강좌: Basic Equation of Chemical Equilibrium (Dissociation Equilibrium of Water)

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Dissociative equilibrium questions for water
- 　The water molecule (H2O), which is the base (solvent) of the solution reaction, dissociates in part into H+ and OH- ions. The reaction equation for the dissociation equilibrium of water is shown below, and the standard Gibbs energy of formation is written below each substance. The standard Gibbs energies of formation are H₂O：－237 (kJ/mol)、H⁺：0 (kJ/mol)、OH^－：－157 (kJ/mol).
  
  　　　　　　　　　　　original form　　　　product form
  
  Reaction equation for dissociation equilibrium of water：　　H₂O　　　　⇆　　　H⁺　 + OH^-　　
  
  　　　　　G_f⁰ (kJ/mol) 　　　　　　　　　　　　　　　　－237                               0       －157
  
  　The total difference in standard Gibbs energy of formation between the product form and the original form (⊿ΣG _{product form－original form} ＝ΣG product form － ΣG original foem) is,
  
  ⊿ΣG _{product form－original form} ＝（－157 + 0）×1000 － (－237)×1000
  
  　　　　　　　　＝ 80×1000   (J/mol)・・・①
  
  ※　I've removed the kilo (k) from the Gibbs energy unit, so I've multiplied by 1000.
  
  The condition equation for the dissociative equilibrium of water is,
  
  ⊿ΣG _{product form－original form} ＝－RT・Ln{【H⁺】【OH^－】／【H₂O】}・・・②
  
  　When dealing with reactions in aqueous solution, the base water molecules may be involved in the reaction. If the amount of water is sufficiently large relative to the solute involved in the reaction, and if the amount of water changes little before and after the reaction, then "the concentration of water is assumed to be 1 (【H₂O】＝ 1 )" in the reaction.
  　(There is also a way to express the dissociation equilibrium constant of water in which the molar amount of water molecules in 1 kg of water is 【H₂O】. Note that this is different from the constant when the concentration of water is set to 1.)
  
  　If the dissociation of water has reached equilibrium, the equilibrium constant (Kw*) is expressed as Kw = 【H⁺】×【OH^-】. Transform the equation ② to obtain Kw. Assume, however, that 【H₂O】 = 1, R = 8.3, and T = 300 (K).
  
  Kw = 【H⁺】×【OH^-】= exp (⊿ΣG _{product form－original form} ／ (－RT) )
  
     = exp( 80×1000／(－8×300 )　 =　 1.1×10^－¹⁴ ・・・③
  
     Remember that at a water temperature of 25°C, Kw ≒ 10^－¹⁴.
  
  The dissociation equilibrium constant Kw of water is called the "ionic product of water".
  
  　Many chemical reactions involve H⁺and OH^-, and hydrogen ion concentrations range from about 1 to 10^-16 (mol/L). In order to display the concentration in an easy-to-read manner, the hydrogen ion concentration is made logarithmic and a minus sign is added.
                              －Log₁₀【H⁺】= pH
  
  　This is called the hydrogen ion index (pH).
  For example, when the water temperature is 25°C and Kw = 10^-14,
  
  【H⁺】＝ 5×10^－¹⁰(mol/L)→　pH=9.301        → 【OH^-】＝ 2×10^－⁵ (mol/L)
  
  【H⁺】＝ 10^－¹⁰(mol/L)　        →　pH=10      → 【OH^-】＝ 10^－⁴ (mol/L)
  
  【H⁺】＝ 5×10^－¹¹(mol/L)→　pH=10.301      → 【OH^-】＝ 2×10^－⁴ (mol/L)
  
  【H⁺】＝ 10^－¹¹(mol/L)　        →　pH=11       → 【OH^-】＝ 10^－³ (mol/L).
  
  Equation ② shows that Kw is highly dependent on temperature.
  
  For example,
  
  H⁺】 and 【OH^-】 at 5°C (Kw = 0.185 x 10-14) and at the following pH,
  
  pH = 5 → 【H⁺】= 10^－⁵ (mol/L) 、【OH^-】＝ 1.85×10^－¹⁰ (mol/L)
  
  pH = 6               → 【H⁺】= 10^－⁶ (mol/L) 、【OH^-】＝ 1.85×10^－⁹ (mol/L)
  
  pH = 7               → 【H⁺】= 10^－⁷ (mol/L) 、【OH^-】＝ 1.85×10^－⁸ (mol/L)
  
  pH = 7.37 → 【H⁺】= 0.43×10^－⁷ (mol/L) 、【OH^-】＝ 0.43×10^－⁷ (mol/L)
  
  pH = 8    → 【H⁺】= 10^－⁸ (mol/L) 、           【OH^-】＝ 1.85×10^－⁷ (mol/L).
  
  At pH 7.37, 【H⁺】=【OH^-】.

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Dissociative equilibrium questions for water