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…

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter20: Applications Of Oxidation/reduction Titrations

Section: Chapter Questions

Problem 20.6QAP

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A 30.00 mL sample of a 0.064 M weak acid solution is titrated with 0.068 M strong base. What is the pH after the addition of 12.65 mL of the base? Ka for the acid is 9.16x106. REPORT YOUR ANSWER TO 2 DECIMAL PLACES. DO NOT INCLUDE UNITS. Type your answer...
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.
Questions 11-14 refer to the same strong base/strong acid (SB/SA) titration. A 15.00 mL solution of 0.100 M sodium hydroxide (NaOH) is being titrated with 0.250 M hydrobromic acid (HBr). What is the total solution volume (in mL) at the equivalence point? (Nearest whole number) Type your answer...
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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
What volume of 0.1951 M NaOH is needed to titrate 19.5100 m L of 0.2130 M acetic acid to the phenolphthalein endpoint? Answer with two decimal places. Your Answer: Answer View hint for Question 3. units H
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.
Exactly 10.66 ml of 0.066 M strong acid is added to a 30.00 ml. sample of a 0.068 M weak base solution. What is the pH at this point in the titration? K for the base is 2.51x10-5. REPORT YOUR ANSWER TO 2 DECIMAL PLACES. DO NOT INCLUDE UNITS. Type your answer....
1. A student failed to carry out all of the procedural steps when doing the experiment. Would the following procedural variations result in an experimentally determined molarity of NaOH that is too high or too low? Briefly Explain. (a) The student didnt clean the buret before the beginning the titration. After completeing the titration, the student noticed that droplets of the titrant were clinging to the inside surface of the barel. (b) The buret tip was not completely filled with NaOH solution when the titration was begun. (c) The student forgot to add phenolphthalein indicator solution to the KHP solution before doing the titration
based on the data tables provided, what is % CH3COOH in vinegar
Calculate the accurate molarity of the sodium hydroxide solution using date from each titration to determine two values for the molarity. Calculate the average molarity of the sodium hydroxide solution from the individual values, and use the average in subsequent calculations. Given: Molarity of sulfuric acid:0.205 M (Sodium hydroxide with sulfuric acid)Approx Titration- initial buret reading : 0mL final buret reading : 31.8mL titration 1- initial buret reading : 0mL final buret reading : 31.5 mL titration 2- initial buret reading: 0mL final buret reading : 31.7mL Analysis of vinegar Approx titration - initial buret reading : 0mL final buret reading : 20.1 mL titration 1- volume of vinegar solution:5mL initial buret reading : 0mL final buret reading : 20.9mL titration 2- Volume of vinegar solution :5mLinitial buret reading : 0mL final buret reading : 20.0mL
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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...