Experiment_7__Buffer_Lab_Report_rduong_duongExperiment7

Experiment_7__Buffer_Lab_Report_rduong_duongExperiment7 -...

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EXPERIMENT 7: BUFFER LAB (INFORMAL) Pre-Lab Questions ( 5 points each ): __9___ / 10 pts Data: ___0___ / 10 pts Table containing pH data from Experiment 1. This table should include the theoretical pH values, the actual measured pH values, and the pH values after correctly preparing the buffer solutions. (5 points) Table containing all pH values for all 7 acetic acid/acetate solutions. (5 points) Sample Calculations: ____5__ / 5 pts Buffer Capacity Calculation (5 points) Results: ____8__ / 16 pts Plots of measured pH vs. Drops (HCl or NaOH) added for each of the 7 buffer solution. Plots must be properly labeled and constructed. Plots must also contain linear regression and the equation of the lines for each. Conclusion: __7____ / 7 pts ___38___ / 48 pts Question 1: 2 pts Question 2: 2 pts
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Question 3: 5 pts Question 4: 10 pts Question 5: 4 pts Question 6: 5 pts Question 7: 5 pts Question 8: 5 pts Question 9: 10 pts Style (organization, spelling, grammar, neatness, etc…) ___4___ / 4 pts Deductions (i.e., Late Lab Report, incorrect formatting, spelling etc.) - ______ Total Score ___71___ / 100 pts
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Experiment 7: Buffer Lab: Preparation and Investigation of Buffers Richard Duong Lab Partner: Devon West October 2, 2009 Christopher Backlund Chem 241L, Section 420, Chapman Hall 125 The work presented in this report is my own, and the data was obtained by my lab partner and me during the lab period.
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Pre-Lab Questions 1a. How much pure acetic acid(mL) would be needed to prepare 100mL pH 5.00 buffer that is . 200M Acetic Acid (CH CO H)? (molar mass CH CO H=60.05g/mol and density of CH CO H=1.049 g/mL) .200M CH CO H x .100L = .0200mol CH CO H x 60.05g/mol x 1mL/1.049g = 1.14mL CH CO H 1b. Calculate mass(g) of sodium acetate needed to add to make 100mL pH 5.00 buffer solution. (molar mass of NA CH CO ·3H 0=136.08g) Using the formula pH = pKa + log(A¯/HA): Desired pH= 5.00, pKa=4.756; therefore 5.00 = 4.756 + log(A¯/HA) 10^(5.00-4.756) = 10^ log(A¯/HA) Solving for the ratio of (A¯/HA), we get the ratio to be 1.754 Using the calculated moles of CH CO H from part 1a: 1.754 = (A¯/.0200mol CH CO H) A¯=.0351mol of CH CO ¯ needed .0351mol CH CO ¯ x (1mol NA CH CO ·3H 0/1mol CH CO ¯) x (136.08g/1 mol NA CH CO ·3H 0) = 4.77g NA CH CO ·3H 0
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2. Prepare 100mL pH 5.00 solution adding 1.0M NaOH using same amount of CH CO H as in 1, how much NaOH(mL) is needed? CH
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This document was uploaded on 11/04/2011 for the course CHEM 241 at UNC.

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Experiment_7__Buffer_Lab_Report_rduong_duongExperiment7 -...

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