In titration of 35 ml 0.046 M Fe2* with 0.056 M Ce4* what is the cell voltage at the equivalence point? E(calomel) = 0.241 V Fe3+ + e¯ 2 Fe²+ E° = 0.767 V Ce4+ + e- = Ce3+ E° = 1.70 V Buret containing Ce Calomel
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- Titration of 25.0 mL of a 0.0500 M Zn2+ solution with 0.0550 M EDTA in a solution buffered at pH 8. Assume that the temperature is 25 oC and that the formation constant for Zn2+ is 3.13 x 1016 at this temperature. What is the pZn of the solution after 30 mL of titrant have been added?A 0.1093-g sample of impure Na2CO3 was analyzed by the Volhard method. After adding 50.00 mL of 0.06911 M AgNO3, the sample was back-titrated with 0.05781 M KSCN, requiring 27.36 mL to reach the endpoint. Report the purity of the Na2CO3 sample. [Ans. 90.9 % (w /w )]Titration of 50.00 mL of 0.04715 M Na2C2O4 required 39.25 mL of a potassium permanganate solution. 2MnO4- + 5C2O4-2 +16H+ -> 2Mn2+ + 10CO2(g) + 8H2O Calculate the molar concentration of the KMnO4 solution What molecules would interfere with the titrimetric analysis?
- Titration of 25.0 mL of a 0.0500 M Zn2+ solution with 0.0550 M EDTA in a solution buffered at pH 8. Assume that the temperature is 25 oC and that the formation constant for Zn2+ is 3.13 x 1016 at this temperature. What is the conditional formation constant for Zn at this pH?50 mL of a solution of 0.0200 M Zn2+ will be titrated with 0.0100 M EDTA in 0.0100 M NH3 at pH 6.0. Ethylenediaminetetraacetic acid (EDTA) can be considered as a tetraprotic acid (H4Y). The stepwise acid dissociation constants are: K1 = 1.02 x 10-2, K2= 2.14 x 10-3, K3 = 6.92 x 10-7 and K4 = 5.50 x 10-11. The alplia value of the un-deprotonated species in a solution buffered to a certain pH is given by the following equation: a0 = [H+]4/([H+]4 + K1[H+]3 + K1K2[H+]2 + K1 K2K3[H+] + K1 K2K3K4) Calculate the alpha value of the fully deprotonated species (Y4- ) in a solution buffered to a pH of 6.0.Which of the following experiments can be described as a displacement titration? A. Calcium in powdered milk is determined by dry ashing a 1.50 g sample and then titrating the calcium with 12.1 mL of 0.008949 M EDTA. B. 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 0.03610 M EDTA. C. A 3.650-g sample containing bromide was dissolved in sufficient water to give 250.0 mL. After acidification, silver nitrate was introduced to a 25.00-mL aliquot to precipitate AgBr, which was filtered, washed, and then redissolved in an ammoniacal solution of potassium tetracyanonickelate(II): Ni(CN)42- + 2AgBr(s) -> 2Ag(CN)2- + Ni2+ + 2Br- The liberated nickel ion required 26.73 mL of 0.02089 M EDTA. D. A solution contains 1.569 mg of CoSO4 (155.0 g/ mol) per milliliter. 50.00 mL of 0.007840 M EDTA…
- A 20.0 mL sample of commercial bleach was diluted with water to a 250.0 mL volumetric flask. A 50.0 mL aliquot of the diluted sample was treated with excess KI and starch. The liberated iodine was titrated with 0.399 M Na2S2O3, requiring 13.9 mL to reach the end point. (a) Identify the type of redox titration method used. (b) Identify the indicator and endpoint color. (c) Calculate the % w/v NaOCl in the sample.Calculate the pSr value for 10 mL of EDTA in the titration of 50 mL of 0.02 M Sr2 + in an environment buffered at pH 11 with a titre of 0.02 M EDTA. A. 1.89B. 3.78C. 7.56D. 12.11Forming Ba(I03)2, 500 mL of 0.5000 M Ba(NO3)2 was mixed with 500 mL of 0.0500 M NaIO3. Ksp=1.57x10^-9 What is the concentration of the excess reagent? What is the molar solubility of Ba(IO3)2 in this solution?
- What mass of ZnS (Ksp = 2.5 x 10-22) will dissolve in 325.0 mL of 0.050 M Zn(NO3)2? Ignore the basic properties of S2-.Titration of 25.0 mL of a 0.0500 M Zn2+ solution with 0.0550 M EDTA in a solution buffered at pH 8. Assume that the temperature is 25 oC and that the formation constant for Zn2+ is 3.13 x 1016 at this temperature. What is the pZn at the equivalence point of the titration?A 20 ml aliquot of malonic acid solution was treated with 10.0 ml of 0.25M Ce4+ leading to the reaction CH2(COOH)2 + 6Ce4+ + 2H2O ® HCOOH + 2CO2 + 6Ce3+ + 6H+ After standing for 10 minutes at 60oC, the solution was cooled and the x’ss Ce4+ was titrated with 0.1M Fe2+, requiring 14.4 ml to reach the ferroin end point. Calculate the M of the malonic in the sample.