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    • 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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