A 0.1017-g sample of KBro, (MM 167.0) was dissolved in dilute HCl and treated with an unmeasured excess of KI. The liberated 12 required 39.75 ml of a sodium thiosulfate solution. Calculate the molar concentration of the Na,S,0; 61 + Bro, + 6H* → 31, + Br + 3H,0 25,0,2 + 1, → 21 + S,0¿²
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- 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-50.00 mL remaining solution was analyzed for its Br- content by potentiometry using a metallic electrode of the second kind. a) Write the cell notation of the potentiometric set-up with SCE as the reference electrode. b) Write the Nernst equation that describes the indicator electrode set-up. Ecell recorded in running the solution using the potentiometric set-up was Ecell = 0.0286 V. (E0Ag/AgBr = 0.095 V) c) Compute for Eind. d) Compute pBr in the 50.00 mL aliquot. e) Compute for % NaBr ( in the potentiometric technique).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 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.A 0.1017 g sample of KBrO3 (MM=166.1) was dissolved in dilute HCI and treated with an unmeasured excess of Kl. The liberated iodine required 39.75 mL of Na₂S₂O3. Calculate the molar concentration of sodium thiosulfate. 0.2561 M 0.08512 M 0.09242 M 0.1041 MA 1.000 g sample containing chlorides, iodides and inert materials was treated with dilute nitric acid followed by AgNO3. A precipitate of AgCl (143.32) and AgI (234.77) was produced and weighs 0.9238 g. On heating in a current of Cl2, the AgI is converted to AgCl, and the resulting product weighs 0.7238 g. Find the percentage of a) NaI (149.89) and b) NaCl (58.44) in the sample
- 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.Caustic potash that has been exposed to air is found on analysis to contain 90.00% KOH, 2.38% K2CO3 and 7.62% H2O. What weight of residue will be obtained if 1.00 g of this sample is added to 46.00 mL of 1.00 N HCl and the resulting solution, after neutralization with 1.070N KOH is evaporated to dryness?Exactly 6.00 ml of acetic acid (density = 1.05 g/mL), MW =60.05) was added to a 1 –liter volumetric flask, and the flask was filled to the mark with distilled water. A portion of the resulting solution was added to a conductance cell (k = 1.25 cm-1), and the conductance was found to be 416 μS. Calculate the dissociation constant Ka of acetic acid. ΛoH+= 349.8 Scm2/mole and ΛoCH3COO- = 41 Scm2/mole
- Given that the Ksp for silver chloride is 1.76 x 10-10, and the Kf of silver(I) diammine cation [Ag(NH3)2 +] is 1.7 x 107, calculate the maximum number of grams of silver chloride that will dissolve in 250.0 mL of a 3.25 M solution of ammonia.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.To measure the iron content of runoff from a ranch a 25.0 mL sample of run off was acidified with HNO3 and treated with excess KSCN to form a red complex. (KSCN itself is colorless.) The solution was then diluted to 100 mL and put into a variable pathlenght cell. For comparison, a 10.0 mL reference sample of 6.80 x 10-4 M Fe3+ was treated with HNO3 and KSCN and diluted to 50.0 mL. The reference was placed in a cell with a 1.00 cm pathlenght. The runoff had the same absorbance as the reference when the pathlenght of the runoff cell was 2.48 cm. What wa the concentraton of iron in the runoff?
