In the experiment, the equivalence point for 0.391g Benzoic Acid was observed when 23.56 mL of 0.0987M NaOH was titrated. Calculate the equivalent(molecular) weight (g/mol) for benzoic acid. Show your result with the proper number of significant figures.Type your answer...

Given:Volume of NaOH used = 23.56 mL ( 4 significant…

Answered: In the experiment, the equivalence…

World of Chemistry, 3rd edition

3rd Edition

ISBN:9781133109655

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

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

Chapter16: Acids And Bases

Section16.3: Titrations And Buffers

Problem 2RQ

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0.2061 grams of an unknown organic acid was dissolved in about 50 mL of water in a 250 mL Erlenmeyer flask and titrated against 0.1000 M NaOH solution using bromothymol blue indicator. 15.56 mL of titrant was used to achieve end point. Calculate the Neutralization Equivalent. Write answer with two decimal places only. No units please.
In the experiment, the amount of acetic acid in vinegar was determined by titrating with 0.1 M NaOH solution. In the analysis, three trials were performed to obtain more precise result. The table shows the initial and final burette readings for three titrations: Initial Burette Reading (mL) Final Burette Reading (mL) Trial 1 2.9 10.6 Trial 2 10.6 18.1 Trial 3 18.1 25.7 According to the given data, calculate molarity of acetic acid in the commercial vinegar. Please be sure that your answer includes followings; a) moles of NaOH for each trial b) average mole of NaOH c) average mole of acetic acid in 50 mL solution d) average mole of acetic acid in 250 mL solution e) molarity of acetic acid in vinegar.
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II only O both I & II ОШ only QUESTION 13 Consider an acid-base titration in which the base is dispensed from a burette into a flask containing an acid. If any drops of the base adhere to the inner walls of the flask, but do not actually mix with the solution, the calculated acid concentration would be O the same as the actual value. O lower than the actual value. O higher than the actual value. Need more information Click Save and Submit to save and submit, Click Save All Answers to save all answers. Save Al
this is analytical chemistry please show solution thanks
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Suppose a student performed a similar standardization titration experiment. Below is the calculated concentration of NaOH from each of their titrations. Titration #1 Titration #2 Titration #3 The concentration of NaOH (M) 0.9554 0.9540 0.9551 Using the 3 concentrations, calculate the average concentration of the student's NaOH solution and calculate the ± error in parts per thousand (ppt). Show the complete calculation. Express your final answer in the form of average (with units) ± error in ppt.
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In the experiment, the equivalence point for 0.391g Benzoic Acid was observed when 23.56 mL of 0.0987M NaOH was titrated. Calculate the equivalent (molecular) weight (g/mol) for benzoic acid. Show your result with the proper number of significant figures. Type your answer...