Titration of Acids and BasesPart B. Determination of the molar concentration of an unknown acid.Molarity of N₂OH = .09M☆Volume of unknown acid used (mL)Final buret reading (mL NaOH)Initial buret reading (mL NaOH)Volume of NaOH used (mL)Molarity of unknown acidTrial 110mbMoles9.6 mL0.5mL9.1mL(OH)Sample calculations for molarity of unknown acid solution:Trial 210mL18,9Average molarity of unknown acid solution:Trial 310mL28.3mL9.6 ml 18.9ml9,3mL 9.4mLmol NaOHused?M N₂OH =M acid = .1GrilTo multipledecimals

To find out the molarity of the unknown acid solution, we need to use the dilution law. Using that,…

Titration of Acids and Bases Part B. Determination of the molar concentration of an unknown acid. Mollarity of NaOH = .09M Volume of unknown acid used (mL) Final buret reading (mL NaOH) Initial buret reading (mL NaOH) Volume of NaOH used (mL) Molarity of unknown acid Trial 1 10mb 9.6 mL 0.5mL 9.1mL Trial 2 Sample calculations for molarity of unknown acid solution: 10ml 18,9 9.6 ml 9,3mL Average molarity of unknown acid solution: Trial 3 Detal 10mL 28.3mL 18.9ml 9.4mL 002 mol us MA Maci G To mun decin

Titration of Acids and Bases Part B. Determination of the molar concentration of an unknown acid. Mollarity of NaOH = .09M Volume of unknown acid used (mL) Final buret reading (mL NaOH) Initial buret reading (mL NaOH) Volume of NaOH used (mL) Molarity of unknown acid Trial 1 10mb 9.6 mL 0.5mL 9.1mL Trial 2 Sample calculations for molarity of unknown acid solution: 10ml 18,9 9.6 ml 9,3mL Average molarity of unknown acid solution: Trial 3 Detal 10mL 28.3mL 18.9ml 9.4mL 002 mol us MA Maci G To mun decin

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

Please solve this
Acid Base Titrations Given the information below, calculate the acid concentration: show work and use correct number of significant figures Unknown acid ID#: ____HCl_____(concentration unknown) Concentration of NaOH: 0.10 M Initial Buret reading (mL) : 4.0 Final Buret reading (mL): 14.2 Volume of NaOH added (mL): 10.2 Calculated acid concentration (M): ???
Quantity Your Data 1. Grams of vinegar sample used for your titration 25.000 g 2. Initial Buret Reading of Sodium Hydroxide solution 10.00 mL 3. Final Buret Reading of Sodium Hydroxide solution 28.00 mL 4. Amount of Sodium Hydroxide Solution used to neutralize the vinegar sample 5. Concentration of NaOH in the NaOH solution 0.050 g/mL 6. Grams of NaOH used to neutralize the vinegar 7. Grams of acetic neutralized by the amount of NaOH 8. Percent acetic acid in the vinegar Hints: For #4, The amount of NaOH used is the difference between the starting buret value and the ending buret value. For #6, Once you calculate the amount of sodium hydroxide used, multiple that value by the concentration of NaOH in the NaOH solution. For #7, Multiply the value obtained in number 6 by the number 1.5. Remember, we learned that every 1 gram of NaOH neutralizes 1.5 grams of acetic acid. For #8, Divide the grams of acetic acid by the grams of vinegar sample and multiply this value by…
The molarity for trial 1 will be 0.4035 and trial 2 molarity will be 0.3587 I need some assistance on question 2 if possible
Please show full work
please don't provide hand written soluton....
Calculate the number of moles of vitamin C in the tablet based on the stoichimetry from the equation given in the back ground 1000 mg is in each tablet
2 Match the following acid-base titration plots with the correct titrant and sample. Veq = 20.00 mL mea a. b. Hd (1) Titrant: Strong Acid; Sample: Strong Base (2) Titrant: Strong Base; Sample: Strong Acid (3) Titrant: Strong Acid; Sample: Weak Base (4) Titrant: Strong Base; Sample: Weak Acid 3219 11 10 9 8 7 6 5 4 3 2 1 0 14 13 12 11 10 9 8 7 6 5 4 3 2 1 0 0 0 5 5 10 10 15 Volume 15 Volume 20 20 25 25 30 30
A multivitamin sample has a label claim of 500 mg ascorbic acid (176.12 g/mol). According to quality assurance guidelines for stability, it has an acceptable range of 90.0%-110.0% of the label claim. After production, the multivitamin was analyzed through acid-base titration, and the sample needed 25.40 mL of a 0.1101 M NaOH titrant to reach the end point. After 3 months, the same multivitamin was analyzed again, and this time 23.34 mL of the same 0.1101 M titrant was used. After 6 months, the same multivitamin was analyzed again, and this time, 22.48 mL of the same 0.1101 M NaOH titrant was needed to reach the end point.
A multivitamin sample has a label claim of 500 mg ascorbic acid (176.12 g/mol). According to quality assurance guidelines for stability, it has an acceptable range of 90.0%-110.0% of the label claim. After production, the multivitamin was analyzed through acid-base titration, and the sample needed 25.40 mL of a 0.1101 M NaOH titrant to reach the end point. After 3 months, the same multivitamin was analyzed again, and this time 23.34 mL of the same 0.1101 M titrant was used. After 6 months, the same multivitamin was analyzed again, and this time, 22.48 mL of the same 0.1101 M NaOH titrant was needed to reach the end point.
Only typed solution.