============================================================================================================================
Please help! Will rate. A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.5...
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 51.7 mL of 0.0500M EDTA. Titration of the excess unreacted EDTA required 15.8 mL of 0.0190 M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN−. Titration of the newly freed EDTA required 13.7 mL of 0.0190 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 51.7...
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 57.2 mL of 0.0700 M EDTA. Titration of the excess unreacted EDTA required 14.6 mL of 0.0150 M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN“. Titration of the newly freed EDTA required 11.3 mL of 0.0150 M Ca2+. What are the concentrations of Cd2+ and Mn2+ in the original solution? concentration: M Mn2+ concentration: M Cd2+
A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 42.5 mL of 0.0700 M EDTA. Titration of the excess unreacted EDTA required 16.8 mL of 0.0120 M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN–. Titration of the newly freed EDTA required 26.9 mL of 0.0120 M Ca2+. What were the molarities of Cd2+ and Mn2+ in the original solution? Answer: ([Cd2+] = 0.00646 M; and [Mn2+] = 0.0490 M) The...
A sample containing Cd2+ and Mn2+ was treated with 60.1 ml of 0.0400M EDTA. Titration of the excess unreacted EDTA required 16.8ml of 0.013M Ca2+. The Cd2+ was displaced from EDTA by the addition of an excess of CN-. Titration of the newly freed EDTA required 10.4ml of 0.013M Ca2=. What are the concentrations of Cd2+ and Mn2+ in the original solution?
A 50.0 mL sample containing Cd²+ and Mn + was treated with 43.8 mL of 0.0700 M EDTA. Titration of the excess unreacted EDTA required 14.4 mL of 0.0320 M Ca2+ The Cd2+ was displaced from EDTA by the addition of an excess of CN. Titration of the newly freed EDTA required 28.2 mL of 0.0320 M Capt. What are the concentrations of Cd- and Mn in the original solution? concentration: M Mn+ concentration: M Cd2+
A 24.34 mL aliquot of a Pb2 solution, containing excess Pb2+, was added to 11.50 mL of a 2,3-dimercapto-1-propanol (BAL) solution of unknown concentration, forming the 1:1 Pb2+-BAL complex. The excess Pb2+ was titrated with 0.0157 M EDTA, requiring 9.88 mL to reach the equivalence point. Separately, 39.89 mL of the EDTA solution was required to titrate 32.75 mL of the Pb2+ solution. Calculate the BAL concentration, in molarity, of the original 11.50 mL solution concentration:
Please help. I will rate! A 0.4721 g sample of pewter, containing tin, lead, copper, and zinc, was dissolved in acid. Tin was precipitated as SnO 4 H,O and removed by filtration. The resulting filtrate and washings were diluted to a total volume of 250.0 mL. A 15.00 mL aliquot of this solution was buffered, and titration of the lead, copper, and zinc in solution required 34.45 mL of 0.001518 M EDTA. Thiosulfate was used to mask the copper in...
A 1.000-mL aliquot of a solution containing Cu2* and Ni2+ is treated with 25.00 mL of a 0.04152 M EDTA solution. The solution is then back titrated with 0.02204 M Zn2 solution at a pH of 5. A volume of 16.29 mL of the Zn2+ solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and N solution is fed through an ion-exchange column that retains NP. The Cu2" that passed through the column...
A 1.000-mL aliquot of a solution containing Cu2 and Ni2 is treated with 25.00 mL of a 0.04328 M EDTA solution. The solution is then back titrated with 0.02246 M Zn2 solution at a pH of 5. A volume of 21.60 mL of the Zn2 solution was needed to reach the xylenol orange end point. A 2.000-mL aliquot of the Cu2 and Ni2 solution is fed through an ion-exchange column that retains Ni2 . The Cu2 that passed through the...
A 25.0 mL aliquot of 0.0440 M EDTA was added to a 49.0 mL solution containing an unknown concentration of V3+. All of the V3+ present in the solution formed a complex with EDTA, leaving an excess of EDTA in solution. This solution was back-titrated with a 0.0340 M Ga3+ solution until all of the EDTA reacted, requiring 11.0 mL of the Ga3+ solution. What was the original concentration of the V3+ solution?