I got this answer wrong and all calssmates seem to have different answers. There

Question

I got this answer wrong and all calssmates seem to have different answers. There were several parts but the questions marked a and b were incorrect. Please help.

An unknown was titrated after being diluted. The dilution was carried out using 2mL of the stock in a volumetric flask and adding up to the 10 mL mark of the volumetric flask. The diluted solution was then titrated. the initial volume in a burette was 2.00 mL and final volume was 1.14 mL. The total used: 0.86 mL.

a) calculate and report the hardness of the water as ppm CaCO3

b) calculate & report the hardness of your unknown calcium sample as ppm CaCO3 for the unknown

Explanation / Answer

The known volume of sample containing CaCO3 is titrated with MgEDTA of known concentration and volume detected by titration.

Let M1=conc of CaCO3 sample=unknown =measure of hardness of the sample

M2=conc of MgEDTA=0.01M

V1=volume of sample=10ml (diluted sample from stock,however the stock volume taken was 2 ml)

V2 =volume of titrant =0.86 ml

M1V1=M2V2

M1=M2*V2/V1=0.01M*0.86ml/10ml=0.00086 M(diluted)

To find the conc of stock (the actual sample)

2ml*M(stock)=10ml*0.00086M

M(stock)=10ml*0.00086M/2ml=0.0043 M

So hardness=0.0043 mol/L of CaCO3=0.0043 mol/L*100.09 g/mol=0.43039 g/L=430.39 mg/L=430.39 ppm CaCO3

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