Determination of Water Hardness
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Texas A&M University *
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Course
1411
Subject
Chemistry
Date
Apr 3, 2024
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Pages
2
Uploaded by UltraMorning901
Determination of Water Hardness
Purpose:
Define soft and hard water in order to determine how CaCO3 affects water
hardness and how to measure it.
Procedure:
I was able to measure the water hardness by performing an EDTA/EBT
titration on a local water supply. The concentration of Ca2+ ions in the local water
supply sample was then estimated.
Data table 1: EDTA Titration Volume
Initial EDTA
Volume (mL)
Final EDTA
Volume (mL)
Total volume of
EDTA use (mL)
Trial 1
10 mL
7.5mL
2.5mL
Trial 2
11 mL
6.5mL
4.5mL
Trial 3
13mL
10mL
3mL
Average Volume
of EDTA Used
(mL):3.33m
Data Table 2: Water Hardness
Average Volume of EDTA
used (mL)
Concentration of CA2+
ions per liter of water
(mol/L)
Water Hardness (ppm
CaCO3)
3.33mL
0.00333 mol/L
333.2 ppm
Observation:
I observed that each water had a different volume. Different waters were soft and hard
Questions:
A. Because our calculations came out to 333.2ppm for the water hardness I would
classify the water as very hard. Anything above 180 ppm is very hard water.
B. 8oz x 8 oz =64 oz x 29.57mL= 1892mL=1.9Lx333.2mg/L=632.7mg
C. 1Lx333mg/L=333mgCa/1150mgCa=.289 x 100=28.9%
Conclusion:
can now recognize the difference between soft and hard water and
understand that the minerals in a water supply determine whether it is hard or soft. I'm
also more acquainted with titrator and EDTA/EBT solutions.
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Related Questions
A 25.00 mL unknown water sample is titrated with a standardized 0.0140 M EDTA solution. It is determined that 8.54 mL of EDTA is required to reach the end point. The blank titre was determined to be 1.50 mL. Determine the concentration of Calcium in the water sample in ppm. (mwt. of CaCO3 = 100.0892 g/mol).
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Determining the Ksp of Calcium Hydroxide:
Ca(OH)2 (s) ↔ Ca2+ (aq) + 2OH– (aq)
Ksp = [Ca2+][OH–]2
The molarity of the hydrochloric acid used was 0.050 M.15.00 mL of filtered Ca(OH) 2 was measured out for the titration.Calculate the [Ca 2+ ]. Explain, please.
Trial
Initial Buret Reading
Endpoint Buret Reading
Endpoint Volume
Equivalence Point Volume
1
80 ml
9.31 pH
12 ml
12.60ml
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A 15.00 mL sample of hard water was titrated with a 0.0150 M EDTA solution. The titration required 26.72 mL of EDTA to reach the end point. What is the total hardness concentration of the water sample in ppm?
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5. A water supply company, One Team, found out that their water supply had been
contaminated with a sparingly soluble magnesium salt. They requested the help of
Magent Analytical Laboratory to identify the magnesium salt. They chose to identify this
using Ksp determination after narrowing it down to 2 salts:
Salt
MgCO3
Mg(OH)2
Theoretical Ksp
3.5 x 10-8
1.8 x 10-11
For the standardization of the HCl titrant, 5.230 mg Na2CO3 (98.9% purity, MW = 105.99
g/mol) required 24.4 mL of the titrant to reach the phenolphthalein endpoint.
After isolating and precipitating the magnesium salt A, a portion was dissolved in distilled
water in order to make a 250.0-mL saturated solution. The titration of a 50.0-mL aliquot
of the saturated solution required 8.20 mL of HCl titrant to reach the phenolphthalein
endpoint.
Meanwhile, dissolving this magnesium salt in 250.0 mL of 0.0200 M MgCl2 required 1.51
mL of the same HCl titrant to titrate a 100.0-mL solution.
a. What is the molarity of the HCl titrant?…
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UPLE
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Procedure and Parameters
Initial volume of 0.100 M acetic acid added
Volume of distilled water used
Conductivity of distilled water
Conductivity of the prepared acetic acid
Molar conductivity of acetic acid at infinite
dilution
Results
5.00 mL
50.00 mL
8 μµS.cm-¹
165 µS.cm-¹
3.955 x 105 μS.cm-¹-M-¹
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Prelim Rxn:
NH3 +
H3BO3 -->
NH4: H2BO3
Titration Rxn:
NH : H2BO3 + HCI --> H3BO3
CI-
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If a solution of dissolved
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[]-base to be 10mm, what is
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pH =
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Concentration sodium thiosulfate solution used:
0.056 mol L-1
ACCURATE TITRATION VOLUME OF SODIUM THIOSULFATE SOLUTION USED
Volume titration 1, (ml)
Volume titration 2, (ml)
Your Group
15.80
15.75
Group 1
16.05
15.70
Group 2
16.30
16.25
ANALYSIS
Average volume used, (mL)
Standard deviation
Number of moles sodium thiosulfate used, (mol)
Number of moles calcium iodate in solution, (mol)
Volume of calcium iodate used, (mL)
Concentration of saturated calcium iodate solution (mol L-¹)
Unrounded value
15.9750
0.26220
Rounded value
15.97
0.262
10.0
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10.0
pH 8.15, V= 8.6 mL
8.0
pH
6.0
pH= 3.88
D%3D4,3ml.
4.0
2.0
2.0
4.0
6.0
8.0
10.0
Volume (mL)
The figure shows the data obtained in the pH titration of a biochemical substance that is a weak acid. Determine the pK, of this compound.
exact number, no tolerance
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Preparation and Standardization of KMnO4 solution
Experimental data
Complete the table below.
Trial 1
0.2001 g
Trial 2
0.2065 g
Trial 3
Weight of sodium oxalate
(Na2C2O4, MM= 134 g/mol)
Titration data
0.2050 g
Final reading
Initial reading
29.86 mL
0.00 mL
30.66 mL
30.52 mL
0.00 mL
0.00 mL
Total vol. of KMNO4 used
Computed Molarity of KMNO4
solution
Mean Molarity
Computed Normality of KMNO4
Mean Normality of KMNO4
solution
Reaction Involved:
Calculations:
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If you could help with the below pre-lab questions please.
PRELABORATORY ASSIGNMENT
Define the term "salt.'
The solubility of AgCl(s) in water is 1.3 x 10-5 M. Calculate its Ksp.
The solubility of CsF2(s) in water is 2.1 x 10-4 M. Calculate its Ksp.
The of Pb(OH)2 is 4 x 10-15. Calculate the molar solubility of this compound.
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Mass of KHP + Vial
Mass of 'Empty' Vial
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Reading
Initial Burette
Reading
Determination #1 Determination #2 Determination #3
14.1992 g
14.0219 g
14.0982 g
13.3149 g
13.2301 g
13.2779 g
39.21 mL
0.49 mL
35.46 mL
0.86 mL
37.42 mL
1.59 mL
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the correct number of significant figures and only report the numerical value (no
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Redox Titration
2 KMnO4 + 5 (COOH)2 + 3 H2SO4 -------> 10 CO2 + 2 MnSO4 + K2SO4 + 8 H2O
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Data for titration of 15.00 mL of vinegar with approximately 1.0 M NaOH. Note you must use the exact
concentration of the standard NaOH solution to calculate the moles of NaOH.
Quantity
Еxample
Trial 1
Trial 2
Trial 3
M NaOH (exact
concentration)
0.992 M
0.953 M
0.953 M
0.953 M
V initial buret
reading = V,
0.20 mL
1.20 mL
0.52 mL
0.15 mL
V final buret
reading = V,
12.90 mL
14.22 mL
13.71 mL
13.31 mL
%3D
Vep = VNAOH
added = V; - V,
12.70 mL
13.02 mL
13.19 mL
13.16 mL
Vep = VNAOH in L
0.01270 L
0.01302 L
0.01319 L
0.01316 L
moles NaOH =
0.0126 mol
0.0124 mol
0.0126 mol
0.0125 mol
MNAOH X VNAOH
moles AA =
0.0126 mol
0.0124 mol
0.0126 mol
0.0125 mol
moles NaOH
V sample = V
acid
0.0150 L
0.0150 L
0.0150 L
0.0150 L
(15.00 mL)
(15.00 mL)
(15.00 mL)
Actual molarity
0.840 M
0.827 M
0.840 M
0.833 M
of AA
Average molarity for 3 trials =>
of AA
0.833 M
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A. How many mol of EDTA reacted with BaSO₄?
B. How many grams of sulfate is present in the sample?
C. Which of the following chemical reaction does not happen in the analysis?
a. EDTA + Mg²⁺ ⇌ Mg-EDTA
b. SO₄²⁻ + Ba(NO₃)₂ ⇌ BaSO₄ + 2 NO₃⁻
c. BaSO₄ + EDTA ⇌ Ba²⁺ + SO₄-EDTA
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adding titrant past the color change of the analyte solutionrecording the molarity of titrant as 0.1 M rather than its actual value of 0.01 Mspilling some analyte out of the flask during the titrationstarting the titration with air bubbles in the buretfilling the buret above the 0.0 mL volume mark
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Ca(OH)2 (s) ↔ Ca2+ (aq) + 2OH– (aq)
Ksp = [Ca2+][OH–]2
The molarity of the hydrochloric acid used was 0.050 M. 15.00 mLof filtered Ca(OH) 2 was measured out for the titration.Calculate the Ksp for calcium hydroxide. Explain your calculations
Trial
Initial Buret Reading
Endpoint Buret Reading
Endpoint Volume
Equivalence Point Volume
1
80ml
9.31 pH
12ml
12.60ml
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You collected the following data from a titration experiment using a 0.123M standardized NaOH solution to titrate a 27.36 mL solution with an unknown Molarity concentration (M) of
sulfuric acid (H2SO4).
Initial
Final
Vol
Delivered
|(mL)
Burette
Burette
Reading
(mL)
Reading
(mL)
Trial 1
0.56
19.20
??
Trial 2
0.57
19.7
Trial 3
0.92
19.56
For just Trial 1, determine the amount of NaOH delivered for the titration with appropriate significant digits. Do not include units.
Answer:
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Calculate the number of moles of EDTA required to titrate the water sample. (enter your answer with 3 significant figures)
The total hardness is due to one or a combination of Ca2+, Mg2+, and Fe2+ in your sample. It is convenient to express this hardness as though it was entirely due to Ca2+. Making this assumption, determine the number of moles of Ca2+ present in the bottled water sample titrated. (enter your answer with 3 significant figures)
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A student performed three KHP titrations in Part 1 in order to standardize their
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Mass of KHP + Vial
Mass of 'Empty' Vial
Final Burette
Reading
Initial Burette
Reading
Determination #1
16.7538 g
16.2649 g
21.67 mL
0.40 mL
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14.1709 g
13.5560 g
26.98 mL
0.31 mL
Determination #3
15.6625 g
15.0833 g
26.04 mL
0.93 mL
How many moles of KHC8H4O4 did they use in their second titration? Report your
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