Titrated the vinegar sample until the end point was reached Recorded the final

Titrated the vinegar sample until the end point was

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to the flask. Titrated the vinegar sample until the end point was reached. Recorded the final reading and prepared a new sample for the second trial. Results Standardizing a Sodium Hydroxide Solution Data: I. Delivering Water from a Buret II. Preparing the KHP Sample Determination 1 Determination 2 III. Analyzing an Unknown Permanganate Ion Solution IV. Titrating the KHP Initial Buret Reading, mL Final Buret Reading, mL Volume of H2o delivered, mL 0.40mL 11.50mL 11.10mL mass of weighing paper, g mass of weighing paper and KHP, g mass of weighing paper, g mass of weighing paper and KHP, g 2.85g 4.83g 2.86g 4.58g code number of unknown %T absorbance, A molarity, M #1 80.6 0.094 4.6x10^-4 #2 58.8 0.231 1.16x10^-4
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Determination 1 Determination 2 Calculations: 1. mass of KHP titrated, determination 1: 1.98g mass of KHP titrated, determination 2: 1.72g 2. Calculate number of moles of KHP titrated 1.98g(1mol/204.4g)= .0096mol KHP moles of KHP titrated, determination1: 0.009696mol KHP moles of KHP titrated, determination2: 0.008423mol KHP 3. Calculate the number of moles of NaOH required to neutralize the KHP 0.009696(1molNaOH/1molKHP)= 0.009696mol NaOH moles of NaOH required, determination1: 0.009696mol NaOH moles of NaOH required, determination2: 0.008423mol NaOH 4. Calculate the volume (L) of NaOH solution required to neutralize the KHP 19.80mL-0.02mL= 19.78mL 19.78ml(10^3/ml)= 0.01978L volume of NaOH solution, determination1: 0.01978L volume of NaOH solution, determination2: 0.0171L 5. Calculate the molarity of the NaOH solution (0.009696molNaOH)/((19.78ml)(10^3L/mL))= 0.490M initial buret reading, mL 0.02mL 0.00mL finial buret reading, mL 19.80mL 17.10mL
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6. Find the average molarity of the NaOH solution (0.490M + 0.493M)/2=0.4915M average molarity of NaOH solution: 0.4915M Analyzing the Acetic Acid Content of Vinegar Data: Trial 1 Trial 2 Average Calculated Molarity of NaOH from Standardization Lab, mol/L: 0.4915mol/L Calculations: 1. Moles of NaOH delivered, mol 00.4915mL(0.0098mL)= 0.0048167mol Trial1: 0.0048167mol Trial 2: 0.0041286mol 2. Moles of acetic acid (1mol NaOH=1mol CH3COOH) Trial1: 0.0048167mol Initial buret reading, mL 0.00mL 9.8mL Final buret reading, mL 9.8mL 18.2mL Volume of NaOH delivered, mL 9.8mL 8.4mL Volume of acid in titrated sample, mL 10.00mL 10.00mL
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Trial 2: 0.0041286mol 3. Molarity of acetic acid. mol/L (0.0048167molCH3COOH)/((10ml)((1L/1,000mL))= 0.48167M Trial 1: 0.48167M Trial 2: 0.41286M 4. Average molarity of acetic acid, mol/L (0.48167M+0.41286M)/2=0.4473M Average: 0.4473M 5. Mass of acetic acid in sample, g 0.004867molCH3COOH(60.05g/mol)=0.2922g Trial1: 0.2922g Trial2: 0.2479g 6. Mass percent of acetic acid ((0.2922g)/(10.00g))x100= 2.922% Trial1: 2.922% Trial2: 2.479% 7. Average mass percent (2.922%+2.479%)/2=2.7005% Average mass percent: 2.7005% Percent error for the standardized NaOH ((0.5M - 0.4473M)/(0.5))x1005= 10.54%
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