A 0.6006-g sample of Ni/Cu condenser tubing was dissolved in acid and diluted to 100 mL in a volumetric flask. Titration of both ions in a 25.0ml aliquot of this solution required 45.81 mL of 0.052285M EDTA. Mercaptoacetic acid and NH3 were then introduced; production of the Cu complex with the former resulted in the release of an equivalent amount of EDTA, which required 22.85ml titration with 0.07238M Mg2+. Calculate the %Ni (59.693 g/mol) in the alloy.
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A 0.6006-g sample of Ni/Cu condenser tubing was dissolved in acid and diluted to 100 mL in a volumetric flask. Titration of both ions in a 25.0ml aliquot of this solution required 45.81 mL of 0.052285M EDTA. Mercaptoacetic acid and NH3 were then introduced; production of the Cu complex with the former resulted in the release of an equivalent amount of EDTA, which required 22.85ml titration with 0.07238M Mg2+. Calculate the %Ni (59.693 g/mol) in the alloy.
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- A 0.9352g sample of ore containing Fe³+, Al³+ and Sr²+ was dissolved and made up to 500.00 mL. The analysis of metals was performed using complexation volumetry. Initially, an aliquot of 50.00 mL had its pH adjusted to 1.0 and titrated with a standard 0.03145 mol/L EDTA solution, requiring 6.95 mL to reach the end point. Subsequently, another 25.00 mL aliquot was buffered at pH=5 and titrated with the same EDTA solution, requiring 6.24 mL to reach the end point. Finally, a third aliquot of 25.00 mL was titrated at pH=11, requiring 11.10 mL of the same EDTA solution to complete the titration. Given the molar masses: Fe=55.845 g/mol; Al-26.982 g/mol and Sr-87.620 g/mol. a) Determine the percentage of each of the metals in the sample. b) Explain why the change in pH allows the determination of the three ions in this sample.A 0.60004g sample of Ni/Cu condenser tubing was dissolved in acid and diluted to 100.0mL in a volumetric flask. Titration of both cations in a 25.00mL aliquot of this solution required 45.81mL of 0.05285M EDTA. Mercaptoacetic acid and NH3 were then introduced; production of the Cu complex with the former resulted in the release of an equivalent amount of EDTA, which required 22.85mL titration with 0.07238M Mg2+. Calculate the percentages of Cu (At. Mass = 63.55) and Ni (At. Mass = 58.69) in the alloy. Please provide the complete solution.a 10 ml sample containing fe3+ and cu2+ was diluted to 100.0 mL solution. A 20 mL aliquot required 25 mL of 0.00500 M EDTA for complete titration. A separate 20 mL aliquot of the diluted sample was treated with NH4F to protect the Fe3+. Then the Cu2+ was treated with thiourea. Upon addition of 25.0 ml of 0.00500 M EDTA, the Fe3+ was liberated from its fluoride complex and formed an EDTA complex. The excess EDTA required 21.5 ml of 0.0180 M Pb2+ to reach an endpoint, using xylenol orange. (a) identify the masking agent (b) identify the auxiliary complexing agent (c) calculate the molar concentration of the fe3+ in the original sample (d) calculate the molar concentration of the cu2+ in the original sample
- A 1.509-g samle of a Pb/Cd alloy was dissolved in acid and diluted to exactly 250.0 mL in a volumetric flask. A 50.00-mL aliquot of the diluted solution was brought to a pH of 10.0 with an HCN/NaCN buffer, which also served to mask the Cd2+; 11.56mL of the EDTA solution were needed to titrate the Pb2+. Calculate the percentage of Pb and Cd in the sample.3. An antacid tablet, weighing 1.25 g, was dissolved in a 1.0 L volumetric flask to allow for determination of calcium carbonate and magnesium carbonate contents. After preparing the solution, an aliquot of 10.0 ml was transferred to an Erlenmeyer flask containing a buffer solution with pH = 10. This aliquot was then titrated with EDTA 0.0040 mol/L and the average volume spent was 15.01 mL. To measure calcium after precipitation. After the magnesium was fractionated, a second 10.00 mL aliquot of the stock solution was transferred to an Erlenmeyer flask, and the pH of the present solution was adjusted to a value of approximately 13. In EDTA titration, the volume consumed for observation ofthe end point was equal to 13.03 mL. Based on this information,a) Outline the two steps involved, representing the related reactions./5b) Calculate the concentrations of CaCO3 and MgCO3 present in the initialsolution./6c) Calculate the masses of CaCO3 and MgCO3 present in the pellet./6d) Calculate the…A second calibration standard solution of an iron(III) salicylate complex was prepared in two steps. First, 10.0 mL of a 0.100 M stock solution was added to 90.0 mL of solvent to make 100.0 mL of the first calibration standard solution and, secondly, 80 mL of that first calibration solution plus 20.0 mL of solvent were mixed to make the second calibration standard. What is the concentration of the second calibration standard solution?
- A 30-mL portion of a solution containing Ca2+ and Mg2+ was titrated with 28.19 mL of 0.213 M EDTA at pH 10. Another 30-mL aliquot of the same Ca-Mg mixture was treated with NaOH to make the solution strongly alkaline and precipitate Mg(OH)2. This solution was then titrated with the same EDTA solution. What would be the required EDTA volume (in mL) to reach the endpoint of the second aliquot if it was found that there was 0.061 M of Mg2+ in the sample?An antacid tablet, weighing 1.25 g, was dissolved in a 1.0 L volumetric flask to allow for determination of calcium carbonate and magnesium carbonate contents. After preparing the solution, an aliquot of 10.0 ml was transferred to an Erlenmeyer flask containing a buffer solution with pH = 10. This aliquot was then titrated with EDTA 0.0040 mol/L and the average volume spent was 15.01 mL. To measure calcium after precipitation After the magnesium was fractionated, a second 10.00 mL aliquot of the stock solution was transferred to an Erlenmeyer flask. and the pH of the present solution was adjusted to a value of approximately 13. In EDTA titration, the volume consumed for observation of the end point was equal to 13.03 mL. Based on this information, (a) Outline the two steps involved, representing the related reactions. (b) Calculate the concentrations of CaCO3 and MgCO3 present in the initial solution. (c) Calculate the masses of CaCO3 and MgCO3 present in the pellet. (d) Calculate the…. An antacid tablet, weighing 1.25 g, was dissolved in a 1.0 L volumetric flask to allow for determination of calcium carbonate and magnesium carbonate contents. After preparing the solution, an aliquot of 10.0 ml was transferred to an Erlenmeyer flask containing a buffer solution with pH = 10. This aliquot was then titrated with EDTA 0.0040 mol/L and the average volume spent was 15.01 mL. To measure calcium after precipitation. After the magnesium was fractionated, a second 10.00 mL aliquot of the stock solution was transferred to an Erlenmeyer flask, and the pH of the present solution was adjusted to a value of approximately 13. In EDTA titration, the volume consumed for observation of the end point was equal to 13.03 mL. Based on this information, a) Outline the two steps involved, representing the related reactions./5 b) Calculate the concentrations of CaCO3 and MgCO3 present in the initial solution./6 c) Calculate the masses of CaCO3 and MgCO3 present in the pellet./6 d) Calculate…
- Titration of a 25.00 mL sample of mineral water (containing both Ca2+ and Mg2+) at pH 10 required 19.18 mL of 0.01125M EDTA solution. Another 25.00 mL aliquot of the same mineral water was rendered strongly alkaline to precipitate the magnesium. Titration with a calcium-specific indicator required 14.92 mL of the EDTA solution. Calculate the mass of MgCO3 (FW= 84.314 g/mol) in the mineral water in mg (keep 2 decimals).A cyanide solution with a volume of 10.33 mL was treated with 22.00 mL of Ni2+ solution (containing excess Ni2+) to convert the cyanide into tetracyanonickelate(II): 4CN−+Ni2+⟶Ni(CN)(2−)/4 The excess Ni2+ was then titrated with 11.85 mL of 0.01303 M ethylenediaminetetraacetic acid (EDTA): Ni2++EDTA4−⟶Ni(EDTA)2− Ni(CN)(2−)/4 does not react with EDTA. If 38.80 mL of EDTA were required to react with 30.47 mL of the original Ni2+ solution, calculate the molarity of CN− in the 10.33 mL cyanide sample. [CN-] = ________________________________MThe bismuth (AW 208.98) in 0.7405 g of an alloy was precipitated as BiOCl (FW 260.43) and separated from the solution by filtration. The washed precipitate was dissolved in nitric acid and treated with 19.45 mL of 0.1498 M AgNO3, causing the precipitation of AgCl. The excess AgNO3 required 13.29 mL of 0.1008 M KSCN for titration. Calculate the % Bi in the sample.