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...
At the equivalence point of an experiment where a strong acid is titrated using a strong base, which of the following statements is/are always true? Select as many as appropriate however points will be deducted for incorrect guesses. Phenolphthalein indicator will turn a light pink colour Phenolphthalein indicator will turn an intense dark pink colour number of moles of acid initially = number of moles of base added The volume of acid initially = volume of base added [H3O+] = [OH-] in the final solution
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....
Titration of Acids and Bases fa Vola!! Part B. Determination of the molar concentration of an unknown acid. Volume of unknown acid used (mL) Final buret reading (mL NaOH) Initial buret reading (mL NaOH) Volume of NaOH used (mL) an unknown and Molarity of unknown acid Trial 1 10mL 9.6 mL 0.5mL 9,1mL Trial 2 Sample calculations for molarity of unknown acid solution: 10mL 18,9 Trial 3 Average molarity of unknown acid solution: 10mL 28.3mL 9.6 ml 18,9ml 9,3m² 9.4mL mol NaOH used? M N₂OH = M acid= .1 Gril n.1 To multiple decimals
Table 1. Titration data Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 1.48 1.89 1.95 Molarity of NaOH (M) 0.100 0.100 0.100 Volume of vinegar sample (mL) 5.00 5.00 5.00 Final burette reading (mL) 49.32 49.06 49.93 Table 2. Titration data Table view List view Table 2. Titration data Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 1.48 1.89 1.95 Molarity of NaOH (M) 0.100 0.100 0.100 Volume of vinegar sample (mL) 5.00 5.00 5.00 Final burette reading (mL) 49.32 49.06 49.93 Expected color at end point Choose... Choose... Choose... Volume of NaOH used (mL) 1. Average Volume of NaOH used in liters 2. Average moles of NaOH used (mol) 3. Average moles of acetic acid (mol) 4. Average molarity of acetic acid (M) 5. Average mass of acetic acid (g)…
4) Students were asked to perform a lab to identify an unknown acid.The experiment has several parts. First the students had to make a sodium hydroxide solution with a molarity of approximately 0.2M. Then the students had to perform a titration with potassium hydrogen phthalate to standardize their solution. Finally they had to perform a titration with the unknown acid to find the pKa and the molar mass, then using theirdata they had to determine which of the weak acids they had used.Part 1) Make a sodium hydroxide solutionA student measured 4.03 grams of NaOH in a tared weigh boat. They then dissolved it into approximately 200 mL of deionizedwater in a beaker. Once it was completely dissolved the student poured the solution into a 500.0 mL volumetric flask. The student rinsed the beaker with deionized water and added the rinse water to the volumetric flask. They repeated with process, adding the rinse water to the volumetric flask again, they mixed this thoroughly. Finally, they…
A student carries out a titration to determine the molar mass and structure of a weak acid A. The student follows the method below. Dissolve a weighed mass of A in 100 cm3 of distilled water and make the solution up to 250 cm in a beaker. Add the solution of A to a burette. Titrate the solution of A with a standard solution of sodium hydroxide, NAOH. (a) What is meant by the term standard solution? (b) Sodium hydroxide is an alkali. What is meant by the term alkali? (c) The student carries out a trial, followed by three further titrations. The diagram shows the initial and final burette readings for the three further titrations. The student measures all burette readings to the nearest 0.05 cm3. Titration 1 Titration 2 Titration 3 Initial reading Final reading Initial reading Final reading Initial reading Final reading 27
What volume of 0.85 M NaOH (in mL) is needed to reach the equivalence point in a titration of 83.0 mL of 0.90 M HCIO4? Your Answer: Answer units
The molarity of HC2H3O2, M for trail 1,2,3.

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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...