Lab 4 Final

pdf

School

Kennesaw State University *

*We aren’t endorsed by this school

Course

2800L

Subject

Chemistry

Date

Feb 20, 2024

Type

pdf

Pages

Uploaded by HighnessRamMaster2774

LAB 4 REPORT SHEET: How much hydrogen peroxide is in commercial hydrogen peroxide? Prelab Questions: 1. You have used a graduated cylinder to measure stock MnO 4 - solution and a plastic bottle to prepare a working solution of MnO 4 - that you will later use to determine the amount of hydrogen peroxide. Justify the use of a graduated cylinder and a plastic bottle for preparing MnO 4 - . A graduated cylinder precisely measures volumes to be delivered to another vessel. The plastic bottle will not alter the reagent. 2. Define primary standard, endpoint and equivalence point. Primary Standard: These substances can be obtained in pure form, measured in precise quantities, and won’t chemically change over long periods of time. Endpoint: The point during titration when the indicator shows that the amount of reactant needed to induce color change has been added. Equivalence Point: An acid and base are at an equal or neutral stoichiometric ratio; the reaction comes to an end. 3. In the experiment 25 mL of 3 M H 2 SO 4 is used. Will that volume be measured with a 25.00 mL pipet or a graduated cylinder? 25 ml has 2 significant figures and does not need precision to the decimal places, so a graduated cylinder will suffice. 4. A 25.00 mL H 2 O 2 solution required 22.50 mL of 0.01881 M MnO 4 - for titration to the endpoint. Given that the original H 2 O 2 was diluted 1 in 20 before titration with the MnO 4 - , (i) what is the molarity of the diluted H 2 O 2 ? (ii) What is the molarity of the original H 2 O 2 solution? (iii) What is the % w/w of H 2 O 2 in the original solution, assuming a solution density of 0.9998 g/mL? show work. ?𝑯 ? 𝑶 ? + ?𝑴?𝑶 ? − → ?𝑴? ?+ + ?𝑶 ? + ?𝑯 ? 𝑶 i) 0.2550? × 0.01881? ??? 4 − × 5 ??? 𝐻 2 ? 2 2 ??? ??? 4 − × 1 ??? 𝐻 2 ? 2 .02500 ? = .04232 ? 20 ?𝑎??? = .002166 ? ?𝑖????? ii) 0.2550? × 0.01881? ??? 4 − × 5 ??? 𝐻 2 ? 2 2 ??? ??? 4 − × 1 ??? 𝐻 2 ? 2 .02500 ? = .04232 ? ?𝑟𝑖𝑔𝑖?𝑎? iii) . 20 L 1 20 = 1.25 ml → .9998 g ml × 1.25 ml = 1.250 g → 1.25 ml 1000 ml × .8464 mol 1 L × 34.014 g mol = .03598 g → % w w = .03598g 1.250 g × 100% = 2.88%

A. Calculation of the volume of KMnO 4 used to prepare a roughly 0.02 M solution. Stock solution concentration: 0.2 M (0.2?)(𝑉) = (0.02?)(250 ??) → 𝑉 = (250 ??)(0.02 ?) (0.2 ?) → ?? ?? ?? 𝑲𝑴?𝑶 ? ?????? B. Table 1. Standardization of KMnO 4 using Na 2 C 2 O 4 as the primary standard. Trial 1 Trial 2* Trial 3 Trial 4 Na 2 C 2 O 4 , g 0.2386 g 0.2197 g 0.2387 g Moles Na 2 C 2 O 4 .001781 mol .001639 mol .001781 mol Final buret reading 35.79 ml 32.78 ml 35.71 ml Initial buret reading 0.39 ml 0.49 ml 0.50 ml Volume used, mL 35.40 ml 32.29 ml 35.21 ml Moles KMnO 4 .0007122 mol .0006558 mol .0007125 mol Molarity of KMnO 4 .02012 M .02031 M .02024 M Mean, M .02018 M SD 0.00008485 M If any trial is removed from the calculation of mean and standard deviation explain here: Trial 2 was removed from the calculation of the mean and standard deviation. A Grubbs-Test was performed for each trial to determine outliers. Trial 1: |??.?? − ??.??| ?.?? = . ???? < ?. ??? ; Outlier is kept with 95% confidence. *Trial 2: |??.?? − ??.??| ?.?? = . −?. ??? > ?. ??? ; Outlier is rejected with 95% confidence. Trial 3: |??.?? − ??.??| ?.?? = . ???? < ?. ??? ; Outlier is kept with 95% confidence. Our original first trial was completely invalid because we forgot to add the Sulfuric Acid, and we had no time for a 4 th trial. Sample calculations of Table 1 using Trial 1 using equation editor in Word. (Moles Na 2 C 2 O 4 , Moles KMnO 4 , Molarity of KMnO 4 ) Moles of Na 2 C 2 O 4: 0.2386 𝑔 × 1 ??? ?𝑎 2 ? 2 ? 4 134.0 𝑔/??? = . ?????? ??? ?𝐚 ? 𝑪 ? 𝑶 ? Moles of KMnO 4: . 001781 ??? ?𝑎 2 ? 2 ? 4 × 2 ??? ???? 4 5 ??? ?𝑎 2 𝐶 2 ? 4 = . ??????? ??? 𝑲𝑴?𝑶 ? Molarity of ???? 4 : . 0007122 ??? ???? 4 × 1 ? (35.40 𝑉 𝑑 × 1 𝐿 1000 ?? ) = . ????? 𝑴 𝑲𝑴?𝑶 ?

C. Table 2. Analysis of 25-mL aliquots of commercial solution of hydrogen peroxide using standardized KMnO 4 . Molarity of KMnO 4 : .02018 M ± .00008485 Trial 1 Trial 2 Trial 3 Trial 4 Final buret reading 29.71 ml 29.21 ml 29.73 ml Initial buret reading 0.75 ml 0.31 ml 0.51 ml Volume MnO 4 - used 28.96 ml 28.90 ml 29.22 ml Moles MnO 4 - used .0005844 mol .0005832 mol .0005897 mol Moles H 2 O 2 found .001461 mol .001458 mol .001474 mol Moles/L H 2 O 2 .05045 mol/L .05045 mol/L .05044 mol/L g/L diluted H 2 O 2 1.716 g/L 1.716 g/L 1.716 g/L g/L undiluted H 2 O 2 33.13 g/L 33.06 g/L 33.42 g/L % H 2 O 2 in  % solution 3.31% 3.31% 3.34% Mean, % w/w 3.32% SD .017 If any trial is removed from the mean and standard deviation calculation, explain here: Trial 1: |??.?? − ??.??| .???? = −. ???? < ?. ??? ; Outlier is kept with 95% confidence. Trial 2: |??.?? − ??.??| .???? = −. ???? < ?. ??? ; Outlier is kept with 95% confidence. Trial 3: |??.?? − ??.??| .???? = −?. ??? < ?. ??? ; Outlier is kept with 95% confidence. There was no time for a 4 th trial, but our 3 trials passed the Grubb’s test. Sample calculation of Table 2 using one trial: (Moles KMnO 4 , Moles H 2 O 2 , Moles/L, g/L diluted, g/L undiluted, %H 2 O 2 ) Used Trial 1 Moles of KMnO 4: 0.02018 M × 28.96 mL KMnO 4 1000 mL = . ??????? ??? ???? ? Moles of H 2 O 2: . 0005844 mol KMnO 4 × 5 mol H 2 O 2 2 mol KMnO 4 = . ?????? 𝐇 ? ? ? Moles/L of H 2 O 2: . 001461 mol H 2 O 2 × 1 (28.96 V d × 1 L 1000 ml ) = . ????? ?????/? 𝐇 ? ? ? g/L Diluted H 2 O 2 : . 001461 mol H 2 O 2 × 34.014 g mol H 2 O 2 .02896 L = ?. ??? g/L Diluted 𝐇 ? ? ? Molarity of Diluted H 2 O 2 : .001461 mol H 2 O 2 .02500 L = . ????? ? ?𝐢???? 𝐇 ? ? ?

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

g/L of Undiluted H 2 O 2 : (M)(. 15 L) = (. 05844 M)(. 250 L) → (.05844 M).250 L) (.015 L) = .974 mol L × 34.014 g mol = 33.13 g/L 𝐔??𝐢????? 𝐇 ? ? ? %w/w of H 2 O 2 : 33.13 g L H 2 O 2 .015 L = .4970 g H 2 O 2 → .9998 g ml × 15 ml = 14.997 g Solution → .4970g H 2 O 2 14.997 g Solution × 100% = ?. ??% 𝐇 ? ? ? 𝐢? ?????𝐢?? Post-lab Questions: 1. What changes would you make to the experiment to improve the accuracy and precision? Using a volumetric pipet to measure the Potassium Permanganate, Sulfuric Acid, and Hydrogen would provide a more precise measurement than a graduated cylinder would. This would overall contribute to more precise calculations and increased accuracy. 2. Verify by calculation that a 3.0% hydrogen peroxide solution of a density approximately 1.000g/mL is about 0.88M. ?. ?% ??. ??? ? ??? (?????) = . ???? 𝑴 𝐇 ? ? ? 3. State two possible reasons why the KMnO 4 /H 2 O 2 titration was carried out at elevated temperatures. Titrations requires precision, and heating up a reaction helps speed up the process. This experiment used acids, and the heat prevented precipitate formation because of the Manganese in Potassium Permanganate. 4. Why was an indicator not necessary for these titrations? An indicator wasn’t necessary because of the naturally occurring purple color of Potassium ions. 5. Write out the two reactions involved and indicate which species is the oxidant, which is the reductant, which is oxidized, and which is reduced for both reactions (8 answers total). ?? ? 𝑶 ? ?− (?????𝐢?𝐠 𝐀𝐠???; ?𝐱𝐢?𝐢𝐳??) + ???? ? − (?𝐱𝐢?𝐢𝐳𝐢?𝐠 𝐀𝐠???; ???????) + ??𝑯 + → ??? ?+ + ???? ? + ?𝐇 ? ? ?𝐇 ? ? ? (?????𝐢?𝐠 𝐀𝐠???; ?𝐱𝐢?𝐢𝐳??) + ???? ? − (?𝐱𝐢?𝐢𝐳𝐢?𝐠 𝐀𝐠???; ???????) + ?𝐇 + → ??? ?+ + ?? ? + ?𝐇 ? ?

Related Documents

Ch 1 Worksheet -3.docx

cation evaluation.pdf

Ice caloremetry Lab.docx

Practice Exam 4(2).pdf

TEMPLATE Acid Base Titration.docx

Lab Report 4.docx

Research Project Lab Report .pdf

Recommended textbooks for you

Chemistry

ISBN:9781305957404

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

Publisher:Cengage Learning

Chemistry: An Atoms First Approach

Chemistry

ISBN:9781305079243

Author:Steven S. Zumdahl, Susan A. Zumdahl

Publisher:Cengage Learning

Chemistry

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Cengage Learning

Chemistry by OpenStax (2015-05-04)

Chemistry

ISBN:9781938168390

Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser

Publisher:OpenStax

Chemistry for Engineering Students

Chemistry

ISBN:9781337398909

Author:Lawrence S. Brown, Tom Holme

Publisher:Cengage Learning

Introductory Chemistry: A Foundation

Chemistry

ISBN:9781337399425

Author:Steven S. Zumdahl, Donald J. DeCoste

Publisher:Cengage Learning

SEE MORE TEXTBOOKS

Recommended textbooks for you

Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Chemistry
Chemistry
ISBN:9781133611097
Author:Steven S. Zumdahl
Publisher:Cengage Learning
Chemistry by OpenStax (2015-05-04)
Chemistry
ISBN:9781938168390
Author:Klaus Theopold, Richard H Langley, Paul Flowers, William R. Robinson, Mark Blaser
Publisher:OpenStax
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning