Question

CALCIUM HARDNESS DETERMINATION

Fill a 50 mL burette with 0.01 M EDTA solution, making sure the tip is full and free of air bubbles. Set Start volume.

Add 50.00 mL of an unknown hard water solution into a 100 mL beaker.

Obtain Magnetic Stirrer from Equipment menu. Place beaker on stirrer. Increase Stir from the context menu.

Add 4 mL of 1.0 M Sodium Hydroxide.

Add 100 mg of Murexide indicator.

Titrate with the 0.01 M EDTA until the color changes from salmon pink to orchid purple. Record all pertinent data on data sheet. Read burette to +/- 0.10 mL.

Repeat the titration until the final volumes agree to +/- 0.20 mL.

Calcium Hardness determination: unknown #1

Trial	Color of Beaker after Indicator (2 drops)	Color after Titration	Final volume after Titration	Initial Buret Reading	Total Volume Titrated (Initial - Final)
Trial #1	Salmon Pink	Orchid Purple	49.50 mL	50.00 mL	0.50 mL
Trial #2	Salmon Pink	Orchid Purple	49.00 mL	50.00 mL	1.00 mL
Trial #3	Salmon Pink	Orchid Purple	49.20 mL	50.00 mL	0.80 mL

Calcium hardness determination: unknown #2

Trial	Color of Beaker after Indicator (2 drops)	Color after Titration	Final volume after Titration	Initial Buret Reading	Total Volume Titrated (Initial - Final)
Trial #1	Salmon Pink	Orchid Purple	49.60 mL	50.00 mL	0.40 mL
Trial #2	Salmon Pink	Orchid Purple	49.80 mL	50.00 mL	0.20 mL
Trial #3	Salmon Pink	Orchid Purple	49.80 mL	50.00 mL	0.20 mL

TOTAL HARDNESS DETERMINATION

Fill a 50 mL burette with O.1 M EDTA solution, making sure the tip is full and free of air bubbles. Set Start volume.

Add 50.00 mL of an unknown hard water solution into a 100 mL beaker.

Obtain Magnetic Stirrer from Equipment menu. Place beaker on stirrer. Increase Stir from the context menu.

Add 10 mL of Ammonia to the beaker.

Add 100 mg of Eriochrome Black T indicator.

4. Titrate with the 0.01 M EDTA until the color changes from wine red to pure blue. Record all pertinent data on data sheet. Read burette to +/- 0.10 mL.

Repeat the titration until the final volumes agree to +/- 0.20 mL.

Total hardness determination: unknown #1

Trial	Color of Beaker after Indicator (2 drops)	Color after Titration	Final volume after Titration	Initial Buret Reading	Total Volume Titrated (Initial - Final)
Trial #1	Wine Red	Pure Blue	49.00 mL	50.00 mL	1.00 mL
Trial #2	Wine Red	Pure Blue	49.30 mL	50.00 mL	0.70 mL
Trial #3	Wine Red	Pure Blue	49.60 mL	50.00 mL	0.40 mL

Total hardness determination: unknown #2

Trial	Color of Beaker after Indicator (2 drops)	Color after Titration	Final volume after Titration	Initial Buret Reading	Total Volume Titrated (Initial - Final)
Trial #1	Wine Red	Pure Blue	49.50 mL	50.00 mL	0.50 mL
Trial #2	Wine Red	Pure Blue	49.80 mL	50.00 mL	0.20 mL
Trial #3	Wine Red	Pure Blue	49.40 mL	50.00 mL	0.60 mL

Conclusions

1. According to your observations, in the calcium hardness determination part of the experiment, did your buret values agree with one another? If not, how would you explain the difference?

2. According to your observations, in the total hardness determination part of the experiment, did your buret values agree with one another? If not, how would you explain the difference?

3. For the calcium hardness determination part, determine the Molarity of Calcium (Ca⁺²) in both unknowns. Show your calculations.

Unknown #1: ____________

Unknown #2: ____________

4. After completing Question #3, determine the ppm of Calcium (Ca⁺²) in both unknowns. Show your calculations.

Unknown #1: ____________

Unknown #2: ____________

5. Given the total volume titrated from the calcium hardness experiment and the total volume titrated from the total hardness experiment, calculate the volume titrated for Magnesium (Mg⁺¹) and the ppm of Magnesium (Mg⁺¹). Calculate these for both unknowns. Show your calculations.

Unknown #1

Total Volume Titrated: _______________

Magnesium ppm: _______________

Unknown #2

Total Volume Titrated: _______________

Magnesium ppm: _______________

Answer 1

1.

For unknown sample #1 the buret values don’t agree with one another, because it was established that the final volumes (volume titrated) must not differ in +/- 0.20 mL, which was not accomplished between Trial #1 and #3, because they differ in 0.5 mL. For unknown Sample #2 the buret values do agree with one another because the final volumes differ in less than +/- 0.20 mL.

#1 the buret values don’t agree with one another, because it was established that the final volumes (volume titrated) must not differ in +/- 0.20 mL, which was not accomplished between Trial #1 and #3, because they differ in 0.5 mL. For unknown Sample #2 the buret values do agree with one another because the final volumes differ in less than +/- 0.20 mL.

2.

The buret values don’t agree with one another in any of the unknown samples, because it was established that the final volumes (volume titrated) must not differ in +/- 0.20 mL, which was not accomplished among any Trials in unknown sample #1 and only between trial #1 and #3 in unknown sample #2.

3.

unknown #1: Average Total Volume Titrated = (1.00 mL + 0.80 mL)/2 = 0.90 mL

(Note: you should consider that values that don’t differ in more that +/- 0.20 mL).

1000 mL of EDTA solution_____________ 0.01 moles

0.90 mL of EDTA solution____________= ?

0.90 x 0.01/ 1000 = 9 . 10^-6 moles of EDTA = moles of Ca²⁺ in the unknown solution, (because the reaction is 1:1 stoichiometry), which are contained in 50 mL:

50 mL______________9 . 10^-6 moles of Ca²⁺

1000 mL___________= ?

1000 mL x 9 x10^-6/50 = 1.8 .10^-4 M = [Ca⁺⁺] in unknown sample #1

unknown #2:

Average Total Volume Titrated = (0.20 mL + 0.20 mL + 0.40 mL)/3 = 0.26 mL

(Note: you should consider that values that don’t differ in more that +/- 0.20 mL).

1000 mL of EDTA solution_____________ 0.01 moles

0.26 mL of EDTA solution____________= ?

0.26 x 0.01/ 1000 = 2.66 x 10^-6 moles of EDTA = moles of Ca²⁺ in the unknown solution, (because the reaction is 1:1 stoichiometry), which are contained in 50 mL:

50 mL______________2.66 x 10^-6 moles of Ca²⁺

1000 mL___________= ?

1000 mL x   9x10^-6 / 50 = 5.33 x10^-5 M = [Ca⁺⁺] in unknown sample #2

4. 1 ppm = 1 g X/ 1 x 10⁶ g solution

When need to calculate the grams of Ca2+ in each unknown solution.

[Ca⁺⁺] in unknown sample #1= 1.8 x10^-4 M

1 mol of Ca²⁺______________40 g (because this is the atomic weight of Calcium)

1.8 x10^-4 moles of Ca²⁺__________=?

(1.8 x10^-4 x 40 /1 = 0.0072 g of Ca²⁺ are in 1000 mL (because is molarity)

Now we need to calculate the amount in 1x10⁶ mL of solution, to convert in ppm:

1000 mL _______________ 0.0072 g of Ca²⁺

1x10⁶ mL_____________=?

(1x10⁶ x 0.0072 /1000 = 7.2 g in 1x10⁶ mL of solution = 7.2 ppm = [Ca²⁺] in unknown #1

Same for unkown #2:

[Ca⁺⁺] in unknown sample #2= 5.33 x10^-5 M

1 mol of Ca²⁺______________40 g (because this is the atomic weight of Calcium)

5.33 x10^-5 moles of Ca²⁺__________=?

(5.33 x10^-5 x 40 /1 = 0.0021 g of Ca²⁺ are in 1000 mL (because is molarity)

Now we need to calculate the amount in 1x10⁶ mL of solution, to convert in ppm:

1000 mL _______________ 0.0021 g of Ca²⁺

1x10⁶ mL_____________=?

(1x10⁶ x 0.0021 /1000 = 2.132 g in 1x10⁶ mL of solution = 2.132 ppm = [Ca²⁺] in unknown #2

5.

Unknown #1

Total Volume Titrated:

volume titrated for Magnesium (Mg⁺¹)=

total volume titrated from the calcium hardness experiment

-

total volume titrated from the total hardness experiment

Average Total Volume Titrated = (1.0 mL + 0.8 mL)/2 = 0.9 mL

Average total volume titrated from the total hardness experiment = (0.4 + 0.7)/2= 0.55 mL

volume titrated for Magnesium (Mg⁺¹)= 0.9 – 0.55 = 0.25 ml

1000 mL of EDTA solution_____________ 0.01 moles

0.25 mL of EDTA solution____________= ?

0.25 x 0.01/ 1000 = 2.5 x 10^-6 moles of EDTA = moles of Mg⁺ in the unknown solution, (because the reaction is 1:1 stoichiometry), which are contained in 50 mL:

50 mL______________2.5x 10^-6 moles of Mg⁺

1000 mL___________= ?

1000 mL x 5.5 x10^-6/50 = 5 x 10^-5 M = [Mg⁺] in unknown sample #1

1 mol of Mg⁺____________24 g (because this is the atomic weight of Magnesium)

5 x10^-5 moles of Mg⁺__________=?

(5 x10^-5 x 24 /1 = 0.0012 g of Mg⁺ are in 1000 mL (because is molarity)

Now we need to calculate the amount in 1x10⁶ mL of solution, to convert in ppm:

1000 mL _______________ 0.0012 g of Mg⁺

1x10⁶ mL_____________=?

(1x10⁶ x 0.0012 /1000 = 1.2 g in 1x10⁶ mL of solution = 7.2 ppm = [Mg⁺] in unknown #1

Unknown #2

Total Volume Titrated:

volume titrated for Magnesium (Mg⁺¹)=

total volume titrated from the calcium hardness experiment

-

total volume titrated from the total hardness experiment

Average Total Volume Titrated = (0.2+0.2 + 0.4)/2= 0.26 mL

Average total volume titrated from the total hardness experiment = (0.5 + 0.6)/2 = 0.55 mL

volume titrated for Magnesium (Mg⁺¹)= 0.55 – 0.26 = 0.29 ml

1000 mL of EDTA solution_____________ 0.01 moles

0.29 mL of EDTA solution____________= ?

0.29 x 0.01/ 1000 = 2.9 x 10^-6 moles of EDTA = moles of Mg⁺ in the unknown solution, (because the reaction is 1:1 stoichiometry), which are contained in 50 mL:

50 mL______________2.9x 10^-6 moles of Mg⁺

1000 mL___________= ?

1000 mL x 2.9 x10^-6/50 = 4.58 x 10^-4 M = [Mg⁺] in unknown sample #2

1 mol of Mg⁺____________24 g (because this is the atomic weight of Magnesium)

4.58 x10^-4 moles of Mg⁺__________=?

(4.58 x10^-4 x 24 /1 = 0.011 g of Mg⁺ are in 1000 mL (because is molarity)

Now we need to calculate the amount in 1x10⁶ mL of solution, to convert in ppm:

1000 mL _______________ 0.011 g of Mg⁺

1x10⁶ mL_____________=?

(1x10⁶ x 0.011 /1000 = 11 g in 1x10⁶ mL of solution = 11 ppm = [Mg⁺] in unknown #2