<Water hardness lab report>Q) My calculation is right?1. I mixedhard water or tap water or soft water 5ml + pH buffer 1ml + 3 drops of indicator + EDTA xml2. Given formula is (_ml EDTA / 1) x (1L / 1000ml) x (0.01 mol EDTA / 1L) x (1mol CaCO3 / 1 mol EDTA) = _ mol CaCO3(_mol CaCO3 / _ml Water sample) x (100.1g CaCO3 / 1mol CaCO3) x (1000mg / 1g ) x (1000ml / 1L) = _mg/L CaCO3_mg/L CaCO3 = _ppm CaCO33. My data is1) hard water Hard Water Trial 1Hard Water Trial 2Hard Water Trial 3Initial Syringe Reading1.0ml1.0ml1.0mlFinal Syringe Reading0.88ml0.84ml0.85mlVolume of EDTA Consumed0.12ml0.16ml0.15mlWater Hardnessppm CaCO3 Average ppm 2) Tap water Tap Water Trial 1Tap Water Trial 2Tap Water Trial 3Initial Syringe Reading1.0ml1.0ml1.0mlFinal Syringe Reading0.65ml0.68ml0.67mlVolume of EDTA Consumed0.35ml0.32ml0.33mlWater Hardnessppm CaCO3 Average ppm 3) Soft water Soft Water Trial 1Soft Water Trial 2Soft Water Trial 3Initial Syringe Reading 1.0ml1.0ml1.0mlFinal Syringe Reading 0.97ml0.95ml0.91mlVolume of EDTA Consumed 0.03ml0.05ml0.09mlWater Hardnessppm CaCO3 Average ppm 4. My result is1) Hard watertrial 1: 24.024ppm CaCO3trial 2: 32.032ppm CaCO3trial 3: 30.03ppm CaCO3Ave : 28.70Ppm CaCO32) Tap watertrial 1: 70.07ppm CaCO3trial 2: 64.064ppm CaCO3Trial 3: 66.066ppm CaCO3ave: 66.73ppmCaCO33) Soft watertrial 1: 6.006ppm CaCO3trial 2:10.01ppm CaCO3trial 3: 18.018ppm CaCO3ave: 11.3653ppmCaCO3 Thank yo. Equation 6:_mL EDTA1L0.01 mol EDTA1 mol CaCO,_mol CaCO,11,000 mL1 L1 mol EDTAmol CaCO,_mL Water Sample100.1 g CaCO,1 mol CaCO,1,000 mg1 g1,000 mL_mg/L CaCO,%D1L_mg/L CaCO,=_ppm CaCO,continued on next

Hard water refers to the water sample that contains soluble salts of calcium and magnesium. It does…

Answered: I mixed hard water or tap water or…

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Related questions

Question

Q) My calculation is right?

1. I mixed

hard water or tap water or soft water 5ml + pH buffer 1ml + 3 drops of indicator + EDTA xml

2. Given formula is

(_ml EDTA / 1) x (1L / 1000ml) x (0.01 mol EDTA / 1L) x (1mol CaCO3 / 1 mol EDTA) = _ mol CaCO3

(_mol CaCO3 / _ml Water sample) x (100.1g CaCO3 / 1mol CaCO3) x (1000mg / 1g ) x (1000ml / 1L) = _mg/L CaCO3

_mg/L CaCO3 = _ppm CaCO3

3. My data is

1) hard water

	Hard Water Trial 1	Hard Water Trial 2	Hard Water Trial 3
Initial Syringe Reading	1.0ml	1.0ml	1.0ml
Final Syringe Reading	0.88ml	0.84ml	0.85ml
Volume of EDTA Consumed	0.12ml	0.16ml	0.15ml
Water Hardness ppm CaCO₃
Average ppm

2) Tap water

	Tap Water Trial 1	Tap Water Trial 2	Tap Water Trial 3
Initial Syringe Reading	1.0ml	1.0ml	1.0ml
Final Syringe Reading	0.65ml	0.68ml	0.67ml
Volume of EDTA Consumed	0.35ml	0.32ml	0.33ml
Water Hardness ppm CaCO₃
Average ppm

3) Soft water

	Soft Water Trial 1	Soft Water Trial 2	Soft Water Trial 3
Initial Syringe Reading	1.0ml	1.0ml	1.0ml
Final Syringe Reading	0.97ml	0.95ml	0.91ml
Volume of EDTA Consumed	0.03ml	0.05ml	0.09ml
Water Hardness ppm CaCO₃
Average ppm

4. My result is

1) Hard water

trial 1: 24.024ppm CaCO3

trial 2: 32.032ppm CaCO3

trial 3: 30.03ppm CaCO3

Ave : 28.70Ppm CaCO3

2) Tap water

trial 1: 70.07ppm CaCO3

trial 2: 64.064ppm CaCO3

Trial 3: 66.066ppm CaCO3

ave: 66.73ppmCaCO3

3) Soft water

trial 1: 6.006ppm CaCO3

trial 2:10.01ppm CaCO3

trial 3: 18.018ppm CaCO3

ave: 11.3653ppmCaCO3

Thank yo.

Equation 6:
_mL EDTA
1L
0.01 mol EDTA
1 mol CaCO,
_mol CaCO,
1
1,000 mL
1 L
1 mol EDTA
mol CaCO,
_mL Water Sample
100.1 g CaCO,
1 mol CaCO,
1,000 mg
1 g
1,000 mL
_mg/L CaCO,
%D
1L
_mg/L CaCO,=_ppm CaCO,
continued on next

Expert Solution

Trending now

This is a popular solution!

Step by step

Solved in 5 steps

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

Similar questions