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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

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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

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

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• 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

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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