A cyanide solution with a volume of 12.99 ml was treated with 30.00 mL. of Ni solution (containing excess Ni) to convert the cyanide into tetracyanonickelate(): 4 CN+ Ni?+ Ni(CN) The excess Ni+ was then titrated with 11.96 ml. of 0.01357 M ethylenediaminetetraacetic acid (EDTA): Nit + EDTA Ni(EDTA)- Ni(CN) does not react with EDTA. If 39.40 ml. of EDTA were required to react with 30.65 ml. of the original Nit solution, calculate the molarity of CN in the 12.99 mL cyanide sample. 0.12055 (CN|= Incorrect
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- Chromel is an alloy composed of nickel, iron and chromium.A 0.6472 g sample was dissolved and diluted to 250 mL. When a50 mL aliquot of 0.05182 M EDTA was mixed with an equal volumeof the diluted sample and all the three ions were chelated, a 5.11 mLback titration with 0.06241 M copper (II) was required.The chromium in a second 50 mL aliquot was masked through theaddition of hexamethylenetetramine, titration of the Fe and Nirequired 36.28 mL of 0.05182 M EDTA.Iron and chromium were masked with pyrophosphate in a third50 mL aliquot and the nickel was titrated with 25.91 mL of theEDTA solution.Calculate the percentage of nickel, chromium and iron in thealloy.Chromel is an alloy composed of nickel, iron and chromium. A 0.6553-g sample was dissolved and diluted to 250.0 mL. When a 50.00-mL aliquot of 0.05173 M EDTA was mixed with an equal volume of the diluted sample, all three ions were chelated, and a 5.34-mL back titration with 0.06139 M copper (II) was required.The chromium in a second 50.0-mL aliquot was masked through the addition of hexamethylenetetramine; titration of the Fe and Ni required 36.98 mL of 0.05173M EDTA. Iron and chromium were masked with pyrophosphate in a third 50.00-mL aliquot, and the nickel was titrated with 24.53 mL of the EDTA solution. Calculate the percentage of Cr in the alloy. Express your answer in 2 decimal places.A cyanide solution with a volume of 12.98 mL was treated with 23.00 mL of Ni²+ solution (containing excess Ni²+) to convert the cyanide into tetracyanonickelate (II): 4 CN- + Ni²+ ->> Ni(CN) The excess Ni2+ was then titrated with 10.60 mL of 0.01231 M ethylenediaminetetraacetic acid (EDTA): Ni2+ + EDTA4 → Ni(EDTA)2- Ni(CN) does not react with EDTA. If 38.70 mL of EDTA were required to react with 30.26 mL of the original Ni²+ solution, calculate the molarity of CN in the 12.98 mL cyanide sample. [CN-]=
- A 25.00 ml of Ni2+ solution was diluted in HCl and treated with 25.00 ml of 0.05283 MNa2EDTA. The solution was neutralized with NaOH followed by addition of acetate buffer untilthe pH 5.5. The solution turns yellow after addition of few drops of xylenol orange indicator.Back titration using standard 0.022 MZn2+ at pH 5.5 requires 17.61 ml until end point, onwhich the solution will turn red. Calculate for the molarity of the unknown.An antihistamine sample, brompheniramine maleate, weighing 5.01234 g was dissolved in alcohol and decomposed with metallic sodium. The resulting solution was treated with 12.00mL of 0.3121M AgNO3 to precipitate all of the liberated bromide ions as AgBr. The excess AgNO3 remaining in the solution was titrated with 9.123mL of 0.2181M KSCN to reach the endpoint. What is the %Br in the sample? Br- + Ag+ --> AgBr(s) Ag+ + SCN- --> AgSCN(s) Br = 79.904 g/molChromel is an alloy composed of nickel, iron, and chromium. A 0.6418-g sample was dissolved and diluted to 250.0 mL. When a 50.00-mL aliquot of 0.05173 M EDTA was mixed with an equal volume of the diluted sample, all three ions were chelated, and a 5.27-mL back-titration with 0.06139 M copper(II) was required. The chromium in a second 50.0-mL aliquot was masked through the addition of hexamethylenetetramine; titration of the Fe and Ni required 35.81 mL of 0.05173 M EDTA. Iron and chromium were masked with pyrophosphate in a third 50.0-mL aliquot, and the nickel was titrated with 25.77 mL of the EDTA solution. Calculate the percentages of nickel, chromium, and iron in the alloy. Percentage of nickel = % Percentage of iron = Percentage of chromium = % %
- Five drops of dichlorofluorescein have been added to a solution containing 0.800 g sample which required 25.30 mL of 0.100 M AgNO3. (a) Identify the type of argentometric titration method used.2. Titration of Ca²+ and Mg²+ in a 50mL sample of hard water required 23.65mL of 0.01205M EDTA. A second 50mL aliquot was made strongly basic with NaOH to precipitate Mg2+ as Mg(OH)2. The supernatant liquid was titrated with 14.53mL of the EDTA solution. Calculate a. concentration in ppm of CaCO3 in the sample. b. concentration in ppm of MgCO3 in the sample.1. A 0.7120 g of iron ore was brought into solution and passed through a Jones reductor. Titration of Fe(II) 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
- A 25.00 mL aliquot of sample containing Hg in dilute nitric acid was treated with 10.00 mL of 0.04882 M EDTA and the pH was adjusted to 10. The excess EDTA was titrated with 24.66 mL of 0.01137 M Mg. What is the molarity of Hg in the original solution?II. 3. Compleximetry required 18.75 mL 0.1357 M EDTA. A 20.00 mL sample is made strongly alkaline to precipitate magnesium as Mg(OH)2. Titration of this sample with a calcium specific indicator required 6.34 mL of the standard EDTA solution. A drinking water sample was analyzed for Mg and Ca. Titration of a 25.00 mL sample at pH 10 a. Calculate the CaCO3 (FM = 100.09) titer of the EDTA. %3D titer = mg/mL b. Calculate ppm Ca2+ (FM = 40.08) in the water. %3D Ca2+ content= mg/L c. Calculate ppm Mg2+ (FM = 24.31) in the water. %3D Mg2+ content = mg/L %3DThe As in a 9.13-g sample of pesticide was converted to AsO43- andprecipitated as Ag3AsO4 with 50.00 mL of 0.02015 M AgNO3. The excess Ag+was then titrated with 4.75 mL of 0.04321 M KSCN. Calculate the % of As2O3in the sample.