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Calculat the Molarity of Iodine per trial using the table
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- If 55 mg of lead(II) sulfate is placed in 250 mL of pure water, does all of it dissolve? If not, how much dissolves?Part A: Standardization of a Sodium Hydroxide Solution Titration 1 Titration 2 Titration 3 Mass of 125 mL flask 45.849g 46.715g 44.953g Mass of flask and KHP 46.849g 47.745g 46.003g Initial buret reading (mL) 0.5 ml 0.5 ml 0.5 ml Final buret reading (mL) 27.8 ml 26.5 ml 26.7 ml Volume of NaOH used (mL) 45.11 ml 45.06 ml 45.14 ml Calculations Titration 1 Titration 2 Titration 3 Moles of KHP Moles of NaOH Molarity of NaOH Average Molarity of NaOH: _______________TITRIMETRIC DATA SAMPLE: CANE VINEGAR % acidity in label: 4.5% %purity of KHP: 99.80% FORMULA WEIGHT of KHP: 204.22 g/mol STANDARDIZATION OF NaOH SOLUTION TRIAL 1 TRIAL 2 TRIAL 3 Weight of KHP, g 0.1012 0.1004 0.09987 Initial volume of NaOH, mL 5.00 10.01 Final volume of NaOH, mL 4.95 9.99 14.93 Molarity of NaOH Average Molarity of NaOH ANALYSIS OF ACETIC ACID IN A VINEGAR SAMPLE TRIAL 1 TRIAL 2 TRIAL 3 Volume of vinegar, mL Initial volume of NaOH, mL 1.00 1.00 1.00 14.98 22.90 30.87 Final volume of NaOH, mL 22.84 30.77 38.75 Molarity of acetic acid Average molarity of acetic acid ANALYSIS OF CARBONIC ACID IN A SODA SAMPLE TRIAL 1 TRIAL 2 TRIAL 3 Volume of soda, mL 20.0 20.0 20.0 Initial volume of NaOH, mL 20.20 22.05 23.93 Final volume of NaOH, mL 22.03 23.89 25.8 Molarity of carbonic acid Average molarity of carbonic acid
- Calculate the mass of acetic acid in 100ml of both samples in vinegar. given: NaOH vs CH3COOH Burette solution is NaOH and the pipette solution is 5.0 mL of Vinegar Titration Initial burette reading Final burette reading Volume of NaOH consumed Average volume of NaOH Approximate 0.0 mL 20.1 mL 20.1 mL 20.3 mL Titration 1 0.0 mL 20.9 mL 20.9 mL Titration 2 0.0 mL 20.0 mL 20.0 mL To find the average volume Average volume = 20.1 mL + 20.9 mL + 20.0 mL320.1 mL + 20.9 mL + 20.0 mL3 = 20.3 mL Concentration of Vinegar = Volume of NaOH * Concentration of NaOHVolume of vinegarVolume of NaOH * Concentration of NaOHVolume of vinegar = 20.3 mL * 0.0647 M5.0 mL20.3 mL * 0.0647 M5.0 mL = 0.2627 M Concentration of NaOH = Volume of H2SO4 * Concentration of H2SO4Volume of NaOHVolume of H2SO4 * Concentration of H2SO4Volume of NaOH = 10.0 mL…Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 2.29 1.41 1.95 Molarity of NaOH (M) 0.100 0.100 0.100 Volume of vinegar sample (mL) 5.00 5.00 5.00 Final burette reading (mL) 50.37 49.39 49.84 Table 2. Titration data Trial 1 Trial 2 Trial 3 Initial burette reading (mL) 2.29 1.41 1.95 Molarity of NaOH (M) 0.100 0.100 0.100 Volume of vinegar sample (mL) 5.00 5.00 5.00 Final burette reading (mL) 50.37 49.39 49.84 Expected color at end point Volume of NaOH used (mL) 48.08 47.98 47.89 Compute for the ff: a. Average moles of acetic acid (mol)? b. Average molarity of acetic acid (M)? c. Average molarity of acetic acid (M)?Molarity of Na2S2O3 (stock solution). 0.100 M Volume of diluted bleach titrated (taken from the volumetric flask). 10ml Initial Burette Reading. 14.80 Final Burette Reading 25.60 Volume of Na2S2O3 used in the titration (mL). 10.80 Moles of Na2S2O3 used in the titration. 1.71 Moles of S2O3–2 used. 1.66 Moles of I2 initially present in titration mixture 0.83 Moles of NaOCl in the diluted bleach titrated. 0.415 Volume of bleach titrated. 1ml Mass of commercial bleaching solution titrated ?
- Molarity of Na2S2O3 (stock solution). 0.100 M Volume of diluted bleach titrated (taken from the volumetric flask). 10ml Initial Burette Reading. 14.80 Final Burette Reading 25.60 Volume of Na2S2O3 used in the titration (mL). 10.80 Moles of Na2S2O3 used in the titration. 1.71 Moles of S2O3–2 used. 1.66 Moles of I2 initially present in titration mixture 0.83 Moles of NaOCl in the diluted bleach titrated. 0.415 Moles of OCl– in the diluted bleach titrated. ?Molarity of Na2S2O3 (stock solution). 0.100 M Volume of diluted bleach titrated (taken from the volumetric flask). 10ml Initial Burette Reading. 14.80 Final Burette Reading 25.60 Volume of Na2S2O3 used in the titration (mL). 10.80 Moles of Na2S2O3 used in the titration. 1.71 Moles of S2O3–2 used. 1.66 Moles of I2 initially present in titration mixture 0.83 Moles of NaOCl in the diluted bleach titrated. 0.415 Volume of bleach titrated. 1ml Mass of NaOCl in commercial bleach titrated. ?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…
- Standardization of 1N H2SO4 Titration 1 Titration 2 Titration 3 Titration 4 Mass of flask and Sodium Carbonate 1.5067 g 1.5098 g 1.5076 g 1.5077 Final buret reading (mL) 22 mL 22mL 21.5 mL 22.5 mL Volume of Sulphuric acid used (mL) 30 mL 30 mL 30 mL 30 mL What is the result of my titration?What is the percentage of total acid expressed as acetic acid in a sample of vinegar if 4.0g of vinegar requires 29.5ml of 0.153N KOH for an endpoint with phenolphthalein indicator? (molar mass of acetic acid: HC2H3O2 = 60.05g/mol a. 3.995% b. 6.776% c. 8.650% d. 12.68%Which calcium solution is saturated? 6.0x10-5M CaCO3 (Ksp=4.5x10-9) 0.010 M Ca(OH)2 (Ksp=6.5x10-6) 7.2x10-7M Ca3(PO4)2 (Ksp=2.0x10-29) 2.0x10-4M CaF2 (Ksp=3.9x10-11)