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
You May Also Find These Documents Helpful
-
3. Molarity of HCl = 5.26 x 10-3 mol / 0.010 L = 0.526 M…
- 561 Words
- 3 Pages
Satisfactory Essays -
4. A 10.00 mL solution is diluted to 680.0 mL. Calculate the original concentration if it now has [ ] = 0.055 M?…
- 907 Words
- 5 Pages
Satisfactory Essays -
6. What is the concentration, in m/v %, of a solution prepared from 50. g NaCl and 2.5 L…
- 1080 Words
- 5 Pages
Good Essays -
Trial 3: .1010 M of NaOH = moles of NaOH / 0.0158 L = 0.0015958 mols of NaOH…
- 304 Words
- 2 Pages
Satisfactory Essays -
Given the mass of NaCl we can easily calculate moles using the fence method. First find the mass of one mole of NaCl, which is equal to 58.44 g. Then take the amount of moles you have been given, 1.7, and divide that by 58.44 g. The total number of moles in this problem is 0.03 moles. Given the mL of solution we can easily calculate liters, again using the fence method. Take the amount of mL given and use a conversion factor of 1 mL/1000 L to calculate that there is 0.75 liters in the final solution. To find Molarity you divide 0.03 moles/ 0.75 liters to end up with 0.04 Molarity.…
- 1243 Words
- 5 Pages
Satisfactory Essays -
- to find the molar mass of a compound add up all individual compounds. Ex. Na2 Cl2 = Na = 2 × 22.99 g/mol…
- 1725 Words
- 7 Pages
Powerful Essays -
| How many grams of CaCl2 (molar mass = 111.0 g/mol) are needed to prepare 4.44 L of 0.500 M CaCl2 solution?…
- 1884 Words
- 8 Pages
Satisfactory Essays -
Abstract: The purpose of this experiment is to find out the structure of the compound of alum through two tests. By determining the melting point of alum, and by determining the water of hydration in alum crystals by comparing the molar ratio between moles of AlK(SO4)2 and moles of H2O, the structure of the compound could be found. When a compound is created, it is essential to confirm whether or not the right compound has formed. There are many ways of determining to verify the identity of a compound. In this lab, two tests are used to determine whether or not the clear crystal is indeed, aluminum potassium sulfate (alum). The results are shown to support the literature value of the melting point of alum. The average experimental value of the melting point of aluminum potassium sulfate (alum) is 91.2 oC, approximately 1.8% error. In procedure #2, the ratio between alum and water is determined by dehydrating the water molecules from the AlK(SO4)2.12H2O. The experimental value of the ratio between alum and water is approx 10.4:12, so 10:12, approximately 13.3%…
- 277 Words
- 2 Pages
Satisfactory Essays -
0.00075 moles of metal * (32.192 g/ 1 mol) * (1000 mg/ 1 g) = 24.144 mg…
- 1876 Words
- 8 Pages
Better Essays -
The reagents are to be in a 1:1 molar ratio. The limiting reagent is 2-methylcyclohexanol because it required the least mol amount to complete the reaction.…
- 896 Words
- 4 Pages
Powerful Essays -
8. What is the molarity of a solution that contains 14.92 grams magnesium oxalate in 3.65 ml of solution?…
- 448 Words
- 2 Pages
Satisfactory Essays -
of the compound. From those masses one finds the number of moles of each element present. The…
- 498 Words
- 2 Pages
Good Essays -
A. Calculate the initial and final concentrations as needed to complete Tables 1 and 2.…
- 459 Words
- 2 Pages
Satisfactory Essays -
The mass of the weighing dish was measured as 0.6 g. The mass of the filter paper was measured as 1.0 g. Initially it was calculated that 0.68 g of CaCO3 was needed for a full reaction. The net mass of Na2CO3 (reactant) was 0.72 g and the net mass of CaCO3 was 0.7 g. The dried calcium carbonate measured at 0.7 g (net mass).…
- 578 Words
- 3 Pages
Good Essays -
The precipitated calcium carbonate is then filtered, dried, and weighed. The moles of calcium carbonate, CaCO3, are equal to the moles of Group 1 metal carbonate, M2CO3, added to the original solution. Dividing the mass of the unknown carbonate by the moles of calcium carbonate yields the formula weight, and thus the identity, of the Group 1 metal carbonate.…
- 1140 Words
- 5 Pages
Good Essays