0.050mol = 0.075 moles

    QH⁺ = 0.050 moles

    Replacing in H-H equation:

    pOH = 5.10 + log₁₀ [0.050] / [0.075]

    pOH = 4.92

    As pH = 14 - pOJ

    pH = 9.08
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