1.14.15: Degree of Dissociation
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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 \(\mathrm{K}_{\mathrm{A}}\).
\[\mathrm{K}_{\mathrm{A}}=\alpha^{2} \, \mathrm{c}_{\mathrm{A}}^{0} /(1-\alpha)\]
If
\[1-\alpha \cong 1, \alpha^{2}=\mathrm{K}_{\mathrm{A}} / \mathrm{c}_{\mathrm{A}}^{0}\]
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
\[=2 \,\left(\mathrm{n}_{\mathrm{A}}^{0}-\xi_{1}\right)+\xi_{1}+\xi_{2}+\xi_{1}-\xi_{2}=2 \, \mathrm{n}_{\mathrm{A}}^{0}\]
Total amount of \(\mathrm{A}\) in the system
\[=\mathrm{n}_{\mathrm{A}}^{0}-\xi_{1}+\xi_{1}-\xi_{2}+\xi_{2}=\mathrm{n}_{\mathrm{A}}^{0}\]
If the thermodynamic properties of the solution are ideal,
\[\mathrm{K}_{1}=\mathrm{c}_{\mathrm{A}}^{0} \,\left[\alpha_{1}+\alpha_{2}\right] \,\left[\alpha_{1}-\alpha_{2}\right] /\left[1-\alpha_{1}\right]\]
If
\[\mathrm{K}_{2}=0, \alpha_{2}=0, \mathrm{~K}_{1}=\mathrm{c}_{\mathrm{A}}^{0} \, \alpha_{1}^{2} /\left(1-\alpha_{1}\right)\]
But
\[\mathrm{K}_{2}=\left(\alpha_{1}+\alpha_{2}\right) \, \alpha_{2} \, \mathrm{c}_{\mathrm{A}}^{0} /\left(\alpha_{1}-\alpha_{2}\right)\]