Substituting NaOH for Ca(OH)2 in Buffering Capacity Experiment: Will it Work?

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Discussion Overview

The discussion revolves around the feasibility of substituting NaOH for Ca(OH)2 in an experiment designed to test the buffering capacity of soils. Participants explore the implications of this substitution in terms of equivalent hydroxide ion concentration.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Technical explanation

Main Points Raised

  • One participant calculates the equivalent amount of NaOH needed to match the hydroxide concentration of a 0.01M Ca(OH)2 solution, arriving at 0.8g per liter.
  • Another participant challenges the initial calculation, questioning the logic behind the mass of hydroxide derived from the given amount of Ca(OH)2.
  • A correction is made regarding the percentage of hydroxide in the original calculation, clarifying that it should be 34% rather than 0.34g.
  • One participant references the concept of moles and equivalents, suggesting a deeper understanding of the chemical principles involved.
  • A later reply confirms the initial calculation of 0.8g/L NaOH as correct and questions the effectiveness of the substitution in soil analysis.

Areas of Agreement / Disagreement

Participants generally agree on the calculation of the equivalent amount of NaOH needed, but there is some confusion and correction regarding the initial understanding of the hydroxide content in Ca(OH)2. The discussion does not reach a consensus on whether the substitution will work effectively in soil analysis.

Contextual Notes

There are unresolved aspects regarding the assumptions made in the calculations and the potential impact of substituting NaOH for Ca(OH)2 in the experimental context.

Who May Find This Useful

This discussion may be useful for students and researchers involved in soil chemistry, buffering capacity experiments, or those transitioning from microbiology to chemistry-related fields.

iross75
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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.


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

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