Molarity of EDTA (M): O.003 Blank Titration Volume (mL) tration of unknown water

Question

Molarity of EDTA (M): O.003 Blank Titration Volume (mL) tration of unknown water sample Unknown Number: 12 Trial Initial burette Final burette Total Molarity of hardness value (mL) value (mL) corrected Unknown Water 5 6 22.50 22.40 4 (if needed) Titration of Municipal Tap Water Sam Tap Water Sample Number Molarity of water hardness Total known Sample Trial value (mL) value (mL) corre volume I. 6.00 to 3 19.30 25.90 lo 4 (if needed) Observations: (Submit via LON-CAPA) Data Analysis Unknown Water Sample unknown sample) Experimental Mean (mg CaCO3 /L, original Standard Deviation mg/L) 95% confidence value for the mean (value, value (mg/L) Reporting value for mean based on 95% confidence Pag Chez 309 v163.0 Prelab Score Data Analysis Score Complete Incomplete Excel Score (calc)

Explanation / Answer

I am showing calculations for the first trial for the unknown water sample:

Total corrected volume of EDTA used = 22.7 mL = 0.0227 L

Molarity of EDTA solution used = 0.003366 M

So, moles of EDTA used in the titration = Molarity * Volume in liters = 0.003366 * 0.0227 = 0.0000764

Since 1 mole EDTA chelates only 1 mole of Ca2+ ions, so moles of Ca2+ ions present = 0.0000764

Now, these many moles are present in the diluted water sample whose volume is 259 mL

So, molarity of diluted unknown sample = moles/volume in liters = 0.0000764/0.259 = 0.000295 M

Since 25 mL was diluted to 259 mL, so molarity of undiluted unknown sample = 0.000295*(259/25) = 0.00305 M

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