Adv ance Study Assignment: Determination of the Hardness of Water 1. A 0.394 6 g

Question

Adv ance Study Assignment: Determination of the Hardness of Water 1. A 0.394 6 g sample of CaCO, is dissolved in 12 M HCI and the resulting solution is diluted to 250.0 ml. in a volumetric flask How many moles of CaCO, are used (formula mass = 100.1)? a. 0003942 moles b. What is the molarity of the Ca" in the 250 mL of solution? 001576 M c. How many moles of Ca" are in a 25.00-mL aliquot of the solution in lb? 0.000394 moles 2. 25.00-mL aliquots of the solution from Problem 1 are titrated with EDTA to the Eriochrome Black T end point. A blank containing a small measured amount of Mg requires 2.77 mL. of the EDTA to reach the end point. An aliquot to which the same amount of Mg is added requires 39.04 mL of the EDTA to reach the end point. a. How many milliliters of EDTA are needed to titrate the Ca ion in the aliquot? 6.27 mL b. How many moles of EDTA are there in the volume obtained in Part a? 0-0003167 moles c. What is the molarity of the EDTA solution? O.ol094M (continued on following page)

Explanation / Answer

a. Volume of EDTA required is 31.84 mL to titrate 100ml sample.

b. we need to get the moles of EDTA in 31.84ml

in 1000ml =0.011 moles

c. 31.684ml = 31.84/1000 *0.011 moles = 3.534 *10^-4

reacting ratio of EDTA AND Ca2+ = 1:1

moles of caco3 that reacted = 3.534 *10^-4 moles

d. this moles are in 100ml of water

in 1000ml moles = 1000/100 * 3.534 *10^-4 moles = 3.534 *10^-3 moles

there are 3.534 *10^-3 moles of CaCO3 per litre of water

mass in grams = moles * molar mass

mass in grams = 3.534 *10^-3 * 100g/ mole

=0.3534 gram of CaCO3 per litre of water

e. ppm = mg/litre

0.3534*1000mg = 353.4 mg/liter = 353.4 ppm

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