Calculate the molalities of H^+ and OH^– in water at 298 K.

Question

Calculate the molalities of [latex]H^{+} ext{ and } OH^{–}[/latex] in water at 298 K.

Accepted Answer

In Table 14.1, the standard reduction potential for the half-cell reaction

Table 14.1 Standard Electrode Potentials at 298 K, 1 atm
Electrode Reaction	[latex]ε^{o,x}[/latex] volts
Acid Solutions
[latex]F_{2} + 2 e^{-} = 2 F^{-}[/latex]	2.65
[latex]S_{2}O_{8}^{2-} + 2e^{-} = 2SO_{4}^{2-}[/latex]	1.98
[latex]Co^{3+} + e^{-} = Co^{2+}[/latex]	1.82
[latex]Ce^{4+} + e^{-} = Ce^{3+}[/latex]	1.61
[latex]½Cl_{2} + e^{-} = Cl^{-}[/latex]	1.3595
[latex]Cr_{2}O_{7}^{2-} + 14H^{+} + e^{-} = 2Cr^{3} + 7H_{2}O[/latex]	1.33
[latex]MnO_{2} + 4H^{+} + 2e^{-} = Mn^{2+} + 2H_{2}O[/latex]	1.23
[latex]Br_{2}(l) + 2e^{-} = 2Br^{-}[/latex]	1.0652
[latex]2Hg^{2+} + 2e^{-} = Hg_{2}^{2+}[/latex]	0.92
[latex]Hg^{2+} + 2e^{-} = Hg[/latex]	0.854
[latex]Ag^{+} + e^{-} = Ag[/latex]	0.7991
[latex]Fe^{3+} + e^{-} = Fe^{2+}[/latex]	0.771
[latex]I_{2} + 2e^{-} = 2I^{-}[/latex]	0.5355
[latex]Fe(CN)_{6}^{3-} + e^{-} = Fe(CN)_{6}^{4-}[/latex]	0.36
[latex]Cu^{2+} + 2e^{-} = Cu[/latex]	0.337
[latex]S_{4}O_{6}^{2-} + 2e^{-} = 2S_{2}O_{3}^{2-}[/latex]	0.17
[latex]Cu^{2+} + e^{-} = Cu^{+}[/latex]	0.153
[latex]Sn^{4+} + 2e^{-} = Sn^{2+}[/latex]	0.15
[latex]S + 2H^{+} + e^{-} = H_{2}S[/latex]	0.141
[latex]2H^{+} + e^{-} = H_{2}[/latex]	0.000
[latex]Fe^{3+} + 3e^{-} = Fe[/latex]	-0.036
[latex]Pb^{2+} + 2e^{-} = Pb[/latex]	-0.126
[latex]Sn^{2+} + 2e^{-} = Sn[/latex]	- 0.136
[latex]Cd^{2+} + 2e^{-} = Cd[/latex]	– 0.403
[latex]Cr^{3+} + e^{-} = Cr^{2+}[/latex]	– 0.41
[latex]Fe^{2+} + 2e^{-} = Fe[/latex]	– 0.440
[latex]Zn^{2+} + 2e^{-} = Zn[/latex]	– 0.763
[latex]Al^{3+} + 3e^{-} = Al[/latex]	– 1.66
[latex]½ H_{2} + e^{-} = H^{-}[/latex]	-2.25
[latex]Mg^{2+} + 2e^{-} = Mg[/latex]	– 2.37
[latex]Na^{+} + e^{-} = Na[/latex]	– 2.714
[latex]Ca^{2+} + 2e^{-} = Ca[/latex]	– 2.87
[latex]Ba^{2+} + 2e^{-} = Ba[/latex]	– 2.90
[latex]Cs^{+} + e^{-} = Cs[/latex]	–2.923
[latex]K^{+} + e^{-} = K[/latex]	-2.925
[latex]Li^{+} + e^{-} = Li[/latex]	– 3.045
Basic Solutions
[latex]O_{3} + H_{2}O + 2e^{-} = O_{2} + 2OH^{-}[/latex]	1.24
[latex]Fe(OH)_{3} + e^{-} = OH^{-} + Fe(OH)_{2}[/latex]	– 0.56
[latex]Ni(OH)_{2} + 2e^{-} = Ni + 2OH^{-}[/latex]	– 0.72
[latex]2H_{2}O + 2e^{-} = H_{2} + 2OH^{-}[/latex]	– 0.828
[latex]SO_{4}^{2-} + H_{2}O + 2e^{-} = 2OH^{-} + SO_{3}^{2-}[/latex]	– 0.93
[latex]CNO^{-} + H_{2}O + 2e^{-} = 2OH^{-} + CN^{-}[/latex]	– 0.97
[latex]ZnO_{2}^{2-} + 2H_{2}O + 2e^{-} = Zn + 4OH^{-}[/latex]	– 1.216
[latex]Cr(OH)_{3} + 3e^{-} = Cr + 3OH^{-}[/latex]	– 1.3
[latex]Ca(OH)_{2} + 2e^{-} = Ca + 2OH^{+}[/latex]	– 3.03
Note : S tandard state is 1 molal.

[latex]\mathrm{H}_{2}\mathrm{O}_{(l)}+e^{-}={\frac{1}{2}}\mathrm{H}_{2(g)}+\mathrm{OH}_{(m)}^{-}[/latex]

is – 0.828 volts, and the standard reduction potential for the reaction

[latex]{\bf H}_{(m)}+e^{-}=\frac{1}{2}{\bf H}_{2(g)}[/latex]

is zero. Summing gives

[latex]\mathrm{H}_{2}\mathrm{O}_{(l)}=H_{ (m)}^{+} +\mathrm{OH}_{( m)}^{-}[/latex]

for which

[latex]\begin{array}{r l}{\Delta G_{298 K}^{\circ}}&{{}}&{{}=-{f {\varepsilon}}^{\circ}=-96487 imes(-0.828)}\end{array}[/latex]

= 79,900 J

[latex]=-8.3144 imes298\,\ln\,(\gamma_{\pm}m_{\mathrm{H+}}m_{\mathrm{OH-}})[/latex]

Thus, presuming that [latex]\gamma_{\pm}=1[/latex],

[latex]m_{\mathrm{H}^{+}}m_{\mathrm{OH}^{-}}=0.97 imes10^{-5}[/latex]

or, from the stoichiometry of the dissociation, as [latex]m_{\mathrm{H^{+}}}=m_{\mathrm{OH^{-}}}[/latex],

[latex]m_{\mathrm{H}^{+}}=m_{\mathrm{OH}^{-}}=10^{-7}[/latex]

At a molality of [latex]10^{– 7}[/latex] , the assumption that [latex]γ_{±} = 1[/latex] is reasonable.

Question 14.9.1: Calculate the molalities of H^+ and OH^– in water at 298 K.

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