When a Reaction Proceeds from Chemical Equilibrium (Conditions Under Which Manganese Hydroxide Dissolves)

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• Determine precipitation conditions for manganese hydroxide in the Winkler method (review)
  • Exercises (review)

    　Find the condition (pH) under which Mn(OH)₂ precipitates when sodium hydroxide is added to seawater when processing samples by the Winkler method.

    
    

                           【Original form】                                    【Product form】
    

                                Mn(OH)₂                         ⇆           Mn²⁺      +          2OH^－

    G_f⁰ (kJ/mol)        －615.4                     　　　        －228                  －157.2×2

    
    

    The total difference in standard Gibbs energy of formation for this reaction is,　⊿∑G_f⁰ ＝ ΣG_f⁰_{【Product form】}－ΣG_f⁰_{【Original form】} = 73×10³ ( J/mol )

    Substitute a numerical value into the conditional equation for chemical equilibrium (ΔG_f⁰ ＝ －R・T・lnK).

    
    

     73×10³ ( J/mol )  = －8.314・298.15・ln（【Mn²⁺ mol/L】・【OH^－ mol/L】²／【Solid activity (=1) 】）

    ln【Mn²⁺】・【OH^－】² ＝ －29.45 , then 【Mn²⁺】・【OH^－】² ＝ e^－29.45 ＝ 1.6×10^－13.

     

    If the pH of seawater is 8,  【OH^－】= 10^－6 mol/L, so 【Mn²⁺】= 1.6・10^－1 mol/L
    If we set alkaline pH 10, 【OH^－】= 10^－4 mol/L, so 【Mn²⁺】= 1.6・10^－5 mol/L
    If pH 12 is set, 【OH^－】= 10^－2 mol/L, so 【Mn²⁺】= 1.6・10^－9 mol/L　(Almost insoluble. Precipitates as manganese hydroxide)
    
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• Example of the direction in which a chemical reaction proceeds (precipitation and dissolution of manganese hydroxide)
  • In the Mn(OH)₂ dissolution reaction above (Mn(OH)₂→ Mn²⁺ + 2OH^-), given the value of the equilibrium constant (K), we can easily calculate the conditions under which the reaction proceeds (whether dissolution proceeds or precipitation increases).

     
    

    Since the equilibrium constant K = 1.6 x 10^-13,
    

    If [Mn²⁺][OH^-]²   <  1.6×10^－13, dissolution proceeds (original form → product form).

    If [Mn²⁺][OH^-]²   >  1.6×10^－13, increased precipitation (product form → original form).

     

    Example 1) Will manganese hydroxide precipitate at a temperature of 25°C, pH 9, and [Mn²⁺] = 0.16 mol/L?

    
    

    Since [OH^-] = 10^-5 mol/L, 

    [Mn²⁺][OH^-]² = 1.6×10^－1×10^－10　=  1.6×10^－11 (> K), so precipitation of Mn(OH)₂ occurs.

     

    Example 2) Will manganese hydroxide precipitate at a temperature of 25°C, pH 9, and [Mn²⁺] = 1.6 x 10-5 mol/L?

    
    

    
    

    Since [OH^-] = 10^-5 mol/L, 

    [Mn²⁺][OH^-]² = 1.6×10^－5×10^－10　=  1.6×10^－15 (< K),  if Mn(OH)₂ precipitates, dissolution proceeds.

     

    Example 3) Will manganese hydroxide precipitate at a temperature of 25°C, pH 11, and [Mn²⁺] = 0.016 mol/L?

    
    

    
    

    Since [OH^-] = 10^-3 mol/L, 

    [Mn²⁺][OH^-]² = 0.016×10^－6　=  1.6×10^－8 (> K), so precipitation of Mn(OH)₂ occurs.

     

    Example 4) Will manganese hydroxide precipitate at a temperature of 25°C, pH 11, and [Mn²⁺] = 1.6 × 10^－7 mol/L?

    
    

    Since [OH^-] = 10^-3 mol/L,

    [Mn²⁺][OH^-]² = 1.6×10^－7×10^－6　=  1.6×10^－13 (=  K), so Equilibrium.
    

  • 【Supplement to the Winkler Method】

    　From Examples 3) and 4), if only 1 mL of concentrated Mn²⁺ solution (1.6 mol/L) is added to the sample water (100 mL) for dissolved oxygen measurement (equivalent to 1.6 × 10^-3 mol of Mn added), the Mn²⁺ concentration in the sample water is approximately 0.016 mol/L.

    　If 1 mL of NaOH solution (0.1 mol/L) is added here, the OH^- concentration in the sample water becomes about 10^-3 mol/L (pH 11). The equilibrium concentration of Mn²⁺ is then 1.6×10^-7 mol/L, so Mn(OH)₂ precipitation occurs until the concentration drops to this level. The amount of Mn added to the sample water is 1.6 x 10^-3 mol, and the amount of precipitation that occurs is 1.59998 x 10^-3 mol (= 1.6 x 10^-3 mol - 1.6 x 10^-7 (mol/L)x0.1 (L)). It can be seen that almost all Mn is precipitated.
    

    　Since the NaOH solution added for the dissolved oxygen measurement is 9 mol/L, a considerable excess of OH- is added. In the analysis, 3 mL of HCl (6 mol/L) is added, which is sufficient to counteract the excess OH-, so the sample water becomes acidic (pH < 1) and all Mn(OH)₂ and MnO(OH)₂ precipitates are dissolved.