How Do You Calculate the mmole of Ca2+ Exchanged by Zeolite in a Lab Titration?

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SUMMARY

The discussion focuses on calculating the millimoles (mmole) of Ca2+ ions exchanged by zeolite during a lab titration involving CaCl2 and EDTA. In titration #1, 1.00 ml of 0.15 M CaCl2 was titrated with 22.60 ml of EDTA, while in titration #2, a 1.00 ml filtrate solution containing 2g of zeolite in 50ml of CaCl2 was titrated with 18.4 ml of EDTA. The participants successfully calculated the initial mmole of Ca2+ but required guidance on incorporating titration data to determine the mmole exchanged per gram of zeolite.

PREREQUISITES
  • Understanding of titration techniques and calculations
  • Knowledge of molarity and millimole conversions
  • Familiarity with EDTA as a chelating agent
  • Basic principles of ion exchange in zeolites
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  • Calculate the mmole of Ca2+ exchanged using the formula: (Volume of EDTA titrated) x (Molarity of EDTA)
  • Research the ion exchange capacity of zeolites to understand the relationship between zeolite mass and Ca2+ exchange
  • Learn about the stoichiometry of the reaction involving zeolites and calcium ions
  • Explore advanced titration methods for more accurate ion concentration measurements
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Chemistry students, laboratory technicians, and researchers involved in ion exchange studies and titration methodologies.

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Homework Statement


Firstly, I have done the bulk of the lab but me and my lab partner cannot determine somethings and we are confused!

- Calculating the mmole of Ca2+ ion present in each sample titration
In titration #1 the beaker contained 1.00ml .15 M CaCl2 stock solution (with EDTA Buffer and Indicator). The Volume of EDTA titrated was 22.60 ml.

In titration #2 we had a 1.00ml (2g zeolite/50ml CaCL2) filtrate solution (with EDTA Buffer and Indicator). The Volume of EDTA titrated was 18.4 ml.

Then the second question is how do we calculate the mmole of Ca2+ exchanged by zeolite and also in per gram zeolite used!

(General directions would help so much)

Homework Equations


2NaAlO2 + 2Na2SiO3 ∙ 5H2O → Na2Al2Si2O8 + 4.5H2O + 4NaOH + 3.5H2O



The Attempt at a Solution


We found the initial mmole of Ca2+ but we do not know how to fit in the titration data.

We did .15 mol X .001L =.00015 M x 1000ml/L = .15 mmol/Ca2+

If anyone could clue us in the correct direction that would be great!
 
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