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1.14.15: Degree of Dissociation

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    374477
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    A given aqueous solution is prepared using \(\mathrm{n}_{1}^{0}\) moles of water and \(\mathrm{n}_{\mathrm{A}}^{0}\) moles of a weak acid \(\mathrm{HA}\). The composition of the solution at equilibrium (at fixed \(\mathrm{T}\) and \(\mathrm{p}\)) is described as follows.

      \(\mathrm{HA}(\mathrm{aq})\) \(\leftrightarrow\) \(\mathrm{H}^{+}(\mathrm{aq}) +\) \(A^{\prime}(\mathrm{aq})\)
    At \(t=0\) \(n_{A}^{0}\)   \(0\) \(0 \mathrm{~mol}\)
    At equilibirium, \(n_{A}^{0}-\xi^{\mathrm{eq}}\)   \(\xi^{\mathrm{eq}}\) \(\xi^{\mathrm{eq}}\mathrm{~mol}\)
    or, \(\mathrm{n}_{\mathrm{A}}^{0} \,\left[1-\frac{\xi^{e q}}{\mathrm{n}_{\mathrm{A}}^{0}}\right]\)   \(\xi^{\mathrm{eq}}\) \(\xi^{\mathrm{eq}} \mathrm{~mol}\)
    or. \(\mathrm{n}_{\mathrm{A}}^{0} \,(1-\alpha)\)   \(\alpha \, \mathrm{n}_{\mathrm{A}}^{0}\) \(\alpha \, \mathrm{n}_{\mathrm{A}}^{0}\mathrm{~mol}\)

    If the volume of the system is \(1 \mathrm{~dm}^{3}\) then, \(\mathrm{c}_{\mathrm{A}}^{0} \,(1-\alpha) \quad \alpha \, \mathrm{c}_{\mathrm{A}}^{0} \quad \alpha \, \mathrm{c}_{\mathrm{A}}^{0} \quad \mathrm{~mol} \mathrm{dm}^{-3}\)

    By definition, the degree of dissociation, \(\alpha=\xi^{\mathrm{eq}} / \mathrm{n}_{\mathrm{A}}^{0}\); \(\alpha\) is an intensive variable describing the ‘degree’ of dissociation. If the total volume of the solution is \(\mathrm{V}\), the concentration \(\mathrm{c}_{\mathrm{A}}^{0}=\mathrm{n}_{\mathrm{A}}^{0} / \mathrm{V}\). If the thermodynamic properties of the solution are ideal, the composition of the solution can be described by an equilibrium acid dissociation constant \(\mathrm{K}_{\mathrm{A}}\).

    \[\mathrm{K}_{\mathrm{A}}=\alpha^{2} \, \mathrm{c}_{\mathrm{A}}^{0} /(1-\alpha) \nonumber \]

    If

    \[1-\alpha \cong 1, \alpha^{2}=\mathrm{K}_{\mathrm{A}} / \mathrm{c}_{\mathrm{A}}^{0} \nonumber \]

    If the acid is dibasic, the analysis is a little more complicated.

    For an acid \(\mathrm{H}_{2}\mathrm{A}\), there are two extents of acid dissociation.
      \(\mathrm{H}_{2}\mathrm{A}\) \(\rightleftarrows\) \(\mathrm{H}^{+} +\) \mathrm{HA}^{-}\)
    At \(t=0\), \(\mathrm{n}_{\mathrm{A}}^{0}\)   \(0\) \(0 \mathrm{~mol}\)
    At equilibrium, \(\mathrm{n}_{\mathrm{A}}^{0}-\xi_{1}\)   \(\xi_{1}+\xi_{2}\) \(\xi_{1}-\xi_{2}\mathrm{~mol}\)
    Or, \(n_{A}^{0} \,\left[1-\frac{\zeta_{1}}{n_{A}^{0}}\right]\)   \(\mathrm{n}_{\mathrm{A}}^{0} \,\left[\frac{\xi_{1}}{\mathrm{n}_{\mathrm{A}}^{0}}+\frac{\xi_{2}}{\mathrm{n}_{\mathrm{A}}^{0}}\right]\) \(\mathrm{n}_{\mathrm{A}}^{0} \,\left[\frac{\xi_{1}}{\mathrm{n}_{\mathrm{A}}^{0}}-\frac{\xi_{2}}{\mathrm{n}_{\mathrm{A}}^{0}}\right]\mathrm{~mol}\)

    By definition \(\mathrm{c}_{\mathrm{A}}^{0}=\mathrm{n}_{\mathrm{A}}^{0} / \mathrm{V}\) where \(\mathrm{V}\) is the volume of solution expressed in \mathrm{dm}^{3}\). Also by definition \(\alpha_{1}=\xi_{1} / \mathrm{n}_{\mathrm{A}}^{0}\) and \(\alpha_{2}=\xi_{2} / \mathrm{n}_{\mathrm{A}}^{0}\)

    Hence from equation (d)

      \(\mathrm{H}_{2}\mathrm{A}\) \(\rightleftarrows\) \(\mathrm{H}^{+} +\) \mathrm{HA}^{-}\)
    At equilibrium, \(\left(\mathrm{n}_{\mathrm{A}}^{0} / \mathrm{V}\right) \,\left[1-\alpha_{1}\right]\)   \(\left(\mathrm{n}_{\mathrm{A}}^{0} / \mathrm{V}\right) \,\left[\alpha_{1}+\alpha_{2}\right]\) \(\left(\mathrm{n}_{\mathrm{A}}^{0} / \mathrm{V}\right) \,\left[\alpha_{1}-\alpha_{2}\right]\mathrm{~mol}\)
    Or, \(\mathrm{c}_{\mathrm{A}}^{0} \,\left[1-\alpha_{1}\right]\)   \(\mathrm{c}_{\mathrm{A}}^{0} \,\left[\alpha_{1}+\alpha_{2}\right]\)

    \(\mathrm{c}_{\mathrm{A}}^{0} \,\left[\alpha_{1}-\alpha_{2}\right]\mathrm{~mol}\)

      \(\mathrm{HA}^{-}\) \(\rightleftarrows\) \(\mathrm{H}^{+} +\) \mathrm{A}^{-2}\)
    At \(t=0\), \(0\)   \(0\) \(0 \mathrm{~mol}\)
    Also At equilibrium, \(\xi_{1}-\xi_{2}\)   \(\xi_{1}+\xi_{2}\) \(\xi_{2}\mathrm{~mol}\)
    Or, \(\mathrm{n}_{\mathrm{A}}^{0} \,\left[\frac{\xi_{1}}{\mathrm{n}_{\mathrm{A}}^{0}}-\frac{\xi_{2}}{\mathrm{n}_{\mathrm{A}}^{0}}\right]\)   \(\mathrm{n}_{\mathrm{A}}^{0} \,\left[\frac{\xi_{1}}{\mathrm{n}_{\mathrm{A}}^{0}}+\frac{\xi_{2}}{\mathrm{n}_{\mathrm{A}}^{0}}\right]\) \(n_{A}^{0} \,\left[\frac{\zeta_{2}}{n_{A}^{0}}\right]\mathrm{~mol}\)
      \(c_{\mathrm{A}}^{0} \,\left[\alpha_{1}-\alpha_{2}\right]\)   \(\mathrm{c}_{\mathrm{A}}^{0} \,\left[\alpha_{1}+\alpha_{2}\right]\) \(\mathrm{c}_{\mathrm{A}}^{0} \, \alpha_{2}\)

    Total amount of \(\mathrm{H} in the system

    \[=2 \,\left(\mathrm{n}_{\mathrm{A}}^{0}-\xi_{1}\right)+\xi_{1}+\xi_{2}+\xi_{1}-\xi_{2}=2 \, \mathrm{n}_{\mathrm{A}}^{0} \nonumber \]

    Total amount of \(\mathrm{A}\) in the system

    \[=\mathrm{n}_{\mathrm{A}}^{0}-\xi_{1}+\xi_{1}-\xi_{2}+\xi_{2}=\mathrm{n}_{\mathrm{A}}^{0} \nonumber \]

    If the thermodynamic properties of the solution are ideal,

    \[\mathrm{K}_{1}=\mathrm{c}_{\mathrm{A}}^{0} \,\left[\alpha_{1}+\alpha_{2}\right] \,\left[\alpha_{1}-\alpha_{2}\right] /\left[1-\alpha_{1}\right] \nonumber \]

    If

    \[\mathrm{K}_{2}=0, \alpha_{2}=0, \mathrm{~K}_{1}=\mathrm{c}_{\mathrm{A}}^{0} \, \alpha_{1}^{2} /\left(1-\alpha_{1}\right) \nonumber \]

    But

    \[\mathrm{K}_{2}=\left(\alpha_{1}+\alpha_{2}\right) \, \alpha_{2} \, \mathrm{c}_{\mathrm{A}}^{0} /\left(\alpha_{1}-\alpha_{2}\right) \nonumber \]


    This page titled 1.14.15: Degree of Dissociation is shared under a Public Domain license and was authored, remixed, and/or curated by Michael J Blandamer & Joao Carlos R Reis via source content that was edited to the style and standards of the LibreTexts platform.