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Hard Water Pre Lab Questions

By gurilfa Apr 28, 2014 391 Words
PRE-LAB QUESTIONS:

1) What cations are responsible for water hardness?

Ca2+ and Mg2+ are responsible for water hardness

2) Calculate the mass of disodium Ethylenediaminetetraacetate required to prepare 250 mL of a 0.010 M solution.

Mass = M x L x MM  0.010 M x 0.250 L x 372.24 g/mol = 0.93g of Na2H2Y

3) What is the molar concentration of the Na2H2Y solution?

mol Ca = M x L  0.0107 M x 0.025 L = 2.65 x 10-4 mol
if 1 mol of Ca = 1 mol Na2H2Y then: M of Na2H2Y = mol / L  2.67 x 10-4 / 0.0257 = 0.0104 M Na2H2Y

4) a. Which hardening ion binds more tightly to the EBT indicator used for today’s analysis?

Mg2+ has higher lattice energy so it will form a stronger complex and bind more tightly to EBT than Ca2+

b. What is the color change at endpoint?

The color change at endpoint is sky blue/light blue.

5) A 50.0 ml water sample requires 16.33 ml of 0.0109 M Na2H2Y to reach EBT endpoint. a. Calculate the moles of hardening ions in the water sample.

1 mol of Na2H2Y = 1 mol of CaCO3
If mol of Na2H2Y = 0.0109 M x 0.01633 L = 1.78 x 10-4 mol then there are 1.78 x 10-4 mol of CaCO3

b. Express the hardness concentration in mg CaCO3 / L sample.

(1.78 x 10-4mol x 100.1 x 1000) / (50ml / 1000) = 17.8 / 0.5  356 mg CaCO3 / L

c. What is the hardness concentration express in ppm CaCO3?

The ratio is 1mg CaCO3 / L = 1 ppm  356.3 ppm of CaCO3

d. Classify the hardness of this water according to table 9.1

The sample is considered to be “VERY HARD WATER”

6) a. Determine the number of moles of hardening ions present in a 100 mL volume sample that has a hardness of 58 ppm CaCO3.

58mg x 1g x 1 mol / L x 1000mg x 100.1g = 5.8 x 10-4 mol / L

convert to mol: 5.8 x 10-4 mol / L x 1 L x 100ml / 1000 ml  5.8 x 10-5 mol of hardening ions b. What volume of 0.100 M Na2H2Y is needed to reach the EBT endpoint for the analysis of the solution?

If ratio is 1 mol CaCO3 = 1 mol Na2H2Y and M = mol / L then: L = mol / M

5.8 x 10 -5 mol / 0.100 M = 5.8 x 10-4 L x 1000 ml / 1 L  0.58 ml

c. Express the hardness of this slightly hard water sample in grains / gallon.

Water hardness is 58 ppm and if 1 grain / gallon = 17.1 ppm then:
58ppm x 1 g/g / 17.1 ppm  3.39 grains / gallon

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