1. A 10.00 mL aliquot of 0.010 M CaCO3 is titrated with 18.22 mL of EDTA solutio
ID: 518846 • Letter: 1
Question
1. A 10.00 mL aliquot of 0.010 M CaCO3 is titrated with 18.22 mL of EDTA solution. What is the EDTA molarity?
2. A 250.0 mL water sample requires 30.85 mL of the EDTA solution from question 1 to reach the calmagite endpoint. What was the molarity of the hard metal ions in the water sample?
3. If the metal ions in the water sample of question 2 are assumed to be Ca2+ from CaCO3, express the concentration in ppm CaCO3. This is the same as the concentration in units of mg CaCO3 per liter of sample.
Explanation / Answer
1. M1V1 = M2V2
Where, M1 = Molarity of CaCO3
V1 = Volume of CaCO3
M2 = Molarity of EDTA
V1 = Volume of EDTA
thereby,
0.010 M CaCO3 * 10.00 ml = M2 * 18.22ml
(0.010 M CaCO3 * 10.00 ml) / 18.22ml = M2
Molarity of EDTA is 0.0054 M .
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2. The same formula will be applied
M1V1 = M2V2
Where, M1 = Molarity of Hard metal ions in water sample
V1 = Volume of water sample
M2 = Molarity of EDTA
V1 = Volume of EDTA
thereby,
M1 * 250.0 ml = 0.0054 M EDTA * 30.85 ml
M1 = ( 0.0054 M EDTA * 30.85 ml ) / 250.0 ml
Molarity of Hard metal ions is 0.00066M.
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3. This means that there are 0.00066 mol / l of Ca2+ ions in the water sample.
thus,
( 0.00066 mol / l ) * ( 40.08 + 12.01 + 16 *3)g/ mol * ( 1000)mg / g = 66.05 mg CaCO3 per litre or 66.05 ppm.
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