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Date Posted: 4/6/2008 12:48:51 PM  Status: Live
pH
Course Textbook Chapter Problem
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Question Details:
You prepare 100mL of 0.1M Na2CO3 solution.
a. Calculate the pH of this solution.
b. You add 100 mL of 0.18M HCl to the solution.  Calculate the pH.

Assuming a closed system.
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Date Posted: 5/16/2008 9:23:55 AM  Status: Live
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Response:
   Given that
   The volume of Na2CO3 = 100.0 ml
   The molarity of Na2CO3 = 0.1 M
                                          = C
  a) We know that
    Kb for CO32- = 2.1*10-4
    [OH-] = √Kb*C
              = √ 2.1*10-4 *0.1
              = 4.58*10-3 M
    pOH = - log(4.58*10-3 M)
             = 2.34
      pH = 14 - 2.34
            = 11.66
  b) The number of moles of Na2CO3 = 0.1M *0.1 L
                                                          = 0.01 mol
    The number of moles of HCl = 0.18M * 0.1 L
                                                = 0.018 mol
  We know that two moles of HCl convert one mole of Na2CO3 to one mole of NaHCO3
  ∴ The number of NaHCO3 formed = 0.018mol/2
                                                         = 0.009 mol
  The number of moles of Na2CO3 remaining = 0.01 mol -0.009 mol
                                                                      = 0.001 mol
     Ka of NaHCO3  = Kw / Kb
                                 = 1*10-14 / 2.1*10-4
                                 = 4.8*10-11
                         pKa = 10.319
  
   We know that
     pH = pKa + log [ NaHCO3 ]/[Na2CO3 ] 
           = 10.319 + log[0.001]/[0.009]
           = 9.3647



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