A neutralization titration was performed to determine the benzoic acid (C6H5COOH) in a 500 mL sample. In order to prepare the titrant, 5 grams of NaOH was weighed and dissolved in 1 L of distilled water. 10 mL of the sample was titrated with the NaOH solution after adding an appropriate indicator to the erlenmeyer. Four trials were done and 2.5 mL, 2.7 mL, 2.6 mL and 2.4 mL of NaOH were consumed. What is the mass of benzoic acid as gram in the sample. (C:12 g/mol, H:1 g/mol, O:16 g/mol, Na: 23 g/mol)

A neutralization titration was performed to determine the benzoic acid (C6H5COOH) in a 500 mL sample. In order to prepare the titrant, 5 grams of NaOH was weighed and dissolved in 1 L of distilled water. 10 mL of the sample was titrated with the NaOH solution after adding an appropriate indicator to the erlenmeyer. Four trials were done and 2.5 mL, 2.7 mL, 2.6 mL and 2.4 mL of NaOH were consumed. What is the mass of benzoic acid as gram in the sample. (C:12 g/mol, H:1 g/mol, O:16 g/mol, Na: 23 g/mol)

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter5: Errors In Chemical Analyses

Section: Chapter Questions

Problem 5.10QAP

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Data for titration of 15.00 mL of vinegar with approximately 1.0 M NaOH. Note you must use the exact concentration of the standard NaOH solution to calculate the moles of NaOH. Quantity Еxample Trial 1 Trial 2 Trial 3 M NaOH (exact concentration) 0.992 M 0.953 M 0.953 M 0.953 M V initial buret reading = V, 0.20 mL 1.20 mL 0.52 mL 0.15 mL V final buret reading = V, 12.90 mL 14.22 mL 13.71 mL 13.31 mL %3D Vep = VNAOH added = V; - V, 12.70 mL 13.02 mL 13.19 mL 13.16 mL Vep = VNAOH in L 0.01270 L 0.01302 L 0.01319 L 0.01316 L moles NaOH = 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol MNAOH X VNAOH moles AA = 0.0126 mol 0.0124 mol 0.0126 mol 0.0125 mol moles NaOH V sample = V acid 0.0150 L 0.0150 L 0.0150 L 0.0150 L (15.00 mL) (15.00 mL) (15.00 mL) Actual molarity 0.840 M 0.827 M 0.840 M 0.833 M of AA Average molarity for 3 trials => of AA 0.833 M 1. Report the acetic acid (AA) concentration of vinegar in units of molarity (M). 2. Use your average AA concentration to calculate the mass…
3. Titration: Strong Acid / Weak Base. (I warned you to be ready!) Here's the cool chart! Analyte: 20.00 mL 0.100 M Hydrazine (N₂H4) Hd A: initial Vertical (Value) Axis Title 0 B: half equivalence C: equivalence 5 10 15 20 Titrant: mL of 0.100 M HCI added 25 20.00 mL of 0.100 M hydrazine (N₂H4, pKb = 5.77) is titrated with 0.100 M HCl. a) Find the initial pH (point A) and the pH at the equivalence point (point C). b) Find the pH at the half equivalence point (point B).
Acid-Base titration: Standardization of 0.1 M NAOH Trial 1 Mass KHP used, g 0.297 g Final reading Volume NaOH, mL 15.59 mL Initial reading Volume NaOH, mL 0.23 mL Volume NaOH consumed, mL 15.36 mL Moles KHP used Moles NaOH used Molarity of NaOH solution
5. A 300.0 mg sample containing Na,CO3, NaHCO, and NaOH and inert material either alone or in some combination was dissolved and titrated with 0.1000 M HCI the titration required 24.41 ml to reach the phenolphthalein endpoint. And an additional 8.67 mL to reach the methyl red endpoint. Determine the composition of the sample and calculate the percent of each titrated component.
A drug is ordered to be administered IV at the initial rate of 4 mU/min, and may increase by 2 mu/min q 15 min to a max of 10 mU/min. The IV strength is 5 mU/mL. Construct the first two lines of the titration table for this order. O A. Dose Rate (mU/min) IV rate (mL/h) 4 mU/min (initial) 8 mU/min 36 mL/h 48 mL/h O B. Dose Rate (mU/min) 4 mU/min (initial) 6 mU/min Oc. Dose Rate (mU/min) 4 mU/min (initial) 6 mU/min O D. Dose Rate (mu/min) 4 mU/min (initial) 6 mu/min IV rate (mL/h) 24 mL/h 48 mL/h IV rate (mL/h) 48 mL/h 72 mL/h IV rate (mL/h) 36 mL/h 36 mL/h
5. A 300.0 mg sample containing Na,CO3, NaHCO3 and NaOH and inert material either alone or in some combination was dissolved and titrated with 0.1000 M HCI the titration required 24.41 mL to reach the phenolphthalein endpoint. And an additional 8.67 mL to reach the methyl red endpoint. Determine the composition of the sample and calculate the percent of each titrated component.
0.02 M NaOH reagent was standardized using 0.1000 g Potassium hydrogen phthalate (204.22 g/mol) and it required 25.8 mL to reach end point.
A double indicator analysis of a soda ash sample was performed. The student weighed 0.4375 g of the soda ash sample and dissolved it to make 100.0 mL solution. The students then took a 25 mL aliquot of the sample solution, then added 25 mL H2O and performed double indicator titration. The phenolphthalein endpoint required 10.27 mL 0.04027 M HCl titrant. The sample was further titrated using Methyl Orange and took a total of 15.97 mL of the titrant to reach the endpoint. Given that the molar weight of Na2CO3 - 105.99 g/mol, NaHCO3 = 84.007 g/mol, and NaOH - 39.997 g/mol answer the following: a. What are the components of the soda ash sample? (The constituents of the sample came from Na2CO3, NaHCO3, and NaOH). b. What are the moles of each component in the titrated 25 mL aliquot? c. What is the percent composition in (% w/w) of each component in the soda ash sample?
1. A 0.1250-g sample of pure (100%) of K2Cr2O7 was dissolve in 250-mL volumetric flask, and dilute to volume, producing a concentration of 1.700 × 10-3 M. The resulting solution is an example of a standard solution prepared from a primary standard. Select one: a. True b. False 2. Which of the following is not an example of a standard solution? a. 1000 ppm Lead Nitrate solution b. 1.000g pure sodium carbonate dissolved in 1L of solution c. 3:4 nitric acid: hydrochloric acid solution d. NaOH solution standardized using KHP e. 0.104 M NaOH
1. Molarity of the NaOH solution 0.238 mol/L Trial 1 Trial 2 Trial 2. Volume of H;PO4 added to flask 22.0 mL 22.0 mL 22.0 mL 3. Initial NaOH volume _0.25_ mL _0.75_ mL 0.55 mL 4. Final NaOH volume 17.60_ mL 18.20_ mL _17.90_ mL 5. NAOH volume used for titration to reach green end point mL mL mL 6. NaOH volume used for titration 7. Moles of NaOH used for titration mol mel mel 8. Moles of H;PO4 that reacted mol mel mol 9. Volume of H;PO4 added to flask L 10. Molarity of H3PO4 mol/L mol/L mol/L
An unknown sample containing mixed alkali (NAOH, NaHCO3, or NazCO3) was analyzed using the double flask method. A 250 mg sample was dissolved in 250 mL CO, free water. A 20.0 mL aliquot of this sample required 11.3 mL of 0.009125 M HCI solution to reach the phenolphthalein end point. Another 29.0 mL aliquot of the sample was titrated to the bromocresol green endpoint using 311 mL of the standard acid. What is/are the component/s of the sample? O Cannot be determined O NAHCO3 only O NAOH and Na,CO3 O NazCO3 only O NAHCO, and Na2CO3
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