Chem1011GeneralChemistryII_TitrationforAceticAcidinVinegar_134692
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Date
Feb 20, 2024
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Student Name
Student ID
Lesson
Titration for Acetic Acid in Vinegar
Institution
University of New England
Session
Fall 2023
Course
Chem 1011 - General Chemistry II
Instructor
David Henderson
Final Report
Test Your Knowledge
Chem 1011 - General Chemistry II
Titration for Acetic Acid in Vinegar
Completely ionizes in water
g/mol
(Grams of solute/grams of
solution) x 100%
Moles of analyte = moles of
titrant
Known concentration in a
titration
Moles of solute/L of solution
Only partially ionizes in water
Unknown concentration in a
titration
A quantitative, volumetric
technique used to determine
the concentration of an
unknown analyte
Match each term with the appropriate definition.
Strong acid
Molar mass
Mass percent
Equivalence point
Titrant
Molarity
Weak acid
Analyte
Titration
1
2
3
4
5
6
7
8
9
Copyright 2024 - Science Interactive | https://scienceinteractive.com
Exploration
No Color Change
Turns Pink
Determine whether each chemical substance would remain the same
color or turn pink in the presence of phenolphthalein.
1
Tap Water
pH = 7
Battery Acid
pH = 1
Mashed Avocados
pH = 6.5
Lime Juice
pH = 2
2
Ammonia
pH = 11
Sea Water
pH = 8.5
CH
COOH(aq) +
→ H
O(l) +
Complete the chemical equation.
3
NaOH(aq)
2
CH COONa(aq)
3
A weak acid completely ionizes in water to produce both a conjugate base
and a H
O
ion.
3
+
True
False
A titration is useful for determining the concentration of a given
substance.
True
False
1
2
Copyright 2024 - Science Interactive | https://scienceinteractive.com
Exercise 1
During a titration, a known concentration of _____ is added to a _____ of an
unknown concentration.
analyte; titrant
titrant; analyte
The equivalence point is the moment in a titration where exactly enough
titrant has been added to completely react with the analyte.
True
False
Phenolphthalein is an effective pH indicator because equivalence points in
titrations are marked by the analyte changing in color from _____ in acidic
and neutral solutions, to _____ in basic solutions.
light yellow; purple
bright pink; colorless
colorless; bright pink
white; bright purple
What are the coefficients in the chemical equation?
1, 1, 2, 1
2, 2, 1, 3
1, 1, 1, 1
1, 3, 1, 1
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Related Questions
What is the molarity of an aqueous solution that contains 0.0720
g C2H6O2 per gram of solution . The density of the solution is 1.04 g/mL
X
STARTING AMOUNT
ADD FACTOR
ANSWER
RESET
*( )
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What is the reason for washing the precipitate with water in Step 9?
Define precipitate. Define filtrate.
In Step 2, what is the purpose for rinsing the stirring rod?
read the Procedure to answer the questions
Using a balance, mass between 1.50 – 2.00 grams of sodium carbonate in a pre-massed 150mL beaker.
Add 20 mL of distilled water and stir thoroughly to make sure all the crystals are dissolved. Rinse the stirring rod with a little distilled water after stirring.
Using a balance, mass between 1.50 – 2.00 grams of calcium chloride dihydrate in a pre-massed 50 mL beaker.
Repeat Step 2 for the solution in the 50 mL beaker.
Pour the calcium chloride solution into the 150mL beaker containing the sodium carbonate solution and stir.
Mass a piece of filter paper. Fold the filter paper and place it into the funnel. Wet it with a little distilled water to ensure that it is stuck to the sides of the funnel.
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Solid ammonium bromide is slowly added to 75.0 mL of a 0.283 M silver nitrate solution until the concentration of
bromide ion is 0.0583 M. The percent of silver ion remaining in solution is
%.
Submit Answer
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STARTING AMOUNT
X
=
0.1
A student dissolves 24.3 g Na2S₂O3 in enough water to make a 0.302
solution. How many mL solution did the student make?
ADD FACTOR
x( )
0.302
24.3
509
1
DELETE
6.022 x 10²3 0.509
L solution g Na₂S₂O3
A
158.1
ANSWER
1000
0.001
1.965
0.01
M
RESET
5
3.311
100
M Na2S₂O3 kg Na₂S₂O3 mol Na2S₂O3 mL solution
10
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16
You need to make an aqueous solution of 0.237 M cobalt(II) fluoride for an experiment in lab, using a 250 mL volumetric flask. How much solid cobalt(II)
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Review Topics]
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Quantity
Your Data
1.
Grams of vinegar sample used for your titration
25.000 g
2.
Initial Buret Reading of Sodium Hydroxide solution
10.00 mL
3.
Final Buret Reading of Sodium Hydroxide solution
28.00 mL
4.
Amount of Sodium Hydroxide Solution used to neutralize the vinegar sample
5.
Concentration of NaOH in the NaOH solution
0.050 g/mL
6.
Grams of NaOH used to neutralize the vinegar
7.
Grams of acetic neutralized by the amount of NaOH
8.
Percent acetic acid in the vinegar
Hints: For #4, The amount of NaOH used is the difference between the starting buret value and the ending buret value.
For #6, Once you calculate the amount of sodium hydroxide used, multiple that value by the concentration of NaOH in the NaOH solution.
For #7, Multiply the value obtained in number 6 by the number 1.5. Remember, we learned that
every 1 gram of NaOH neutralizes 1.5 grams of acetic acid.
For #8, Divide the grams of acetic acid by the grams of vinegar sample and multiply this value by…
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When excess water is added to a mixture of citric acid and sodium
Expe
bicarbonate present in a bath bomb, the following reaction takes place
C6H&O7(s) + 3 NaHCO3(s) + H2O(1) → Na3C6H;07(aq) + 4 H20(1) + 3
CO2(g)
Calculate the volume of CO2 gas collected over water at 38.0 °C when
a mixture of 55.5 g of citric acid and excess sodium bicarbonate is
added to a tub of water if the total pressure is 725 torr. The vapor
pressure of water at 38.0 °C is 49.7 torr.
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A student performed titration to find the concentration of acetic acid. Titrant was sodium hydroxide. He went beyond the indicator color change by
adding too much titrant. How will his experimental error affect the calculated value for the concentration of unknown acetic acid?
calculated value
actual value.
calculated value < actual value
calculated value = actual value.
more information is necessary
this is not a source of error
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1. Solubility Test
Table 1. Results for the solubility test.
OBSERVATION*
INFERENCE
SAMPLE
Ethyl acetate
**
Нехаne
Methanol
Water
Coconut oil
Vitamin E
Beef fat
Miscible (soluble) or immiscible (insoluble)
** polar or nonpolar
Sudan IV Test
Table 2. Results for Sudan IV test.
OBSERVATION*
INFERENCE
SAMPLE
Нехаne
Ethyl acetate
Methanol
Water
**
Coconut oil
Vitamin E
Beef fat
Homogeneous or heterogeneous mixture
** polar or nonpolar
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1. 500.0 mL 0.2 M KI2. 500.0 mL 0.2 M KCl3. 500.0 mL 0.1 M K2S2O84. 500.0 mL 0.1 M K2SO45. 500.0 mL 4.0 mM Na2S2O3 (from Na2S2O3∙5H2O)pls compelete the 2nd table given the data
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A 21.47-mL sample of hydrochloric acid solution requires 35.00 mL of 0.109 M sodium hydroxide for complete neutralization. what is the concentration of the original
hydrochloric acid solution?
Submit Answer
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Pre-lab question #8: Suppose you were dissolving a metal such as zinc with
hydrochloric acid. How would the particle size of the zinc affect the rate of its
dissolution? As the particle size of the zinc increases, the rate of dissolution
AA (decreases/increases).
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31
Name
Date
Lab Section
Results and Discussion - Experiment 3 - Acids and Bases: Analysis
A. Analysis of vinegar
1.
Unknown vinegar number
$1- 10716 feR=103.63
fl:112.2 f2=108.4
2. Mass of flask
%3D
3. Mass of flask + 5.00 mL of vinegar
%3D
4. Mass of 5.00 mL of vinegar
5. Density of vinegar
Oo\M
6. Molarity of sodium hydroxide solution
Titrations
flask 1
flask 2
flask 3
41.7
7. Final buret reading
8. Initial buret reading
41.7
41.7
9. Volume of NAOH soln.
10. Moles of NaOH
11. Moles of acetic acid
12. Molarity of acetic acid
13. Average molarity of acetic
acid in vinegar
14. Deviation of each
molarity from average
15. Average deviation of molarity (see Expt. 4)
16. Mass percent of acetic acid in vinegar
(show calculations below)
O 2015 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
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Question 22 of 25
Submit
What is the molarity of an aqueous solution that
contains 0.0720 g C2H,O, per gram of solution. The
density of the solution is 1.04
g/mL
STARTING AMOUNT
ADD FACTOR
ANSWER
RESET
*( )
1.21
0.0720
g C2H,O2/mol
1
1.21 x 10-3
mL solution
1.12
g C2H,O2/g solution
4650
mol C2H,O2
62.08
g solution/mL
0.001
g solution
6.022 x 1023
M C2H,O2
1000
g C2H,O2
1.04
L solution
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Determination of known acid concentration
Trial 1
Trial 2
Trial 3
Volume of acid used
(mL)
25 mL HCl
25 mL HCl
25 mL HCl
Initial buret reading
(mL)
50 mL
50 mL
50 mL
Final buret reading
(mL)
30 .7 mL
31.1 mL
29.3 mL
Volume of NaOH used
(mL)
50 mL
50 mL
50 mL
Molarity of NaOH
Calculated Molarity of HCl
Show a sample calculation of HCl molarity
Average concentration of HCl from Trials 1,2, and 3
The “true” concentration of HCl is 0.10 M. The calculated average concentration of HCl is ______________M.
The % error is = (???? − ?????)/????? ? 100
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Solubility of Ionic Compounds
Classify each of the compounds as soluble or not soluble:
soluble
silver acetate
soluble
sodium nitrate
not soluble
chromium(II) sulfate
a
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Math 122 Quiz 5.docx
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Based on the data given answer questions. 1,3, and 5 please ??
Question 2 is already answered to solved 1 and 3
1. Based on your three trials, discuss the precision of your data
2. The actual concentration of the unknown is 0.200 M. (Check with instructor to ensure that this value is accurate.) Calculate the percent error based on the average molarity of your three trials.
Answer: Actual Molarity= 0.200M
Mcalculated = 0.1755M
% error = ( [actual - calculated value]/actual ) *100
Putting values;
= ((0.200 - 0.1755)/0.200 )*100
=
12.25%
3. Based on your answer to (2), discuss the accuracy of your data.
5. 80.0 mL of 0.30 M NaOH and 80.0 mL of 0.30 M HCl are mixed together. What is the approximate pH of the solution —acidic, basic, or neutral? Explain.
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Name
Chem
Section
Data Sheet
determination
23°C
23°C
25°C
temperature of filtered KHT
solution, °C
0.04m
concentration of NaOH solution,
mol L-
0.04 m
0.04m
KHT solution
14.9
49-5
final buret reading, mL
initial buret reading, mL
49.9
volume of KHT
used, mL
NaOH solution
28
final buret reading, mL
10
49.8
१.१
49.7
initial buret reading, mL
50
40mL
volume of NaOH
used, mL
number of moles of NaOH used
0014
0014
number of moles of HT titrated
.04
[HT] in KHT solution, mol L-
1.42
1.142
1.142
[K*] in KHT solution, mol L-1
solubility of KHT, mol L-1
average solubility of KHT,
mol L-
Ksp
average Ksp
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1-
17.5 mL of a 0.1050 M Na,CO, solution is added to 46.0 mL of 0.1250 M NaCl.
What is the concentration of sodium ion in the final solution?
Select one:
a. 0.165 M
b. 0.148 M
C. 0.119 M
d. 0.205 M
e. 0.539 M
Previous p
Chemistry 101 Exam
Jump to.
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Mg3(PO4)2 BaCO3
MgF₂
Most soluble
Least soluble
Fe(OH)3
Answer Bank
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DATA AND CALCULATIONS
Show all calculations neatly on an attached sheet.
Trial 1
Trial 2
Trial 3
70.6849.
68.0749
2.61g
23.25ML
O.65ML
22.6mL
Mass of flask + vinegar
Mass of empty flask
Mass of vinegar used
Final buret reading
Initial buret reading
Volume of NaOH used
Moles of NaOH used
0.00226 mol
Moles of açetic acid titrated
Mass of acetic acid titrated
Mass % acetic acid in vinegar
5.0%
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Based on this student data, calculate the average molarity of the NaOH solution. Standard: KHP =
potassium hydrogen phthalate, MM 204.4 g/mol
Student Data Trial 1
E KHP
mL initial: NaOH
mL final: NaOH
2.02
0.57
19.25
Trial 2
2.05
19.25
39.80
Trial 3
1.99
1.65
20.75
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A chemist tries to dissolve a solid ionic compound called "X" in water at 25°C. He adds 100 mL of water to
a beaker equipped with a stirrer on a balance. While slowly adding X to the beaker, he measures the mass of
X added and, at the same time, carefully watches to make sure that everything he adds dissolves. Suddenly,
when he has just added a total of 2 g of X, he sees that the mixture is no longer homogeneous (i.e. some of
the X remains on the bottom of the beaker and won't dissolve). He divides the 2 g of X by the 100 mL of
water and obtains the value 0.02 g X/mL water. This number répresents an important physical property of X.
What is it uniquely called?
A) the molarity of X in water at 25°C
B) the concentration of X in water at 25°C
C) the mass of X in water at 25°C
D) the density of X in water at 25°C
E) the solubility of X in water at 25°C
How would you describe the solution that the chemist prepared in Question#1?
A)…
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Determination of unknown acid concentration
Trial 1
Trial 2
Trial 3
Unknown Acid number
Volume of acid used
(mL)
25 mL
25 mL
25 mL
Initial buret reading
(mL)
50 mL
50 mL
50 mL
Final buret reading
(mL)
14.8 mL
15 mL
14.5 mL
Volume of NaOH used
(mL)
50 mL
50 mL
50 mL
Molarity of NaOH
Calculated molarity of HCl (M)
Show a sample calculation of HCl molarity
AVERAGE MOLARITY OF Unknown HCl
Show a sample calculation of NaOH molarity
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Titration of Acetic Acid CH3COOH
1) Record the exact Molarity of the freshly prepared solution of NaOH by
reading the label on the bottle, include all digits provided. This is the
solution that will be in the buret.
MB 0.0500 M
2) Record the exact volume of acid CH,COOH that was pipetted into the Erlenmeyer
10.00
aak
mL. Note the concentration is unknown and what you will solve for.
(This was diluted with 50 mL of water in the Erlenmeyer flask)
Trial 1
Trial 2
Trial 3
Final base
(mL)
30.46
30.98
29.88
Initial basc (mL)
0.45
0.05
0.04
Total base
(mL)
*Molarity (M)
HOOD HO
"Use atoichiometry to determine the molarity of your CIHCOOH solution fromn each of
trial. Show all work Including units for one.
Caleulate Average Molarity
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In the experiment, the amount of acetic acid in vinegar was determined by titrating with 0.1 M NaOH solution. In the analysis, three trials were performed to obtain more precise result. The table shows the initial and final burette readings for three titrations:
Initial Burette Reading (mL)
Final Burette Reading (mL)
Trial 1
2.9
10.6
Trial 2
10.6
18.1
Trial 3
18.1
25.7
According to the given data, calculate molarity of acetic acid in the commercial vinegar.
Please be sure that your answer includes followings;
a) moles of NaOH for each trial
b) average mole of NaOH
c) average mole of acetic acid in 50 mL solution
d) average mole of acetic acid in 250 mL solution
e) molarity of acetic acid in vinegar.
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[References]
Use the References to access important values if needed for this question.
You need to make an aqueous solution of 0.208 M chromium(II) acetate for an experiment in lab, using a 500 mL volumetric flask. How much solid
chromium(II) acetate should you add?
grams
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DELL
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Lab report – investigating solubility of solids in water and
interpreting the solubility curves.
-
Solubility Curves
150
140
130
120
NANO
KNO:
110
100
90
80
HCI
70
NH,CI
60
KCI
NH
50
40
NaCl
30
KCIO,
20
10
SO2
10 20 30 40 50 60 70 80 90 100
Temperature °C
Use the above graph to answer the following questions.
1.What is the solubility of KCI at 80 degree
С?....
2Which solvent is used in this
experiment?..
3. What is the solubility of NH3 at 10 degree C?...
4.Which two solutes have same solubility at 60 degree
C?.
5.At 90 degree C, 10 g of NH3 is dissolved in 100 g of water. Is
this solution saturated , unsaturated or unsaturated ?
6.Which term is used to represent a solution in which more
solute can be dissolved?...
7.---
+
= solution
8.Which is the most soluble solute at 20 degree
C?.
9.Identify the solute which is a gas
?.
10. What happens to the solubility of solids when temperature
increases?..
Grams of solute /100 g H,O
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7
Solid lead (II) nitrate is slowly added to 125 mL of a 0.190 M potassium hydroxide solution until the concentration
of lead ion is 0.0547 M. The percent of hydroxide ion remaining in solution is
%.
Submit Answer
$
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LLO
%
5
[Review Topics)
[References
Use the References to access important values if needed for this question.
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As part of this experiment you will need to determine the concentrations of Fe* and SCN ions
present in test tubes 1-9. These values can be calculated using the volumes and concentrations
provided in Tables 1 and 2 of the Equilibrium Lab - Procedure and the dilution equation (M,V1 =
M2V2).
Note: the solutions provided in the lab, and therefore the tables in the lab document, are the
compounds used to prepare the solution. You must determine the concentration of the ions based on
how these compounds díssociate in solution
Complete the following table by filling the the calculated initial concentrations of each substance. Be
sure to:
Pay attention to significant figures
• Round appropriately
Use decimal notation (not scientific)
Do not include units
Test
Initial [Fe3*] (M)
Initial [SCN'] (M)
Tube
1
3
4.
6.
7
8.
2.
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85.0 mL of water. What is the final
concentration of the acid?
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Based on the analysis of a vinegar solution via titration with 0.1 M NaOH. Results are as follows:
Trial 1
Volume of vinegar solution used (mL)
5.00
Final burette reading (mL)
44.20
Initial burette reading (mL)
0.00
Determine the percentage mass of acetic acid in vinegar for Trial 1. The density of vinegar is 1.01 g/mL and molecular mass for acetic acid is 60 g/mol.
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I had 500 mL solution of 0.300M HCI that I left on the bench. WhenI got back to lab
the next day, the concentration of the solution was 0.5M. What volumne of my
solution evaporated?
200 mL
300 mL
125 mL
500 mL
225 mL
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Hello, I'm having issue with this question:
"Set up the factor label (dimensional analysis) calculation required to determine the concentration of the NaOH solution via titration of a given amount of KHP. Include all numbers except the given mass of KHP. Use “Mass of KHP” in place of your given mass"
I started by converting the mass of KHP to moles, but I'm going wrong somewhere and not getting a correct answer because I think I'm missing numbers somewhere, but I'm not entirely sure. A picture of the titration I'm working from is also attached.
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CALCULATOR
COLOR THEME
O ZOOM
ADD NOTE QUESTI
9. How many mL of stock NH, is needed to make 100.0 mL of 1.00 M NH, if your stock solution is
14.8 M?
1.48 mL
6.76 mL
67.6 mL
1480 mL
CLEAR ALL
Answered
13
14
NEXT >
10
11
12
...
6.
7
8.
3.
( PREVIOUS
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40
5.
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Based on the following SAMPLE 2 data, calculate the concentration (M) of NaOH. Report answer with FOUR decimal places. Do not include units in response.
REPORT FOR EXPERIMENT 22
Neutralization-Titration I
Data Table
Sample 1
Sample 2
Mass of flask and KHP
94.7994 g
95.3800 g
Mass of empty flask
93.6616 g
94.2144g
Mass of KHP
Final buret reading
18.25 mL
36.94 mL
Initial buret reading
0.15 mL
18.24 mL
Volume of base used
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K Question 13 of 43
>
General Chemistry 4th Edition
McQuarrie Rock Gallogly
University Science Books
presented by Macmillan Learning
How many milliliters of 11.5 M HCI(aq) are needed to prepare 255.0 mL of 1.00 M HCl(aq)?
V =
mL
> TOOLS
x10
Question Source: McQuarrie, Rock, And Gallogly 4e - General Chemsitry Publisher: University Science Books
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The following volumes of 0.000300 M SCN- are diluted to 15.00 mL. Determine the concentration of SCN- in each sample after dilution. These values will be used during the experiment.
Sample
0.000300 M SCN- (mL)
[SCN-] (M)
1
1.50
2
3.50
3
7.00
4
10.00
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