The active ingredient in aspirin is acetylsalicylic acid. A 2.51-g sample of acetylsalicylic acid required 27.36 mL of 0.5106 M NaOH for complete reaction. Addition of 13.68 mL of 0.5106 M HCl to the flask containing the aspirin and the sodium hydroxide produced a mixture with pH = 3.48. Determine the molar mass of acetylsalicylic acid and its Ka value. State any assumptions you must make to reach your answer.

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2 Jul 2023

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A student used standard solutions of aspirin in a FeCl₃-KCl-HCl mixture to plot a graph of molarity versus absorbance for diluted aspirin solutions of known concentration. The student determined the slope of the graph to be 4,687 M^-1cm^-1. Next the student measured out 0.288 grams of a headache medicine tablet and dissolved it in 10.0 ml of NaOH and then added enough water to make a 100 ml solution. Five ml of this solution was then added to another 100 ml flask and diluted to the mark with enough FeCl₃-KCl-HCl mixture. The solution was then measured forabsorbency as 0.306

Calculate the molarity of the diluted solution determined from the absorbency and the slope. Then use the molarity of the original solution, its volume and molar mass to calculate the grams of pure acetylsalicylic acid.

The concentration of a hydrochloric acid solution is approximately 0.25 M but the solution must be standardized by titration against sodium hydroxide to determine the precise concentration. If the molarity of the sodium hydroxide solution is 0.2053 M and the volume required to reach the end point in the titration against 25.00 mL of the hydrochloric acid is 29.57 mL, what is the molarity of the hydrochloric acid solution?Molarity of HCl = MUsing an analytical balance, 0.1682 grams of the nickel(II) ammine complex was weighed by difference and placed in an Erlenmeyer flask. Exactly 25 mL of the standardized hydrochloric acid solution was added to neutralize the ammonia in the complex. The excess hydrochloric acid remaining in the Erlenmeyer flask was titrated against the sodium hydroxide. 14.55 mL of sodium hydroxide was required to reach the end point.How many moles of sodium hydroxide were required?Moles of NaOH required = molThis equals the number of moles of excess HCl remaining in the flask. How many moles of HCl are there in 25.00 mL of the hydrochloric acid solution, and therefore how many moles of HCl were required to neutralize the ammonia in the nickel(II) ammine complex?Moles of HCl in 25.00 mL = molMoles of HCl required to neutralize the ammonia = molThis equals the number of moles of ammonia in your sample of the complex. Multiply by 17.031 to obtain the mass of ammonia. Subtract this from the mass of the complex (0.1682 grams) to obtain the mass of nickel ions and nitrate ions in the sample.Mass of ammonia in the sample = gMass of Ni(NO3)2 in the sample = gDivide this mass by 182.70 to obtain the number of moles of Ni(NO3)2 in the sample and then calculate the mole ratio of Ni(NO3)2 to ammonia in the complex.Moles of Ni(NO3)2 in the sample = molMole ratio of ammonia to Ni(NO3)2 in the complex =