post lab 6 and 7

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Feb 20, 2024

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POST-LAB REPORT #6 Standardization of a Base and Titration of a Vinegar Solution I. Calculations Part A · Volume of NaOH (aq) used in first completed Trial: = Final burette volume - Initial burette volume = 19.00 mL - 0.20 mL (Note: Not necessary for the initial burette reading exactly 0.0 mL) = 18.80 mL · Molarity of NaOH (aq) calculated from first completed Trial: 0.1459 M (influence of air bubble on overall concentration calculation: Assume that it has air bubble- air bubble usually takes 1.0 mL) in this volume of NaOH 18.80 mL we have 1.00 mL of air bubble Without air bubble do this math and place 0.01780 L for NaOH volume and 0.1540M With air bubble we are getting less concentration (0.1459< 0.1540 ) · Average molarity of NaOH (aq) : Part B · Volume of NaOH (aq) used in first completed Trial: 29.65 mL = Final burette volume - initial burette volume = 29.75 mL - 0.10 mL = 29.65 mL · Molarity of CH3COOH (vinegar) calculated from first completed Trial: 0.872 M At the equivalent point, Moles of acid/ na= moles of base/ nb Moles of acid/1 = moles of base/1 M1V1/ 1= M2V2/1 M1= M2V2/V1
M acetic acid = · Average molarity of CH3COOH (vinegar) : 0.872 M · Mass percent CH3COOH in the vinegar: 5.14% (Ave. Molarity CH 3 COOH) × (Molar Mass) ÷ (Density of CH 3 COOH) × 100 = Mass % CH 3 COOH in vinegar solution % II. ADDITIONAL QUESTIONS 1. What is the percent error in your mass % CH3COOH compared to the label-reported value? % 2. Does your calculated mass % CH3COOH support selling the vinegar as labeled? Why or why not? My calculated mass of 5.14% of CH 3 COOH supports selling the vinegar as labelled (5%). Although it is not exactly 5%, it represents only an error of 2.8% from the accepted labelled value of 5%. Some inevitable errors that may have occurred to produce this minimal error include reading of initial and final volumes of the NaOH titrant in the burette, as well as possibly a volume
of vinegar that is not exactly 5.00 mL. Overall, this represents very minimal error; hence, can be used to sell the vinegar as labelled. 3. A student, Pat, forgot to add phenolphthalein indicator before starting to add titrant to the diluted vinegar. After titrating a while Pat added three drops of indicator; the solution remained colorless. Pat asks you whether it’s okay to continue or whether to start over. What’ s your advice? Why? In this case since Pat’s solution remained colorless after the addition of phenolphthalein indicator, it should be okay to continue the experiment. Since no color change was observed yet, this means that the equivalence point has not yet been reached. Thus, Pat will still need to add more NaOH titrant to the vinegar solution hence he can proceed with this after adding the indicator. 4. If dil. sulfuric acid H 2 SO 4 had been used in Part B, instead of CH3COOH (vinegar) , calculate the molarity of the H2SO4 (aq) . Assume the sample volume (5.00 mL) and your titration volumes were as in Part B. (1) H 2 SO 4 (aq) + NaOH (aq) → NaHSO 4 (aq) + H 2 O (l) (2) NaHSO 4 (aq) + NaOH (aq) → Na 2 SO 4 (aq) + H2O (l) add 1 and 2 (3) H 2 SO 4 (aq) + 2 NaOH (aq) → Na 2 SO 4 (aq) + 2 H 2 O (l) H2A+ 2NaOh= Na2A+ 2H2) 2MH2SO4*0.005L= 0.1470M*0.02965L M= Molarity= Moles/Liter. Moles= Molarity* Volume= M*V At the equivalent point, 2 moles of H2SO4= 1 moles of NaOH 2 M1V1= M2V2 M1= (M2V2)/(2V1) M2= MOLARITY OF NAOH= 0.14720577 V2= 28.85 mL V1= AMOUNT OF SULFURIC ACID USED= 5 ml
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5. If you omitted any Trials in calculating your averages, please briefly give your rationale for having done so. Although my 3 trials for Part A resulted in % differences that were all less than 1%, out of these the best 2 trials were trials 2 and 3 as these gave the smallest % difference (0.2%). Thus, I used 0.1469 M (Trial 2) and 0.1471 M (Trial 3) to calculate for average molarity of NaOH (0.147 M). % The same rationale goes for Part B where I used the molarity of acetic acid from Trials 1 (0.872 M) and 3 (0.873 M) to calculate for average molarity of acetic acid (0.872 M) as molarities from these 2 trials gave the smallest % difference (0.12%). %
POST-LAB REPORT #7 Titration of an Unknown Acid - Determination of Molar Mass I. Show clearly how calculations were done. Include units and correct significant figures. Part A Volume of NaOH (aq) used in first chosen Trial: 16.20 mL = Final Volume - Initial Volume = 17.10 mL - 0.90 mL = 16.20 mL Molarity of NaOH (aq) calculated from first chosen Trial: 0.1670 M 0.5984 g KHP* 1 mol KHP/204.22 g * 1 mol NaOH/ 1 mol KHP * 1/ 0.01949 L NaOH = 0.15034 0.5487 g KHP* 1 mol KHP/204.22 g * 1 mol NaOH/ 1 mol KHP * 1/ 0.01771 L NaOH = 0.15171 0.6020 g KHP* 1 mol KHP/204.22 g * 1 mol NaOH/ 1 mol KHP * 1/ 0.01440 L NaOH = 0.204708 Average NaOH= (0.15034+0.15171)/2 = 0.151025 Show how you determined that your best three Trials were within 1%: Best 3 Trials Trial 3: 0.1670 M Trial 4: 0.1686 M Trial 5: 0.1673 M To find out the best trials from the given values 1. 0.1155 2. 0.1273
3. 0.1098 Best two trials are- the first and third trial 0.1155 and 0.1098 Difference= 0.0057 Avg = 0.11265 % agreement= difference/ avg *100 0.0057/0.11265 * 100 5.059% % % % Average molarity (best three trials) of NaOH (aq) : 0.1676 M Part B
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Volume of NaOH (aq) used in first chosen Trial: 13.70 mL = Final Volume - Initial Volume = 14.20 mL - 0.50 mL = 13.70 mL Molar mass calculated for the unknown acid from first chosen Trial: 127.6 g/mol 0.1593 g acid* 2 mol NaOH/ 1 mol Acid* L NaOH/ 0.151025 mol NaOH* 1 / 0.01665 L = 126.70 0.1503 g acid* 2 mol NaOH/ 1 mol Acid* L NaOH/ 0.151025 mol NaOH* 1 / 0.01555 L = 127.99 0.1393 g acid* 2 mol NaOH/ 1 mol Acid* L NaOH/ 0.151025 mol NaOH* 1 / 0.01430 L = 128.72 Average molar mass= (126.70+127.99+128.72)/3 = 127.80 Average molar mass (best three trials) for the unknown acid: 126.5 g/mol Trial 1: 129.2 g/mol Trial 2: 122.6 g/mol Trial 3: 127.6 g/mol Percent error in the unknown acid’s molar mass, assuming you chose the correct known acid: 0.34%
% II. ADDITIONAL QUESTIONS 1. A student is given a 2.002 g sample of unknown acid and is told that it might be butanoic acid, a monoprotic acid (HC4H7O2, equation 1), L-tartaric acid, a diprotic acid (H2C4H4O6, equation 2), or ascorbic acid, a diprotic acid (H2C6H6O6, equation 3). If it requires 39.55 mL of 0.570 M NaOH (aq) to neutralize the unknown acid, what is the identity of the unknown acid? Justify your answer. HC 4 H 7 O 2 (aq) + NaOH (aq) → NaC 4 H 7 O 2 (aq) + H 2 O (l) (1) H 2 C 4 H 4 O 6 (aq) + 2NaOH (aq) → Na 2 C 4 H 4 O 6 (aq) + 2H 2 O (l) (2) H 2 C 6 H 6 O 6 (aq) + 2NaOH (aq) → Na 2 C 6 H 6 O 6 (aq) + 2H 2 O (l) (3) A. butanoic acid (88.12 g/mol) B. L-tartaric acid (150.10 g/mol) C. ascorbic acid (176.14 g/mol) D. butanoic or L-tartaric acid E. butanoic or ascorbic acid F. L-tartaric or ascorbic acid G. None of these (A-F) Answer:
The identity of the unknown acid should be butanoic acid . Based on the given concentration and volume of NaOH used to neutralize 2.002 g of the unknown acid, calculations show that 0.0225 mol of NaOH was used. This value, together with the stoichiometric relationship of acid and NaOH from each equation, as well as the corresponding molar mass of each acid were used to determine the mass of unknown acid that should be used initially. Among these, the stoichiometric relationship in equation 1 and the molar mass of butanoic acid resulted in exactly 2.002 g of Acid, which matches the mass of unknown acid used as stated in the problem. 2. L.X., a classmate, started the titration but forgot to add phenolphthalein indicator to the solution of unknown acid at the start. On adding three drops of indicator later, the solution turned pink. L.X. asks you for help, “Is it OK to continue with the titration?” What would you advise L.X. to do? Explain. Given that the solution turned pink upon the addition of indicator, it is not okay to continue with the titration. This means that the equivalence point (equimolar amounts of titrant and solution with unknown concentration) has already been reached or more likely that the experiment has gone past this point. Since the point of doing a titration is to determine the exact volume of titrant needed to neutralize an unknown solution and therefore identify its concentration, there is no way for L.X. to find that out because it is likely that he has added too much titrant to the solution as indicated by its pink color upon addition of the indicator. I would advise L.X. to repeat the titration and make sure to add the indicator before starting. Figure 1. Three Trials from Titration of an Unknown Acid (Part B)
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Happy Titrating! :)