A 28.93 mL aliquot of trimethylamine that has a concentration of 0.661 M will be titrated with 0.752 M HCl. Calculate the pH of the solution upon the addition of 32.6 mL of HCl. The Kb at the temperature of the experiment is equal to 6.11 x 10-5. Kw = 1.00 x 10-14
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A 28.93 mL aliquot of trimethylamine that has a concentration of 0.661 M will be titrated with 0.752 M HCl. Calculate the pH of the solution upon the addition of 32.6 mL of HCl. The Kb at the temperature of the experiment is equal to 6.11 x 10-5. Kw = 1.00 x 10-14
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- The student added universal indicator to the alkali solution at the start of the experiment. The indicator changed colour when the acid was added to the alkali. Explain how the student used this information and the pH scale to explain that this is a neutralisation reaction.A 28.7 mL aliquot of hypochlorous acid that has a concentration of 0.516 M will be titrated with 0.345 M NaOH. Calculate the pH of the solution upon the addition of 8.14 mL of NaOH. The Ka of the acid at the temperature of the experiment is 3.34 x 10-8.A 31.6 mL aliquot of hypochlorous acid that has a concentration of 0.204 M will be titrated with 0.783 M NaOH. Calculate the pH of the solution upon the addition of 33 mL of NaOH. The Ka at the temperature of the experiment is equal to 2.54 x 10-8. Kw = 1.00 x 10-14.
- The solubility of the related salt hydroxyapatite, Ca5(PO4)3OH (Ksp = 1 x 10-29), is affected by the pH of the water to which it is added. a) Calculate the number of grams of hydroxyapatite that will dissolve in 0.250L of neutral water. b) Calculate the number of grams of hydroxyapatite that will dissolve in 0.250L of water having a pH = 5.5Treatment of ammonia with phenol in the presence of hypochlorite yields indophenol, a blue product that absrobs light at 625 nm, which çan be used for the spectrophotometric determination of ammonia. gan OCF OH + NH3 indophenol anion To determinè the ammonia concentration in a sample of lake water, three samples are prepared. In sample A, 10.0 mL of lake water is mixed with 5 mL of phenol solution and 2 mL of sodium hypochlorite solution, and diluted to 25.0 mL in a volumetric flask. In sample B, 10.0 mL of lake water is mixed with 5 mL of phenol solution, 2 mL of sodium hypochlorite solution, and 2.50 mL of a 5.50 x 10-4 M ammonia solution, and diluted to 25.0 mL. Sample C is a reagent blank. It contains 10.0 mL of distilled water, 5 mL of phenol solution, and 2 mL of sodium hypochlorite solution, diluted to 25.0 mL. The absorbance of the three samples is then measured at 625 nm in a 1.00 cm cuvette. The results are shown in the table. Sample Absorbance (625 nm) A 0.361 B 0.608 C…The sample is dissolved in a mixture containing 10ml of distilled water and 20 ml of dilute HCL.After dilution, add 50 ml of water and 1g of potassium bromide.The mixture is titrated with 0.1N NaNO2 solution . 4 drops of atropine are used as an indicator .At the last point of the titration the color change from red to yellow. Write the reaction that takes place in the nitrometer and the sulfatiazole sample indicator?
- A 1.43 x 10-4 L solution of 45.7 mg/mL protein was analyzed using the Kjeldahl procedure. After digestion of the protein, the liberated NH3 was collected in 9.80 mL of 0235 M HCl. The unreacted acid required 10.75 mL of a standard solution of 0.0149 M NaOH for complete titration. Calculate the weight percent of nitrogen in the protein. (Molar mass N = 14.00674 g/mol)A mixture containing only KCl and NaBr is analyzed by the Mohr method. A 0.2125-g sample is dissolved in 25 mL of water and titrated to the Ag2CrO4 endpoint, requiring 25.63 mL of 0.1110 M AgNO3. A blank titration requires 1.15 mL of titrant to reach the same end point. Report the %w/w KCl and NaBr in the sample. KCl = 74.551 NaBr = 102.89A 500.0-mg sample of waste ground coffee beans used as fertilizer was digested and analyzed for nitrogen using the Kjeldahl method. After digestion, the distilled ammonia was collected in 100 mL of 0.5500 M boric acid. This solution required 19.60 mL of 0.5880 M HCI for titration to the methyl red end point. Calculate the %N in the fertilizer. Prelim Rxn: NH3 + H3BO3 --> NH4: H2BO3 Titration Rxn: NH : H2BO3 + HCI --> H3BO3 CI-
- A mixture containing only KCl and NaBr is analyzed by the Mohr method. A 0.2125- g sample is dissolved in 25 mL of water and titrated to the Ag2CrO4 end point, requiring 25.63 mL of 0.1110 M AgNO3. A blank titration requires 1.15 mL of titrant to reach the same end point. Report the %w/w KCl and NaBr in the sample.KCl = 74.551 NaBr = 102.89In the titration of 25.00 mL of a water sample, it took 19.990 mL of 3.480x 10−3 M EDTA solution to reach the endpoint. Calculate the number of moles of EDTA required to titrate the water sample. (enter your answer with 3 significant figures) The total hardness is due to one or a combination of Ca2+, Mg2+, and Fe2+ in your sample. It is convenient to express this hardness as though it was entirely due to Ca2+. Making this assumption, determine the number of moles of Ca2+ present in the bottled water sample titrated. (enter your answer with 3 significant figures) The total hardness is always listed in parts-per-million (ppm) of CaCO3 (or mg CaCO3 / Kg H2O). Since the density of water is 1.0 g/mL, one ppm would be the same as the number of mg of CaCO3 per liter of water. Determine the number of moles of CaCO3 present in the titrated sample of water, assuming that all the Ca2+ combines with CO32−. (enter your answer with 3 significant figures) Calculate the number of grams of…Treatment of ammonia with phenol in the presence of hypochlorite yields indophenol, a blue product that absrobs light at 625 nm, which can be used for the spectrophotometric determination of ammonia. ocr SHN + HO- indophenol anion To determine the ammonia concentration in a sample of lake water, three samples are prepared. In sample A, 10.0 mL of lake water is mixed with 5 mL of phenol solution and 2 mL of sodium hypochlorite solution, and diluted to 25.0 mL in a volumetric flask. In sample B, 10.0 mL of lake water is mixed with 5 mL of phenol solution, 2 mL of sodium hypochlorite solution, and 2.50 mL of a 5.50 x 104 M ammonia solution, and diluted to 25.0 mL. Sample C is a reagent blank. It contains 10.0 mL of distilled water, 5 mL of phenol solution, and 2 mL of sodium hypochlorite solution, diluted to 25.0 mL. The absorbance of the three samples is then measured at 625 nm in a 1.00 cm cuvette. The results are shown in the table. Absorbance Sample (625 nm) A 0.401 В 0.648 0.045 What…