Experiment 28: Determination of the Hardness of Water Thank you! Experiment 28 D

Question

Experiment 28: Determination of the Hardness of Water

Thank you! Experiment 28 Data and Calculations: Determination of the Hardness of Water A. Preparation of . solution with a "known" Ica"I Mas of CacOs) usod (should be close to 0.15 g:you Volume of Ca(aq)"sock"olution prepareod, aficr dilution in volumetric flask can just tare the 250-ml. beake Calculated Data: Moles of CaCOys) used to make Ca "ock solution B. Titration of Blank Inisial Buret Reading OmFinal ore Readine au ol Raw Data Calculated Dasa: Volume of EDTA sohution used to siternic the blank (odded MiOF B C. Standardization of EDTA solution Titration Baw Data Initial (EDTA) Buret Reading Final (EDTA) Buret Reading Moles of Ca" in sample (of stock sol's) aouu o.olxo.ou oolr oouu Moles of EDTA used to titrate the Ca ? Total volume of EDTA solution used uneme undne gu. ume of EDSA solution used soltitrate in the sample (of stock sol'n)* the Ca Molarity (mol/L) of the EDTA solution Average (Best Estimate of) EDTA Molarity Recall the purpose of the "blank', some of the EDTA added in the titration was used to complex the added Mg

Explanation / Answer

I am giving you the answer for the first question,where you went wrong.

For your part A,

250 ml CaCO3 solution contains =0.15 gm

So, 1000 ml CaCO3 contains= (0.15*1000)/250= 0.6 gm

Molecular weight of CaCO3= 100 (approx)

So, molarity of solution= 0.6/100 = 6*10-3 (M)

So, Molarity of Ca2+ in the solution =   6*10-3 (M)

[CaCO3 = Ca2++CO32-; so, x (M) CaCO3 produces x (M) Ca2+ and x (M) CO32-]

So, take the molarity to be 0.006 (M) instead of 0.0015 (M).

Rest, you can proceed as usual.

For part B, I do not think, you can perform a blank for tap water. So, Take the total volume of EDTA used as well in the next blank for EDTA used to titrate Ca2+ in aliquot. (where we will assume, only Ca2+? is present in tap water)

Rest, you can proceed with calculations as usual.

Hope, this serves you well.

Best Wishes.

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