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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?
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- A 0.5027 g of KHP (KC8H5O4; MW = 204.22 g/mol) required 11.90 mL of NaOH titrant to reach the phenolphthalein endpoint. Calculate the standardized concentration of NaOH.42.5 mL of 1.3M KOH are required to neutralize 50.0 mL of H2SO4. This sulfuric acid is further used to standardize NaOH solution. 20mL aliquot of the NaOH solution is obtained and 2 drops of phenolphthalein is added. The initial reading on the buret is 13.2 mL. After the titration, the exact normality of the NaOH solution is found to be 1.254N. (a) What is the molarity of H2SO4? (b) How much of the sulfuric acid is used? (c) What is the final reading on the buret?A solution contains NaHCO3 (84.01 g/mol), Na2CO3 (105.99 g/mol), and NaOH (40.00 g/mol), either alone or in permissible combination. Titration of a 25.0-mL portion to a phenolphthalein end point requires 29.75 mL of 0.1205 M HCl. A second 25.0-mL aliquot requires 36.56 mL of the HCl when titrated to a bromocresol green end point. What is the general composition of the solution? Calculate the no. of mg of each solute per mL of solution.
- 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.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 acidA 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. Determine the following: Molarity of NaOH % (w/v) acetic acid
- Titration of a 0.824 g of 99.99% KHP (204.23 g/mol) with phenolphthalein required 18.3 mL of NaOH (40.00 g/mol) solution to reach the end point. The same titrant was used to analyze an impure acetic acid (CH3COOH, 60.06 g/mol) solution. A 10.0 mL aliquot of the sample required 12.9 mL of the titrant. What substance served as the primary standard?Which served as the indicator in the titration?What is the color at the end point?What is the concentration of the titrant?What is the molar concentration of the acetic acid solution?If 20 mL of a hydrochloric acid solution (HCl) required 18 mL of a sodium hydroxide in a titration and 35mL of the said base is equivalent to 0.8323 g of potassium bipthalate (KHC8H4O4), what would be the normality of the sodium hydroxide solution used in the titration? Atomic Weights :Sodium - 22.99Oxygen - 15.999Hydrogen - 1.008Potassium - 39.098Chlorine - 35.45 Enter numerical value only; round off to 4 decimal placesA 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
- Formulate a hypothesis regarding the solubility of aspirin at different pH.The experimentA) Three teaspoons of water (approx. 15 ml) were added to one tablet said to contain 300 mg of aspirin. Fizzingwas observed. Most of the tablet dissolved, but there were some solid particles. By heating the mug in amicrowave for 10 second increments until the water came to the boil (approx. 3x), all of the solid particlesdissolved. The solution was left to cool to room temperature and then placed in a fridge and NOTHINGHAPPENED. Try this yourself if you can spare two aspirin tablets, your results might look different.• Questions to ask:1. What might the fizzing bubbles be?2. Can you give a chemical explanation?3. Can you write a chemical reaction equation with aspirin reacting with something to give a gas and aspirinin another form?4. What might be the formulation (what the manufacturer mixes with aspirin in making the tablet) “trick”for aspirin to improve solubility?5. How does this compare with…It is found that 24.68 mL of 0.1165 M NaOH is needed to titrate 0.2931 g of an unknown acid to the phenolphthalein end point. Calculate the equivalent mass of the acid.A mixture contains Na2CO3, NaOH and inert matter. A sample weighing 1.500 g requires 28.85 ml of 0.5000N HCl to reach the a phenolphthalein endpoint and an additional 23.85 ml to reach the methyl orange end point. What are the percentages Na2CO3 and NaOH?