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- A 35.0 mL of 1.00 x10–2 M Ca2+ was titrated with 0.0250 M EDTA at a pH of 10 and in the presence of 0.0500 M NH3. The formation constant for Ca2+–EDTA is 4.5 x1010. *αY4– = 0.35 at a pH 10; αCa2+ = 0.969 when the concentration of NH3 is 0.0500 M. Calculate the pCa (20pts) a.) Before equivalence point b.) At equivalence point c.) After equivalence pointSHOW SOLUTION A 0.7120 g of iron ore was brought into solution and passed through a Jones reductor. Titration of Fe(I1) produced required 39.21mL of 0.02086M KMnO4. Express the results of analysis in terms of (a) percent Fe (MM 55.85) and (b) percent Fe2O3 (MM 159.69). 5Fe+2 + MnO4 - + 8H+ → 5Fe+3 + Mn +2 + 4H2O Answer: 32.08% Fe, 45. 86% Fe2O325.00 mL aliquots of the solution in problem 1 are titrated with EDTA to the calmagite end point. A blank containing a small measured amount of Mg^2+ requires 2.12 mL of the EDTA to reach the end point. An aliquot to which the same amount of Mg^2+ is added requires 25.88 mL of the EDTA to reach the end point.
- You want to measure the concentration of carbonate (CO32-) in a mildly basic solution by using an EDTA back titration. CaCO3 has a Ksp of 5x10-9. You add 50.00 mL of 0.3484 M CaCl2 to 500.0ml of sample and filter the solution to remove the precipitate. You then take 250.0 mL of the filtered solution and titrate with 0.1786 M EDTA. You require 23.72 mL to reach the endpoint. What is the concentration of carbonate in the original sample?A solution is prepared by mixed 50.0 mL of 0.0400 M Ni2+ with 50.00 mL of 0.0400 M EDTA.Calculate the concentration of Ni 2+ after mixing. The mixture is buffered to a pH of 3.0. KNiY =4.2x1018, alpha4 =2.5x10 -11.A 25.00 mL unknown water sample is titrated with a standardized 0.0140 M EDTA solution. It is determined that 8.54 mL of EDTA is required to reach the end point. The blank titre was determined to be 1.50 mL. Determine the concentration of Calcium in the water sample in ppm. (mwt. of CaCO3 = 100.0892 g/mol).
- 50.00 mL of a solution containing iron (II) and iron (III) when titrated at pH2.0, 12.50 mL 0.01200 M EDTA and when titrated at pH 6.0, 28.50 mL EDTA is spent. Find the concentration of each solute in this solution in ppm. (Fe (AA) = 55,847 g / mol)3. A 50.00 mL solution containing Ni2+ and Fe2+ was treated EDTA to bind all the metal ions. After back-titration, the amount of EDTA used is 2.500 mmol. In another 50.00 mL solution was added pyrophosphate to mask the Fe2+ ions, and the solution required 25.00 mL of 0.04500 M EDTA. Calculate the ppm Fe (55.85 g/mol) in the solution.Given that the titration of Ca2+ and Mg2+ in a 50.00-mL sample of hard water required 22.35 mL of 0.01115 M EDTA. A second 50.00-mL aliquot was made strongly basic with NaOH to precipitate the Mg2+ as Mg(OH)2(s) . The supernatant liquid was titrated with 15.19 mL of the EDTA solution. Calculate the concentration in ppm of CaCO3 in the sample.
- Consider the titration of 25.0mL 0.020M of Co(NO3)2 with .0100M EDTA in a solution buffered to pH 10.00. Calculate pCo^2+ at the following volumes of added EDTA. a. 0.0 mL b. 20.0 mL c. 40.0 mL d. 49.0 mL e. 50.0 mL f. 50.1 mL g. 55.0 mL h. 60.0 mL9. The Tl in a 9.57 g sample of rodenticide was oxidized to the trivalent state and treated with an unmeasured excess of Mg/EDTA solution. The reaction is:Tl3++ MgY2-→TlY-+ Mg2+Titration of the liberated Mg2+required 12.77 mL of 0.03560 M EDTA. Calculate the percentage of Tl2SO4(MM= 504.80 g/mol) in the sample.The amount of calcium in physiologic fluids can be determined by a complexometric titration with EDTA. In one such analysis, a 0.100-mL sample of blood serum was made basic by adding 2 drops of NaOH and titrated with 0.00376 M EDTA, requiring 0.135 mL to reach the end point. Report the concentration of calcium in the sample as miligrams of Ca per 100 mL. (Ca = 40.078 amu).