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Buffering capacity experiment

  • Thread starter iross75
  • Start date
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1. Homework Statement

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. Homework 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??
 

Answers and Replies

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?
 
6
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Damn......its not 0.34g is is 34%.

34% of the 0.07g is from the OH.
 
chemisttree
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Are you familiar with the concept of moles and equivalents?
 
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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.
 
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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