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Acid base titration problem

  1. Oct 22, 2004 #1

    Mag

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    consider the titration of 50.0mL of 0.10 M methyl ammonia with 0.10 M HCl. Calculate the pH after
    15mL of titrant= 11.05
    25mL of titrant= 10.68
    50mL of titrant= 8.98
    60mL of titrant= ?
    At 60 mL there is an excess of 1 mmol of HCl. At this point do I say that the pH is 1 (-log of 0.10)? I'm at a loss.


    Mag
     
  2. jcsd
  3. Oct 23, 2004 #2

    chem_tr

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    You need to know the reaction first:

    [tex]MeNH_2+HCl \longrightarrow MeNH_3^+Cl^-[/tex]

    Now you can place the millimoles of reactants to see which is excessive. You can find the millimole amounts by multiplying milliliters with molarity, but I think you know how to do this calculation.

    Please consider the volume and millimoles together when trying to find the pH of the solution, 0.1 millimoles of HCl is in excess in 110 mL of total solution, where the contribution of [itex]MeNH_3^+[/itex] can be easily omitted.
     
  4. Oct 24, 2004 #3

    Mag

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    This is what you were trying to tell me, correct?
    (0.10M HCl)(60mL)=6mmol
    (0.10M [tex]MeNH_2[/tex])(50mL)=5mmol


    [tex]MeNH_2+HCl \longrightarrow MeNH_3^+Cl^-[/tex]

    initial: 5mmol[tex]MeNH_2[/tex] 6mmol[tex]HCl[/tex]
    [tex]\Delta[/tex]: -5mmol [tex]MeNH_2[/tex] -5mmol[tex]HCl[/tex]
    final: 0mmol[tex]MeNH_2[/tex] 1mmol[tex]HCl[/tex] 5mmol[tex]MeNH_3^++Cl^-[/tex]

    [tex]\frac{1mmol}{50mL (analyte) + 60mL (titrant)}[/tex]
     
  5. Oct 24, 2004 #4

    chem_tr

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    This is it. Congrats. The negative logarithm of the result will be your pH value.
     
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