Calculate pFe at cach of the points in the titration of 25.00 mL of 0.02253 M Fe* by 0.03592 M EDTA at a pH of 6.00. The values for log Kr and ay can be found in the chempendix. 12.00 mL pFe the equivalence point, V, 19.50 ml. pre TOOLS x10
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- The equilibrium constant for the conjugate acid-base pair HIn+H2OH3O++In is 8.00 10-5. From the additional information in the following table, (a) calculate the absorbance at 430 nmand 600 nm for the following indicator concentrations: 3.00 10-4M,2.00 10-4M, 1.00 10-4M, 0.500 10-4 M, and 0.250 10-4M. (b) plot absorbance as a function of indicator concentration.A Fajans titration of a 0.7908-g sample required 45.32 mL of 0.1046 M AgNO3 . Express the results of this analysis in terms of the percentage of BaCl2 * H2O. (Use a MW value in 4 decimal places)1. 1093-g sample of impure Na2CO3 was analyzed by residual precipitimetry. After adding 50.00 mL of 0.06911 M AgNO3, the sample was back-titrated with 0.05781 M KSCN, requiring 27.36 mL to reach the endpoint. The percentage Na2CO3 (MW = 106.0 g/mole) in the tested sample is ________ % ? Note: Express final answer using least number of significant figures. 2. The alkalinity of natural waters is usually controlled by OH- (MW = 17.01 g/mole), CO3-2 (MW = 60.01 g/mole), and HCO3- (MW = 61.01 g/mole), which may be present singularly or in combination. Titrating a 10.0-mL sample to a phenolphthalein endpoint requires 38.12 mL of a 0.5812 M solution of HCl, and an additional 18.67 mL of the same titrant to reach the methyl orange endpoint. The composition of the sample is _________% CO3-2 and ___________ % OH- Note: Express final answers using least number of significant figures.
- 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?Solve the following problem: Calamine, which is used to relieve skin irritations, is a mixture of zinc and iron oxides. A 1.056 g sample of dry calamine was dissolved in acid and diluted to 250.0 mL. Potassium fluoride was added to a 10.00 mL aliquot of the diluted solution to mask the iron. After adjusting the pH, the Zn2+ consumed 38.37 mL of EDTA 0.01133 M. A second aliquot of 50.00 mL was buffered and titrated with 2.30 mL of a solution of ZnY2- 0.002647 M: Fe3+ + ZnY2- → FeY- + Zn2+ Calculate the percentages of ZnO and Fe2O3 in the sample.
- 2,5 ml volume has taken from an “hypothetic” solution which includes (3+) Sb and (3+) Fe and at the titration with 0.1004 N KMnO4, the wasted amount has found as 16,4 mL. The other 2,5 mL that has taken, has reduced with Zn after that, this 2,5 mL solution has titrates with the same KMnO4 solution solution and the wasted amount is 26,5mL. With these datas find the %concentrations of the ions at the solution.A 49.10 mL aliquot from a 0.500 L solution that contains 0.530 g of MnSO4 (MW=151.00 g/mol) required 41.6 mL of an EDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCO3 ( MW=100.09 g/mol) will react with 1.53 mL 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.
