Into a 125 or 250 mL Erlenmeyer flask add the following: 20mL of tap water using
ID: 888544 • Letter: I
Question
Into a 125 or 250 mL Erlenmeyer flask add the following:
20mL of tap water using a 20 mL volumetric pipette – pipettes are typically accurate to two decimal
places – so 20.00 mL is added.
5 mL of ammonia buffer (1.5m NH3/0.3 M NH4Cl)
2 mL of 0.001 M Na2MgEDTA solution (added to enhance the color change)
10 drops of calmagite indicator solution
Prepare a burette with 0.001 M Na2EDTA (in 0.1 M NH3) – the introduction describes how to prepare a burette.
Titrate the water sample with the EDTA until the sample changes from red to light blue.
Initial Volume
Final Volume
Trial 1 (Tap Water)
0 mL
4.55 mL
Trial 2 (Tap Water)
4.55 mL
11.09 mL
Trial 3 (Tap Water)
11.09 mL
15.20 mL
Unknown Water Sample
15.20 mL
37.00 mL
Calculate the hardness in ppm of water for each of your three trials for tap water. Report the average hardness of the tap water.
Initial Volume
Final Volume
Trial 1 (Tap Water)
0 mL
4.55 mL
Trial 2 (Tap Water)
4.55 mL
11.09 mL
Trial 3 (Tap Water)
11.09 mL
15.20 mL
Unknown Water Sample
15.20 mL
37.00 mL
Explanation / Answer
To prepare standard EDTA SOLUTION
Determine the moles of EDTA Required
Conc of edta sol= 0.001 mol/l
Vol of EDTA sol= 250 ml or 0.25 l
Moles of edta= conc of edta in mol/l x volume in litre
= 0.001 x 0.25= 2.5 × 10-4 mol/ litre
Mass of edta required to make 250 ml of 0.001 mol/ l solution
Molar mass= 372.24 g/ mol
Mass of edta required= moles × molar mass = 2.5 × 10-4 × 372.24 = 0.09306 g
Calculations
Volume of edta used for titration
So average titre used is 15.2 ml ÷ 3= 5.1 ml
Conc of edta= 0.001 mol/ l; vol of edta sol = average titre = 5.1 × 10-3 l
Moles of edta= conc × volume = 0.001 × .1 × 10 -3= 5.1 × 10 -6 mol
Write standard equations
So 1 mole M+2= ( calcium and magnesium ions) react with one mole edta
So 5.1 ×10-6 moles of edta react with 5.1 × 10-6 moles of M+2
Total hardness ( calcium and magnesium)
Concentration M+2 moles M+2 ÷ vol in litre
Vol of water used = 20 ml or 20 × 10-3 l
= 5.1 × 10-6 × 20 × 10-3= 102 × 10-9= 1.02 × 10-7 mol/l or ppm
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