Part A) An unknown water sample is tested to determine its hardness as measured

Question

Part A) An unknown water sample is tested to determine its hardness as measured in ppm of calcium carbonate.  It is found that in a titration with the disodium salt of EDTA, 45.8 mL of a 0.00105 M EDTA solution is required to bind all of the calcium and magnesium ions in 25.0 mL of the water sample.

How many moles of EDTA were used in the titration?

Part B) If all of the moles (as determined in the previous problem) of calcium and magnesium were moles of calcium carbonate, what mass of calcium carbonate is in the 25.0 mL sample of water?

Part C) Assuming the density of water is 1.00 g/mL, calculate the ppm of calcium carbonate in the water sample.

Part D) Based on your result, would the sample be classified as soft, medium, or hard water?

Explanation / Answer

A)

we know that

moles = molarity x volume (L)

so

moles of EDTA used = 0.00105 x 45.8 x 10-3

moles of EDTA used = 4.809 x 10-5

so

4.809 x 10-5 moles of EDTA were used in the titration

B)

in the titration

moles of EDTA used = moles of CaC03 present

so

moles of CaC03 = 4.809 x 10-5

now

we know that

mass = moles x molar mass

so

mass of CaC03 = 4.809 x 10-5 x 100

mass of CaC03= 4.809 x 10-3 g

C)

now

mass of CaC03 in mg / volume in L = 4.809 / 25 x 10-3

= 192.36

so

ppm of CaC03 is 192.36 mg/L

D)

SO the water is very hard

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