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Acid-Base Stoichiometry

  1. Jul 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Calculate the volume of 0.108 M NaOH solution required to react completely with 145 mL of phosphoric acid solution that is 1.15x10^-2 M.


    2. Relevant equations
    n=MV


    3. The attempt at a solution
    [tex]NaOH \rightarrow Na^{+}+OH^{-}[/tex]
    [tex]H_{3}PO_{4} \rightarrow 3H^{+} + PO_{4}^{-}[/tex]

    [tex]1.15x10^{-2} mol/L[/tex] [tex]H_{3}PO_{4} * .145L[/tex] [tex]H_{3}PO_{4} * 3 mol H^{+}/1 mol H_{3}PO_{4} = 5.003 mol H^{+} [/tex]

    [tex]5.003 mol OH^{-} / .108 mol OH^{-} L = 46.324 L[/tex]

    I clearly went wrong somewhere. Why in the world would I ever need 46 L to react with a 145 mL solution.

    PS. I am bad with the latex reference codes, sorry.
     
  2. jcsd
  3. Jul 4, 2009 #2
    You multiplied by 145 instead of 0.145 (or .145 as you had written) in your first line of work.
     
    Last edited: Jul 4, 2009
  4. Jul 4, 2009 #3

    Borek

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    Staff: Mentor

    No need for latex.

    Simple math error.
     
  5. Jul 4, 2009 #4

    Borek

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    Staff: Mentor

    Hard to say - s/he wrote .145 which can be short for 0.145.
     
  6. Jul 4, 2009 #5
    It should have been 5.003 x10^-3, not just 5.003.

    Answer is 46.32mL. Thanks for the help.
     
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