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- A pharmacist is conducting an analysis of 2.7 g sodium bicarbonate and titrated with 26.05 mL of 0.9987 N HCl. What is the potency or percent assay of sodium bicarbonate? Did it pass the USP requirement? ROUND-OFF TO 2 DECIMAL PLACES. Answer = _% Disposition (Passed or Failed) = _A mixture of NaOH (M.wt = 40) and Na2CO3 (M.wt = 106) is titrated with 0.3 M HCl, requiring 30 mL for phenolphthalein end point and an additional 15 mL to reach the modified methyl orange end point, How many milligrams of NaOH and Na2CO3 are in the mixture *A local company sent you their green alternative for window cleaner to be tested for percent (w/v) acetic acid content. For your experiment, you first standardized your NaOH titrant with 0.8053 g of (99.80 % purity) KHP. You used 40.60 mL of NaOH for your standardization. After that you then analyzed a 50.00 mL sample and found that you needed 33.20 mL NaOH to reach the end point. Find the % (w/v) acetic acid.
- Q3/ A) 0.63 g of a sample containing Na CO,, NaHCO; and inert impurities is titrated with 0.2 M HCI, requiring 17.2 ml. to reach the phenolphthalein end point and a total of 43.5 mL to reach the modified methyl orange end point, How many grams Na CO, and NaHCO, are in the mixture?What is the mass of aectylsaliyclic acid (HC9H7O4, MW = 180.154 g/mol) from an aspirin tablets by titrating with NaOH. One tablet is dissolved in 25 mL of 50% ethanol solution. The resulting solution needed 13.41 mL of 0.2069 standardized NaOH titrant in order to achieve the phenolphthalein endpoint. What is the mg of aectylsaliyclic acid in the asiprin tablet?A 0.8250 g of soda ash sample was dissolved in distilled H O and diluted to 250-mL using a volumetric flask. A 50.00 mL aliquot portion of the dilute soda ash solution needed 4.50 mL of 0.1200 M HCl to reach the phenolphthalein end point and 10.00 mL to reach the methyl orange end point. What is the %Na CO (%w/w) present in the soda ash sample? FW Na CO = 105.99 g/mol (Hint: Use dilution factor)
- A local company sent you their green alternative for window cleaner to be tested for percent (w/v) acetic acid content. For your experiment, you first standardized your NaOH titrant with 0.8053 g of (99.80 % purity) KHP. You used 40.60 mL of NaOH for your standardization. After that you then analyzed a 10.00 mL sample and found that you needed 43.20 mL NaOH to reach the end point. Summary of results: Standardization Sample analysis KHP Weight (g) 0.8053 g Volume of sample 50.00 mL Purity 99.80% NaOH (mL) used 33.20 mL NaOH (mL) used 40.60 mL Determine the following: Molarity of NaOH % (w/v) acetic acid5. 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 local company sent you their green alternative for window cleaner to be tested for percent (w/v) acetic acid content. For your experiment, you first standardized your NaOH titrant with 0.8053 g of (99.80 % purity) KHP. You used 40.60 mL of NaOH for your standardization. After that you then analyzed a 50.00 mL sample and found that you needed 33.20 ml NaOH to reach the end point. Summary of results: Standardization Sample analysis KHP Weight (g) Purity NaOH (mL) used Determine the following: 0.8053 g Volume of sample 50.00 mL 99.80% NaOH (mL) used 33.20 mL 40.60 mL a. Molarity of NaOH b. % (w/v) acetic acid
- A 0.3005g sample containing Na2CO3 and NaOH requires 11.5 mL of 0.1055 M HCl to reach the phenolphthalein end point and another 7.9 mL of HCl to reach the methyl orange end point. Calculate the % composition of Na2CO3 and NaOH in the sample.A mixture containing only KCl and NaBr is analyzed by the Mohr Method. A 0.3172-g sample is dissolved in 50 mL water and titrated to the Ag2CrO4 endpoint, requiring 36.85 mL of 0.1120 M AgNO3. A blank titration requires 0.71 mL of titrant to reach the same endpoint. Report the % (w/w) KCl and NaBr in the sample. [Answer should be 82.41 % (w /w)]Aspirin powder = 0.8110g MW of Aspirin = 180g.mol-1 Volume of 0.5N HCl consumed in back titration = 23.50mL Volume of 0.5N HCl consumed in blank titration = 44.50mL Percent purity (USP/NF) = Aspirin tablets contain NLT 90.0% and NMT 110.0% of the labeled amount of aspirin (C9H8O4) What is the calculated weight (in grams) of pure aspirin?..