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Buffering capacity experiment

  1. Mar 27, 2008 #1
    1. The problem statement, all variables and given/known data

    We are doing a experiement where we are testing the buffering capacity of soils. The text says to use Ca(OH)2, but we only have NaOH, which i plan to substitute. I just wanted to check that the solution i make up will have the equivalent OH.


    2. Relevant equations
    making up 0.01M solution of Ca(OH)2 which is 0.07g per litre

    Of that 0.34g is OH 0.07 x 34/70

    Therefore the equialent NaOH would be 0.8g per litre

    0.34g *40/14


    Would this be correct??
     
  2. jcsd
  3. Mar 28, 2008 #2

    chemisttree

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    By inspection your answer is way off. You are saying that of the 0.07 grams of Ca(OH)2 present in a liter that 0.34 grams of that is due to the counterion, OH? How can you have 0.34 grams out of only 0.07 grams?
     
  4. Mar 30, 2008 #3
    Damn......its not 0.34g is is 34%.

    34% of the 0.07g is from the OH.
     
  5. Mar 31, 2008 #4

    chemisttree

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    Are you familiar with the concept of moles and equivalents?
     
  6. Mar 31, 2008 #5
    Learnt the concepts a long time ago, have been a microbiologist for the last decade, and have now moved into an area where chemistry is required.
     
  7. Mar 31, 2008 #6

    chemisttree

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    Ahh, great! You want 0.01 moles per liter of Ca(OH)2 which is 0.02 equivalents of OH- per liter. To get that from NaOH you need (0.02moles/L OH-) X (40 g NaOH/mole) which is 0.8 g/L NaOH.... same answer as yours.

    Do you think the replacement will work in soils analysis?
     
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