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7/19/2016 Chemistry 2A Laboratory Presentations Post-Lab Student Data Summary Post-Lab Data Summary Note : some questions will display a variable

Part IIB. 
Scoring Scheme: 3-3-2-1. 
Calculate and enter the molarity of your three HCl standardization trials using the volume of standardized NaOH solution required for each and the average molarity of the NaOH solution from the standardization trials with KHP. You should report 3 significant figures, e.g. 0.488 M.
#1: 5.000 9.000
#2: 5.000 9.800
#3: 5.000 9.700


7/19/2016 Chemistry 2A Laboratory Presentations Post-Lab Student Data Summary http://chemelements.ucdavis.edu/CHE2A/SSI2016/Exam/index.cfm?EOID=7674&fuseaction=DisplayQuestion 1/4 Post-Lab Data Summary Note : some questions will display a variable like "nCount" or "SyInput" instead of an actual number in the data summary. Q# Question Text 4) Part I A. Preparing NaOH solution. Scoring Scheme: 3­3­2­1. How many grams of pure, solid NaOH are required to make 300 mL of 0.1M NaOH solution? Express your answer to 2 significant figures, e.g. 3.3 g. Mass (g) = Your Answer: 1.2 You Scored 3 points out of 3 Possible 5) Part I A. Preparing NaOH solution. Scoring Scheme: 3­3­2­1. How many grams of a solution that is 50% by weight NaOH is required to make 300 mL of 0.1M NaOH solution? Express your answer to 2 significant figures, e.g. 6.6 g. Mass (g) = Your Answer: 2.4 You Scored 3 points out of 3 Possible 6) Part I A. Preparing NaOH solution. Scoring Scheme: 3­3­2­1. The density of a 50% solution of NaOH is 1.525 g/mL. What volume of a solution that is 50% by weight NaOH is required to make 0.3 Liter of 0.1M NaOH solution? Express your answer to 2 significant figures, e.g. 6.6 mL. Volume (mL) = Your Answer: 1.6 You Scored 3 points out of 3 Possible 7) Part I B. Standardizing against KHP Scoring Scheme: 3­2­1­1. Which of the following best explains why weighing by difference eliminates systematic balance errors? Your Answer: b. Because the mass is determined by the difference between two readings, a systematic error in the absolute mass on the balance will be removed by subtracting the Fnal weight from the initial weight. You Scored 3 points out of 3 Possible 8) Part I B. Data Entry ­ No Scoring. For each of your THREE ACCEPTABLE trials (not the cursory one) in standardizing your NaOH against KHP, enter the mass in grams of KHP for each of the three samples to one thousandth of a gram, e.g. 0.563 g. Your Answer: 0.265 No Points Possible Your Answer: 0.254 No Points Possible Your Answer: 0.257 No Points Possible 9) Part I B. Data Entry ­ No Scoring. Please enter the volume in mL of NaOH required to reach the endpoint for each of the KHP samples and enter the volume to a hundredth of a mL, e.g. 15.52 mL. Make your volume entries in the order corresponding with the masses of KHP. Your Answer: 13.17 No Points Possible Your Answer: 14.25 No Points Possible
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7/19/2016 Chemistry 2A Laboratory Presentations Post-Lab Student Data Summary http://chemelements.ucdavis.edu/CHE2A/SSI2016/Exam/index.cfm?EOID=7674&fuseaction=DisplayQuestion 2/4 Your Answer: 14.85 No Points Possible 10) Part I B. Scoring Scheme: 3­3­2­1. Please enter the values you calculated for each of your three trials for the molarity of your NaOH solution. Place your molarity entries in the order corresponding with the masses of KHP and the volumes of NaOH required for the trials. You should enter 3 significant figures, e.g. 0.149 M . Your Answer: 0.0983 You Scored 3 points out of 3 Possible Your Answer: 0.087 No Points Possible Your Answer: 0.0845 No Points Possible 11) Part I B. Scoring Scheme: 3­3­2­1. Please enter the value you calculated for the average of the three trials determining the molarity of your NaOH solution. You should report 3 significant figures, e.g. 0.149 M Average (M) = Your Answer: 0.0899 You Scored 3 points out of 3 Possible 12) Part I B. Scoring Scheme: 3­3­2­1. Please enter the value you calculated for the standard deviation of the average molarity. You should report your answer in 2 significant digits. Standard deviation (M) = Your Answer: 0.0074 You Scored 3 points out of 3 Possible 13) Part I B. Scoring Scheme: 3­3­2­1. Please enter the value you calculated for the 90% confidence limits for the molarity of your NaOH solution. You should report 2 significant figures, e.g. 0.0054 M 90% confidence limit (M) = Your Answer: 0.024 You Scored 3 points out of 3 Possible 14) Part I B. Scoring Scheme: 3­3­2­1. Which of the following choices is the best explanation for why it does not matter how much water you added when dissolving the acid or when carrying out the titration? Your Answer: a. The titration equivalence point occurs when the acid present in the sample has been exactly neutralized by the volume of base added. Additional water added to the reaction vessel has no effect on the volume of base added. You Scored 3 points out of 3 Possible 15) Part IC. Data Entry ­ No Scoring. Which commercial vinegar sample did you use? Your Answer: a. White Vinegar You Scored 0 points out of 0 Possible 16) Part IC. Data Entry ­ No Scoring. What volume in mL did you use for each of the three acceptable trial samples of your unknown solution of commercial vinegar sample? Report your volume to 4 significant figures, e.g. 5.000 mL. Your Answer: 12.15 No Points Possible
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Since you have used 0.1 M NaOH and titrated with 5 mL of HCl the molarity of the HCl can be... View the full answer

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Attached is a detailed explanation... View the full answer

findings.docx

1 comment
  • alternatively, Since you have used 0.1 M NaOH and titrated with 5 mL of HCl the molarity of the HCl can be calculated using the formula N1V1=N2V2 NHCl x VHCl = NNaOH x VNaOH Trial 1: 5 x NHCl = 9 x 0.1M NaOH NHCl = 0.180 M HCl Trial 2: 5 x NHCl = 9.8 x 0.1M NaOH NHCl = 0.196 M HCl Trial 3 : 5 x NHCl = 9.7 x 0.1M NaOH NHCl = 0.194 M HCl
    • kalebz21
    • Jul 20, 2016 at 1:37am

The best way to approach your question... View the full answer

aaaaaaaaaaaaaa.docx

Since you have used 0.1 M NaOH and titrated with 5 mL of HCl the molarity of the HCl can
be calculated using the formula
N1V1=N2V2
NHCl x VHCl = NNaOH x VNaOH Trial 1: 5 x NHCl = 9 x 0.1M NaOH...

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