General notes regarding lab reports:

 

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Determination of the Chloride Content in a Soluble Salt
By Ion Exchange and Titration

(Don’t just copy the title from the lab manual.)
(Also, note that “Determination of an Unknown Chloride” makes no sense, scientifically or grammatically.  WHAT did you determine about it?  Note the following appropriate definition of the word “determination:” The ascertaining or fixing of the quantity, quality, position, or character of something: a determination of the ship's longitude; a determination of the mass of the universe.)

 

 

 

Kevin Kolack, Ph.D.

The Cooper Union

Ch111, Section E

Professor Kolack

April 1, 2010

 

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Table of Contents

 

Abstract

3

Introduction

3

Experimental

3

Results and Discussion

5

Acknowledgments (no “e”- see your lab manual)

6

References

6

Appendix I- sample calculations

7

Appendix II- error propagation

8

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Abstract

The percent chloride of an unknown soluble salt (AxCly, number 169) was determined by ion exchange to be 31.6% ± 0.5%.  The fact that the propagated error was 0.3% indicates a low incidence of random error.  The percent error was 43.5%.  The difference between the experimental and actual percent chloride (55.9%) was due to the use of fictitious data.  Hydroxide ions were eluted when a known quantity of the unknown was passed through a column of Amberlite IRA-910.  This basic solution was over-titrated with potassium hydrogen phthalate (KHP) and back-titrated with a solution of NaOH prepared in the lab and standardized with KHP.

PAST TENSE.  BRIEF SUMMARY OF PROCEDURE, RESULTS, ERROR, AND EXPLANATION.  RESULTS ARE REPORTED AS ± THE STANDARD DEVIATION.  IF BEING CALCULATED, THE PROPAGATED ERROR IS ALSO NOTED, SEPARATELY. (THINK OF WHY!)
See your lab manual and a statistics book for a discussion of significant figures, especially with regard to standard deviation.
Note that percent error for this lab and the gravimetric lab is not just a subtraction of the actual and experimental (not “determined”) percent chlorides, but is a statistical percent error.

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Introduction
PRESENT TENSE. BRIEF HISTORY AND TECHNIQUE OVERVIEW, INCLUDING WHY THE EXPERIMENT IS BEING PERFORMED (OR WOULD BE PERFORMED IF THIS WERE NOT JUST FOR A GRADE IN A CLASS).  So, this lab report should include: theory of standard solution preparation including the reasons for the standardization of NaOH and amphoteric nature of KHP; reason for over-titration and back-titration; questions from manual.  MUST INCLUDE REFERENCES TO CURRENT PERIODICALS, NOT JUST TEXTBOOKS.  THIS SECTION IS NOT JUST A VAGUE RESTATEMENT OF THE EXPERIMENTAL SECTION!!!

Determination of chloride content is important to such varied fields as disposal of radioactive waste and restoration of classical paintings.  One method of finding the makeup of an unknown is ion exchange…

Experimental
(PAST TENSE.  WHAT DID YOU DO EXACTLY?  SOMEONE SHOULD BE ABLE TO REPEAT THE EXPERIMENT USING THESE INSTRUCTIONS.  THE LAB MANUAL DOES NOT SAY TO INCLUDE DATA HERE, BUT THIS IS THE METHOD I CHOOSE TO FOLLOW.)

Preparation of standardized NaOH solution (notice that I am capitalizing only the first word of my table captions, SECTIONS AND subsections)

Approximately 0.1M NaOH was prepared by dissolving 20 pellets of solid NaOH (approximately 0.1g each) in 500mL distilled water.  This solution was standardized by titrating it three times against known quantities of the monoprotic acid KHP in 50mL water (Tables 1 and 2).  Roughly 0.5g KHP was used in each of the three trials so that the titrations would be accomplished using approximately 25mL solution.  During the titration of Sample 1, a beaker was broken on a nearby lab bench, and in the resulting confusion, several drops of NaOH were added to the solution past the end point of the titration.

 

Sample 1

Sample 2

Sample 3

Initial mass (g) ± 0.0002g

58.3297

57.8168

57.2967

Final mass (g) ± 0.0002g

57.8168

57.2967

56.7809

Mass of KHP transferred (g) ± 0.0003g

0.5129

0.5201

0.5158

Number of moles of KHP ± 2x10-6

2.512x10-3

2.547x10-3

2.526x10-3

Table 1- Amount of KHP used to standardize the NaOH solution
(NOTE: ALL tables and figures must be numbered and must be referred to in the text)

 

 

Sample 1

Sample 2

Sample 3

Initial volume NaOH (mL) ± 0.05mL

0.00

1.58

2.36

Final volume NaOH (mL) ± 0.05mL

26.89

28.36

29.12

Volume NaOH transferred(mL) ± 0.07mL

26.89

26.78

26.76

Table 2- Volume of NaOH required to neutralize KHP

 

From the data above, the concentration of the NaOH solution was calculated.  This standardized NaOH was stored in a sealed plastic container until the next lab period.

 

Preparation of the ion exchange column

A 50mL buret was cleaned until it was free draining.  A glass wool plug was inserted into the buret to prevent clogging of the stopcock by the ion exchange resin.  Approximately 35mL of resin was added to the buret in small portions, draining off excess liquid as needed.  The buret was inverted several times in order to ensure even packing of the column and to remove air bubbles.  At no point during the experiment was the liquid level allowed to drop below the level of the solid resin.  The resin was recharged by passing 40mL of a stock 4% NaOH solution through the buret at a rate of 5mL/min.  Excess NaOH was rinsed from the resin by passing 60mL of water through the column.  (NOTE ON THIS LAB:  HOW WERE YOU SURE WHEN THE EXCESS HAD BEEN RINSED OFF?  THOSE INSTRUCTIONS SHOULD BE INCLUDED AS WELL.)

 

Ion exchange and acidification of unknown

An aqueous solution of the unknown was prepared by adding approximately 0.2g of the unknown (Table 3) to a 50mL volumetric flask and filling to the mark with water.

Initial mass (g) ± 0.0002g

9.0786

Final mass (g) ± 0.0002g

8.8551

Mass unknown transferred(g) ± 0.0003g

0.2235

Table 3- Mass of unknown used to prepare solution

 

A volumetric pipet was used to pass exactly 10mL of this solution through the column.  To ensure complete exchange, an additional 80mL of water was passed through the column.  The eluent was collected in an Erlenmeyer flask, and 3 drops of phenolphthalein were added.  A second buret was filled with a stock solution of KHP (0.400M), and the purple solution was titrated 3mL beyond the equivalence point (when the solution turned clear).  Three drops of CCl4 were added to the acidified solution as a preservative, and the process was repeated twice more with additional 10mL portions of the solution of the unknown (Table 4) after first regenerating and rinsing the column as above.

 

Sample 1

Sample 2

Sample 3

Initial volume KHP (mL) ± 0.05mL

0.08

1.58

2.36

Final volume KHP (mL) ± 0.05mL

5.09

6.73

7.35

Volume KHP transferred(mL) ± 0.07mL

5.01

5.15

4.99

Amount of KHP (moles) ± 3x10-5

2.04x10-3

2.06x10-3

2.00x10-3

Table 4- Volume of KHP used in over-titration of OH--containing eluent

 

Back titration of acidified solutions

A 50mL buret was filled with the standardized NaOH solution prepared earlier, and the three acidified solutions were titrated until a faint pink color persisted.  The data for these titrations are given in Table 5.

 

Sample 1

Sample 2

Sample 3

Initial volume NaOH (mL) ± 0.05mL

15.97

5.98

26.65

Final volume NaOH (mL) ± 0.05mL

22.86

13.19

33.77

Volume NaOH transferred(mL) ± 0.07mL

6.89

7.21

7.12

Table 5- Volume of standardized NaOH used to titrate acidified eluent

 

 

Results and Discussion
(WHAT DID YOU DISCOVER?)

Since 2.512x10-3mol, 2.547x10-3mol, and 2.526x10-3mol (± 2x10-6 mol) KHP were titrated in three runs by 26.89mL, 26.78mL, and 26.76mL (± 0.07mL) NaOH, and each mole of KHP neutralizes one mole of NaOH, the molarity of the NaOH is calculated to be 0.09342M, 0.09511M, 0.09439M.  However, the relative uncertainty of these numbers is 0.012, so the molarity can only be calculated to a precision of 0.094M ± 0.001M.  This concentration can then be used to calculate the number of moles of NaOH in the final back titration (Table 6).

 

Sample 1

Sample 2

Sample 3

moles NaOH ± 8x10-6

6.48x10-4

6.78x10-4

6.69x10-4

Table 6- Calculation of number of moles of NaOH used in back titration

 

Because the chloride in the unknown displaces the hydroxide from the resin in a 1:1 ratio (SHOULD BE EXPLAINED IN INTRODUCTION), the amount of hydroxide in the eluent of the ion exchange equals the amount of chloride in the unknown.  Since acid (KHP) was added to the eluent past the endpoint, the amount of hydroxide in the eluent is equal to the amount of KHP added minus the amount of NaOH required to neutralize it in the back titration.

 

Sample 1

Sample 2

Sample 3

Number of moles of KHP ± 3x10-5 (from Table 4)

2.04x10-3

2.06x10-3

2.00x10-3

Numbe of moles of NaOH ± 8x10-6 (from Table 6)

6.48x10-4

6.78x10-4

6.69x10-4

Difference  (# moles OH- in eluent =  # moles Cl- ± 3x10-5)

1.39x10-3

1.38x10-3

1.33x10-3

Table 7- Collection of data used in the determination of moles of chloride

 

From the number of moles of chloride and the molecular weight of the chlorine atom, the mass of chloride in the original 0.2235g (± 0.0003g) portion of unknown can be determined.  Since 10mL (± 0.02mL)of the original 50mL (± 0.05mL) prepared was used for each run, a factor of 5 is necessary in the calculation.

 

Sample 1

Sample 2

Sample 3

mass of Cl in 10mL of the 50mL/0.2235g sample

0.0493

0.0489

0.0472

% Cl in the sample ± XXXXXXX

110

109

106

Table 8- % chloride in the three runs

 

The average percent chloride is 108 with a standard deviation of 1.7.  There is a high level of precision, but since each of the three samples contains more than 100% chloride, an error has clearly been made.  The numbers used in this report are fictitious, causing the error here.

In student labs, the t and Q test could be performed…..these tests could ALSO be performed on data used earlier in the lab to decide if a sample run should be thrown out (for instance, in the calculation of the molarity of the standardized NaOH).  CONSIDER THE SIGNIFICANCE (MEANING) OF THE Q, SD and ppt calculations.

The results of this experiment will be compared with that of the gravimetric analysis to be performed on the same unknown.

SPECIFIC SOURCES OF ERROR MUST BE NOTED, NOT GENERALITIES SUCH AS "THE BALANCE MAY HAVE BEEN USED INCORRECTLY"… IF THE BALANCE WAS USED INCORRECTLY, IT SHOULD HAVE BEEN NOTED IN YOUR NOTEBOOK AND THEN MENTIONED IN THE EXPERIMENTAL SECTION… THE RESULTS OF THIS ERROR SHOULD BE DISCUSSED IN THE RESULTS SECTION!!

 

Acknowledgments (note the correct spelling of the word)

DO NOT thank your professor or the stockroom for doing their jobs, your lab partner for being your partner, or your roommate for putting up with you and letting you use their computer….while your gratitude is appreciated, it is unnecessary and unprofessional.  (This, and much of the information presented here, may differ from what is stated in your lab manual or class.  You should consider who is grading your report in deciding how to write it.)  If someone was particularly instrumental in helping you with equations, etc, they should be mentioned here.  If your professor stimulates you into pursuing a graduate career in chemistry (or discourages it, for that matter!) then (s)he may be mentioned here.

 

References

Textbooks, the lab manual, AND CURRENT JOURNALS should be consulted.  The "references" section is for items which you "footnote" in the text.  WEB PAGES SHOULD NOT BE USED (with the exception of the online versions of printed materials) as they are neither static nor peer-reviewed.  MANY OF YOU HAVE NEVER USED ANYTHING OTHER THAN GOOGLE FOR RESEARCH UP TO THIS POINT IN YOUR EDUCATION… THAT IS VERY SAD.  GOOGLE DOES NOT PROVIDE AN EXHAUSTIVE EDUCATIONAL OR SCIENTIFIC SEARCH.  Online research can be useful in locating reference data, but caveat emptor.  An "articles consulted" or bibliography section may be included for reference materials you used but which you do not directly reference in the text of your report.  The references should be in the following format, per the American Chemistry Society Style Guide (however, different journals have different standards- there is no universal agreement on reference format):

book

AuthorLastName, AuthorFirst(andMiddleIfWanted)Initial; OtherAuthorLast, OtherAuthorFirst. TitleOfBook, 1-WordPublisher: CityofPublication, YearOfPublication, PageReferenced.

 

journal

AuthorLastName, AuthorFirstInitial; OtherAuthorLast, OtherAuthorFirst. TitleOfJournalAbbreviated, YearOfPublication, Volume:IssueNumber, PageReferenced.

 

examples

Kahn, O. Molecular Magnetism, VCH: NY, 1993.

Wang, S.; Tsai, H.-L.; Hagen, K.S.; Hendrickson, D.N.; Christou, G. J. Am. Chem. Soc. 1994, 116, 8376.

Brown, H.C.; McDaniel, D.H.; Haflinger, O. in Determination of Organic Structures by Physical Methods; Brand, E.A.; Nachod, F.C., Eds. Academic: NY, 1955, Chapter 14.

 

 

Appendix I- sample calculations
(BE ABSOLUTELY SURE SOMEONE LOOKING AT YOUR PAPER CAN UNDERSTAND AND REPEAT WHAT YOU DID!!!!!)

Preparation of standardized NaOH solution

500mL of 0.1M solution desired
assume each NaOH pellet weighs 0.1g
(0.1mol NaOH/L solution)*(0.5L solution)*(40.01g NaOH/mol NaOH)*(1 pellet NaOH/0.1g NaOH) = 20 pellets NaOH needed

assume titration with approximately 25mL KHP solution
since KHP is a monoprotic acid, assume 25mL of 0.1M KHP (1 mol acid neutralizes 1 mol base)
(0.1mol KHP/L solution)*(0.05L solution)*(204.22g KHP/mol KHP) = 0.51g KHP needed

 

Appendix II- error propagation (not being calculated in Cooper CH111 as of Spring 2002)

Weighing by difference

Subtracting a final mass (±0.0002) from an initial mass (±0.0002),
so the mass transferred has an uncertainty of
[(0.0002)2 + (0.0002)2].5 = 0.0003

Calculation of # moles of KHP

Dividing a final mass (±0.0003) by an exact molecular weight,
so the moles of KHP has a relative uncertainty (Sample 3) of
[(0.0003/0.5158)2 + 02].5 = 0.0006

So the absolute uncertainty for Sample 3 is (2.526x10-3)(0.0006)=2x10-6

Calculation of molarity of NaOH

Dividing moles (2.526x10-3mol ± 3x10-5 mol KHP = moles NaOH)
by milliliters (26.76mL ± 0.07mL NaOH) (Sample 3),
so the result has a relative uncertainty of
[(3x10-5/2.526x10-3)2 + (0.07/26.76)2].5 = 0.0121611

Thus, the absolute uncertainty for Sample 3 is (0.0943946)(0.0121611) = 1.147947x10-3

Since the absolute uncertainty is reported to a lesser degree of precision than our answer, the number of significant figures in the answer must be limited: 0.094 ± 0.001

 


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