5.00 mL of stock solution is diluted to 25.00 mL, peoducing solution ALPHA. 10.00 mL of solution ALPHA is diluted to 25.00 mL resulting in solutin BETA. 10.00 mL of solutionBETA is then diluted 25.00 mL producing solution Gamma. a. Determine the dilution factor in produciing solution Alpha from the stock solution. b. Determine the dilution factor in producing solution Beta from the solution Alpha c. Determie the dilution factor in producing solution Gamma from the stock solution Beta.

5.00 mL of stock solution is diluted to 25.00 mL, peoducing solution ALPHA. 10.00 mL of solution ALPHA is diluted to 25.00 mL resulting in solutin BETA. 10.00 mL of solutionBETA is then diluted 25.00 mL producing solution Gamma. a. Determine the dilution factor in produciing solution Alpha from the stock solution. b. Determine the dilution factor in producing solution Beta from the solution Alpha c. Determie the dilution factor in producing solution Gamma from the stock solution Beta.

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Description: Hello, I do not understand this quiestion or how to do this problem. Can you help me so I can follow along. 7. 5.00 mL of stock solution is diluted to 25.00 mL, peoducing solution ALPHA. 10.00 mL of solution ALPHA is diluted to 25.00 mL resulting in

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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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
DATA AND CALCULATIONS Show all calculations neatly on an attached sheet. Trial 1 Trial 2 Trial 3 70.6849. 68.0749 2.61g 23.25ML O.65ML 22.6mL Mass of flask + vinegar Mass of empty flask Mass of vinegar used Final buret reading Initial buret reading Volume of NaOH used Moles of NaOH used 0.00226 mol Moles of açetic acid titrated Mass of acetic acid titrated Mass % acetic acid in vinegar 5.0%
Preparation of HCI solution a. Measure 20.7 ml of concentrated Hcl (37% wt HCI) using the pipet and transfer into a 250 ml volumetric flask containing 50 ml of CO2-free distilled water. b. Swrl the solution and dilute with CO2-free distilled water up the mark c. Shake the solution A. d. Calculate the concentration of the HCI 4. Preparation of NaOH solution a. Measure 10g of solid reagent grade NaOH pellets using the analytical balance b. Dissolve the NAOH pellets in 100mL of CO2-free distilled water in a beaker c. Transfer the solution into a 250 ml volumetric flask and dilute to the mark d. Shake the solution e. Calculate the concentration of the NaOH Solution
2. Calculate how to prepare the following dilutions, ½, ¼, 1/8 and 1/16 of the initial concentration.There is a helpful equation that is used to determine how much stock solution and how much diluent to combine to get the final solution of a desired concentration:
1. A solution contains a mixture of Na2CO3 and NaOH. Using phenolphthalein indicator 25 mL of mixture required 19.5 mL of 0.995 M HCl for the endpoint. With methyl orange, 25 mL of solution required 25 mL of the same HCl for the endpoint. Calculate grams per liter of Na2CO3 in the mixture. a.23b.36 c.78 d.112. A 1.2040 g sample containing sodium carbonate and inert material was dissolved in water and titrated to methyl orange end point, requiring 32.50 mL of 0.1020 M HCl. Calculate the %Na2CO4 in the sample. a. 93.36% b. 14.59% c. 12.75% d. 23.41%3. A 0.1510 g KHP (MW = 204.22 g/mol) sample required a volume of 57.04 mL of NaOH solution to reach a phenolphthalein end point. Calculate the molarity of NaOH titrant. a. 0.02593 M b. 0.006481 M c. 0.01296 M d. 0.01012 M
Select the best representation for an aqueous ammonium bromide (NH4Br) solution key a. ammonium ion bromide ion box a box b O box c box d O box e b. d.
1. Find the number of millimoles of solute in a. 226 ?? of 0.320 ? HClO4. b. 25.0 ? of 8.05 ? 10−3 ? K2CrO4. c. 6.00 ? of an aqueous solution that contains 6.75 ??? of AgNO3. d. 537 ?? of 0.0200 ? KOH. 2. Describe the preparation of a. 5.00 ? of 0.0500 ? KMnO4 from the solid reagent. b. 4.00 ? of 0.250 ? HClO4, starting with an 8.00 ? solution of the reagent. c. 400 ?? of a solution that is 0.0250 ? in I2, starting with MgI2. d. 200 ?? of 1.00% (?/?) aqueous CuSO4 from a 0.365 ? CuSO4 solution. e. 1.50 ? of 0.215 ? NaOH from the concentrated commercial reagent [50% NaOH (w/w), ?? ?? = 1.525]. f. 1.50 ? of a solution that is 12.0 ??? in K+, starting with solid K4Fe(CN)6. 3. Exactly 75.00 ?? of a 0.3132 ? solution of Na2SO3 were treated with 150.0 ?? of 0.4025 ? HClO4 and boiled to remove the SO2 formed. a. What was the mass in grams of SO2 that was evolved? b. What was the concentration of the unreacted reagent (Na2SO3 or HClO4) after thereaction was complete?
5. Species Y is to be extracted from a 1.00 L aqueous sample using multiple 20.0 mL portions of an extractant (K = 5.1). In order to remove 95% of Y from the sample, what minimum total volume of extractant is required? a) 20.0 mL b) 40.0 mL c) 60.0 mL d) 80.0 mL e) 100.0 mL
1. How do you prepare 500 mL of a ~0.10 M NaOH solution? What mass of NaOH(s) should you use? 2. If you wish to use 20 mL of 0.10 M NaOH to react fully with potassium hydrogen phthalate (KHP), what mass of KHP should you use in each titration trial?, g (formula for KHP is: C8H5KO4.)
The problem with making a standard solution of sodium hydroxide from solid NaOH is that O a. the solid is hygroscopic and strongly absorbs water because of hydrogen bonding. O b. once the sodium hydroxide is made, carbon dioxides from the atmosphere immediately begins to react with the solution thereby changing its concentration. O. the solid is extremely reactive and corrodes most laboratory containers. O d. the solid is extremely caustic and difficult to handle.
1. A 50.00 mL stock CaCl2 solution was diluted with water in a 250.0-mL volumetric flask. A 25.00 mL aliquot of this was further diluted to a final volume of 100.0 mL. Finally, 20.00-mL aliquot of the resulting solution was analyzed and found to contain 0.1500 M CaCh2 Solve for the concentration of the stock solution in M. 2. A 50.00 mL stock CaCl2 solution was diluted with water in a 250.0-mL volumetric flask. A 25.00 mL aliquot of this was further diluted to a final volume of 100.0 mL. Finally, 20.00-mL aliquot of the resulting solution was analyzed and found to contain 0.1200 g CaCl2 Solve for %w/v CaCl2 in the stock solution.
Your protocol calls for a 0.15MM NaCl solution. How would you dilute your 3M NaCl solution to 0.15M as your protocol calls for? You can determine the overall dilution factor needed by dividing the final concentration (0.15mM NaCl), by the starting concentration (3M NaCI) (0.00015M/3M = 1/20,000). !! 11. If you want to do this in 3 steps, what woul be three individual dilution factors you could use? Use as your total volume 500 pl. 12. What would be the concentration (in Molarity) for each of your three dilutions, individually? (Note: remember you started with 3M NaCI)