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    General

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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－

      Gf0 (kJ/mol)        －615.4                     　　　        －228                  －157.2×2


      The total difference in standard Gibbs energy of formation for this reaction is,　⊿∑Gf0 ＝ ΣGf0【Product form】－ΣGf0【Original form】 = 73×103 ( J/mol )

      Substitute a numerical value into the conditional equation for chemical equilibrium (ΔGf0 ＝ －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.