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Titration of strong base + weak acid

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data
    Consider a titration in which 50.00 mL of .450 M benzoic acid, C6H5COOH, is reacted with 0.2000 M Ca(OH)2. What will the pH be when 60.00 mL of Ca(OH)2 has been added?


    2. Relevant equations
    2C6H5COOH + Ca(OH)2 -> 2 H2O + Ca(C6H5COO)2


    3. The attempt at a solution
    I have determined that the equivalence point is at 56.25 mL, so this is past that. The thing that is throwing me off is the mol ratio of acid:base. I try doing an ICF chart and keep getting that there is still acid left and no base. I thought that after the equiv. point, there is no acid left and you determine the pH based on the remaining base.

    Can anyone help me fill in the ICF chart so that I can find the pH?
    I need moles of acid and base to do this.
     
  2. jcsd
  3. Nov 14, 2009 #2

    symbolipoint

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    Your first calculation is correct, that equivalence point needs 56.25 ml. of 0.2000 Ca(OH)2.

    Now think in steps: How many moles of extra Ca(OH)2 was added beyond equivalence?
    How many moles of hydroxide is this? What is the concentration of excess hydroxide and therefore what is the pOH?
     
  4. Nov 14, 2009 #3
    This is the part I am stuck on. When I plug into my ICF chart I get that there are .0225 moles of C6H5COOH.
    Then I calculate the moles of Ca(OH)2 using .060L X 0.200mol/L and get .012 moles of Ca(OH)2.
    Do I divide that by 2 because it is a 2:1 reaction?
    Even if I don't, when I go to subtract the smaller amount of moles, the Ca(OH)2 comes out as the limiting reagent, which means it is all used up and I know this is not right.
    What am I missing?
     
  5. Nov 14, 2009 #4

    symbolipoint

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    Forget your ICF chart. Use fundamental principles. You know you passed the equivalence point for neutralization so you have no unneutralized C6H5COOH. How many moles beyond the equivalence point have you gone, and what is the concentration of the hydroxide (in moles per liter), and from this, what is pOH?
     
  6. Nov 14, 2009 #5

    Borek

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    Think in terms of limiting reagents only. And compare with this pH of acid/base mixture calculation example.

    In a way question is idiotic - there is no such thing as 0.2M calcium hydroxide, its solubility is about 10 times smaller if memory serves me well. No idea about benzoic acid solubilityk, but with bulky phenyl it won't be easily soluble as well.

    --
     
  7. Nov 14, 2009 #6
    I am not sure if I came up with the right answer, but I got a pH of 12.13. I calculated that there were .0015 mols of Ca(OH)2 remaining.
     
  8. Nov 15, 2009 #7

    symbolipoint

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