A. The weight of 25 tablets of morphine sulfate (150 mg) is 3.075 g. If 0.3075 g was used in the assay and it required 10.3 ml of 0.023N NaOH to back titrate 25 ml of 0.021N H2SO4. Calculate the percent labeled amount of morphine sulfate. Each ml of 1N H2SO4 is equivalent to 758.83 mg of morphine sulfate.

A. The weight of 25 tablets of morphine sulfate (150 mg) is 3.075 g. If 0.3075 g was used in the assay and it required 10.3 ml of 0.023N NaOH to back titrate 25 ml of 0.021N H2SO4. Calculate the percent labeled amount of morphine sulfate. Each ml of 1N H2SO4 is equivalent to 758.83 mg of morphine sulfate.

Fundamentals Of Analytical Chemistry

9th Edition

ISBN:9781285640686

Author:Skoog

Publisher:Skoog

Chapter16: Applications Of Neutralization Titrations

Section: Chapter Questions

Problem 16.32QAP

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1.876 g of the sample containing H2C2O4requires 38.84 mL of 0.1032 M of NaOH for titration: H2C2O4+ 2 NaOH →Na2C2O4+ 2 H2OA 1.38 mL of 0.0992 M HCl is used in the back –titration. (a)Calculate the percentage of H2C2O4in the sample.(b)Calculate the percentage expressed as the dihydrate, H2C2O4•2H2O.
A 20-tablet sample of soluble saccharin was treated with 20.00 mL of 0.08181 N AgNO3. After removal of the solid, titration of the filtrate and washings required 2.81 mL of 0.04124 N KSCN. Calculate the average number of milligrams of saccharin (205.17 g/mol) in each tablet.
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2.4414 g sample containing KCl, K2SO4and inert materials was dissolved in sufficient water to give 250 mL of solution. A Mohr titration of a 50 mLaliquot required 41.36 mL of 0.05818 M AgNO3. A second 50mL aliquot was treated with 40 mL of 0.1083M NaB(C6H5)4. The reaction is NaB(C6H5)4+ K=KB(C6H5)4(s)+ Na+. The solid was filtered, redissolved in acetone, and titrated with 49.98mLof the AgNO3solution according to the reaction KB(C6H5)4+ Ag= AgB(C6H5)4(s)+ K+. Calculate the percentage of KCl and K2SO4 in the sample.
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