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

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To calculate the mmole of Ca2+ exchanged by zeolite during titration, start by determining the initial mmole of Ca2+ in the stock solution, which is 0.15 mmol for 1.00 ml of 0.15 M CaCl2. For titration #1, 22.60 ml of EDTA indicates the amount of Ca2+ reacted, while for titration #2, 18.4 ml of EDTA is used for the zeolite filtrate. The difference in EDTA volumes between the two titrations reflects the amount of Ca2+ exchanged by the zeolite. To find the mmole of Ca2+ per gram of zeolite, divide the total exchanged Ca2+ by the weight of zeolite used in the filtrate solution.
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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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