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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- A foot powder sample containing Zn was dissolved on 50.00 mL water and was titrated to the end point color with 22.57 mL of 0.01639 M EDTA at pH=4. (α= 3.61x109, KKzny2 = 3.2 x 1016). The pZn in the sample is:The bismuth (AW 208.98) in 0.6805 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 15.02 mL of 0.1498 M AgNO3, causing the precipitation of AgCl. The excess AgNO3 required 14.04 mL of 0.1008 M KSCN for titration. Calculate %Bi in the sample.The bismuth (AW 208.98) in 0.6805 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 15.02 mL of 0.1498 M AgNO3, causing the precipitation of AgCl. The excess AgNO3 required 14.04 mL of 0.1008 M KSCN for titration. Calculate %Bi in the sample.A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br-Nickel ion required 11.70 mL of 0.002146 M EDTAWhat is the percentage of NaBr (102.894) in the 1.000 g sample?
- A 25.00 mL sample containing Fe3+ was treated with 10.00 mL of 0.03676 M EDTA to complex all the Fe3+ and leave excess EDTA in solution. The excess EDTA was then back-titrated, requiring 2.37 mL of 0.04615 M Mg2+. What was the concentration of Fe3+ in the original solution in ppm Fe3+?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.b. EDTA cannot be used as a primary standard always. When can EDTA be used as a primarystandard ? Give reason why EDTA is used in complexometric titrations.
- The sulfate in a247.1 mg sample was precipitated as BasO4 by addition of 25.00 mL of 0.03992 M BaCl2. The precipitate was removed by filtration and the remaining BaCl2 consumed 36.09 mL of 0.0217 M EDTA for titration to the Camalgite endpoint. Calculate the % SO3 in the sample.The 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.In an unprecedented initiative, the professor chose you to carry out the determination of a special sample in order to verify the inherent indication system. For that, an aliquot of 25.00 mL of a solution containing Fe(III) salts was titrated with EDTA 0.00982 mol/L, using potassium thiocyanate as indicator. In this case, you have found the volume of 31.10 mL of titrant to End Point. Based on the information given in the table, choose the option that best describes/explains the process linked to the indication system of the Final Point of the degree: (a) The indication of the system will be given by the disappearance of the reddish coloration, due to the displacement reaction: Fe(III) complex with EDTA is formed in detriment of the Fe(III) complex with thiocyanate. (b) The indication of the system will be given by the appearance of a reddish color, in function of the displacement reaction: Fe(III) complex with thiocyanate is formed in detriment of the Fe(III) complex with EDTA. (c) The…
- A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br- The liberated nickel ion required 11.70 mL of 0.002146 M EDTA. Calculate mmol Ni in the 50.00 mL aliquot. Calculate mmol Br- in the 50.00 mL aliquot. Calculate the percentage of NaBr (102.894) in the 1.000 g sample.A 1.000-g sample containing bromide was dissolved in sufficient water to give 100.0 mL. A 50.00 mL aliquot was measured and after acidification, silver nitrate was introduced to precipitate AgBr, which was filtered, washed, and then dissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) → 2Ag(CN)2- + Ni2+ + 2Br- The liberated nickel ion required 11.70 mL of 0.002146 M EDTA The other 50.00 mL remaining solution was also analyzed for its Br- content by potentiometry using a metallic electrode of the second kind. Ecell recorded in running the solution using the potentiometric set-up was Ecell = 0.0286 V. (E0Ag/AgBr = 0.095 V) g) Compute for Eind. h) Compute pBr in the 50.00 mL aliquot. i) Compute for % NaBr ( in the potentiometric technique). j) Calculate the error between the obtained % NaBr fr the EDTA titration technique and the % NaBr from the potentiometric technique.The concentration of ammonia in a cleaning product was determined by back titration.Firstly, 10.00 cm3 of the cleaning product was pipetted into a large conical flask,containing 250.00cm3 of 0.50 mol/l HCl to give Solution A.Following a period of reaction and shaking, 50.00cm3 of Solution A was removed anddiluted to 250 cm3 with water in a volumetric flask to give Solution B.20 cm3 samples of Solution B were titrated against 0.05 mol/l Na2CO3 solution, givingan average titre of 12.45 cm3. i) Write equations for the reactions that have taken place.ii) Determine the concentration of NH3 in the original cleaning product in mol/l,g/l, ppm, and % w/v.