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- 1ml of 1mg/ml of diluted aspirin powder solution is titrated by 0.005M NaOH. Before titration, 5 ml ethanol, 14 ml CO2 free water, and 4 drops bromothymol blue indicator is added into diluted aspirin powder solution. Mass of 0.005M NaOH used is obtained by weight titration = 1.1384g . [Molar mass of aspirin =180.15]; for solution, 1g=1mL at 25 degree Celcius. Calculate the purity of the powder.A sample is analyzed for chloride by the Volhard method. From the following data, calculate the percentage of chloride present:Weight of sample = 6.0000 g dissolved and diluted to 200 mLAliquot used = 25.00 mL AgNO3 added = 40.00ml of 0.1234MKSCN for back titration = 13.20ml of 0.0930MA 0.512 g sample of CaCO3 is dissolved in 12 M HCl and the mixture is diluted to 250 mL. A small amount of MgCl2 solution is added to a 25 mL aliquot of the solution., and the mixture is titrated with (ethylenediaminetetraacetic acid) EDTA to the Eriachrome Black T (with MgCl2 is used as indicator) end point. The mixed solution requires 28.55 mL of the EDTA solution to reach the end point. A similar amount of MgCl2 solution requires 2.60 mL to reach the endpoint. A 100-mL sample of hard water is titrated with 22.4 mL of the EDTA solution created above. The same amount of MgCl2 is added as previously and the total volume of EDTA solution required is 22.44 mL. a) Assume all of the Ca2+ in the water comes from CaCO3. How many moles of CaCO3 are in 1 L of water? How many grams of CaCO3 are in 1 L of water? c) If 1 ppm CaCO3 = 1 mg/liter, what is the water hardness in ppm CaCO3 ?
- A 0.512 g sample of CaCO3 is dissolved in 12 M HCl and the mixture is diluted to 250 mL. A small amount of MgCl2 solution is added to a 25 mL aliquot of the solution., and the mixture is titrated with (ethylenediaminetetraacetic acid) EDTA to the Eriachrome Black T (with MgCl2 is used as indicator) end point. The mixed solution requires 28.55 mL of the EDTA solution to reach the end point. A similar amount of MgCl2 solution requires 2.60 mL to reach the endpoint. A 100-mL sample of hard water is titrated with 22.4 mL of the EDTA solution created above. The same amount of MgCl2 is added as previously and the total volume of EDTA solution required is 22.44 mL. a) What volume of EDTA is used in titrating the Ca2+ in the hard water? b) How many moles of EDTA are there in that volume? c) How many moles of Ca2+ are in 100 mL of water?A 0.512 g sample of CaCO3 is dissolved in 12 M HCl and the mixture is diluted to 250 mL. A small amount of MgCl2 solution is added to a 25 mL aliquot of the solution., and the mixture is titrated with (ethylenediaminetetraacetic acid) EDTA to the Eriachrome Black T (with MgCl2 is used as indicator) end point. The mixed solution requires 28.55 mL of the EDTA solution to reach the end point. A similar amount of MgCl2 solution requires 2.60 mL to reach the endpoint. A 100-mL sample of hard water is titrated with 22.4 mL of the EDTA solution created above. The same amount of MgCl2 is added as previously and the total volume of EDTA solution required is 22.44 mL. How many moles of Ca2+ are in 100 mL of water?A 0.512 g sample of CaCO3 is dissolved in 12 M HCl and the mixture is diluted to 250 mL. A small amount of MgCl2 solution is added to a 25 mL aliquot of the solution., and the mixture is titrated with (ethylenediaminetetraacetic acid) EDTA to the Eriachrome Black T (with MgCl2 is used as indicator) end point. The mixed solution requires 28.55 mL of the EDTA solution to reach the end point. A similar amount of MgCl2 solution requires 2.60 mL to reach the endpoint. How many milliliters of EDTA are needed to titrate the Ca2+ ion in the sample? How many moles of EDTA are there in the volume calculated for the question above? What is the molarity of the EDTA solution?
- Chemistry A 20 mL solution containing both Ca2+ and Mg2+ cations is diluted in to 100 mL. When 10 mL of this solution is taken and titrated with 0.05 M EDTA at pH = 10 in the presence of Erio – T indicator, the consumption is found as 12 mL. A new 10 mL was taken from the same solution and (NH4)2C2O4 is added on it and then formed the precipitate is filtered. The filtered solution is titrated with the same EDTA solution and the consumption is found as 3 mL. So find the Ca2+ and Mg2+ amounts in the main sample solution in terms of mg/L.175.0 mL of 0.2015 M nitric acid was added to 1.4213 g of impure calcium carbonate sample. The excessacid was back titrated with 72.5mL of 0.1152 M sodium hydroxide to reach the endpoint. Calculate for thepercent calcium carbonate? (MM of CaCO3= 100.0869 g/mol)2HNO3 + CaCO3 → Ca(NO3)2 + H2O + CO2HNO3 + NaOH → NaNO3 + H2O10mL of a 10% by weight MgCl2 solution (density = 1.1 g / mL) is precipitated as magnesium ammonium phosphate after necessary processes, filtered and washed. The precipitate is dissolved in 50mL 1M HCl and excess acid is titrated with 2.0M NaOH solution in the presence of methyl orange. Find the NaOH consumption (Mg = 24,3g / mol, Cl = 35,5g / mol)
- A 1.067g sample of magnesium oxide of 84.736% were treated with 50mL of 1.017 N Sulfuric Acid, and a 5.195mL volume of sodium hydroxide is required in the back titration. 1. What is the equivalent weight consumed by the acidic titrant? A. 5.281 g-meq B. 55.150 g-meq C. 5.723 g-meq D. 50.850 g-meq 2. What is the difference of milliequivalent weight consumed in the reaction? A. 45.127g-meq B. 56.573g-meq C. 50.850g-meq D. 55.150g-meq 3. What is the amount (in mg) of the analyte that is equivalent to 1 milliliter of the tirant at its equivalence point? A. 22.060mg B. 40.680mg C. 44.12mg D. 20.340mgHow many grams of potassium napthenate (KC11H7O2; MW=210.27 g/mole; Kb=85x10-10) must be added to 50.0 ml of 0.4 M Naphthenic acid (HC11H7O2; Ka=5.4x10-5) to obtain a buffer with pH 5.0?A solution of Na3AsO4 is added dropwise to a solution that is 0.0347 M in Cu2+ and 0.000309 M in Ag+.The Ksp of Cu3(AsO4)2 is 7.95e-36.The Ksp of Ag3AsO4 is 1.03e-22.(a) What concentration of AsO43- is necessary to begin precipitation? (Neglect volume changes.)[AsO43-] = __________ M.(b) What is the concentration of AsO43- when the second cation begins to precipitate?[AsO43-] = _______ M.