A redox titration: 160.00mL of 0.0500M Sn2+ solution was titrated with 0.200M Ce4+ according to the reaction below:2Ce4+ + Sn2+ ------> 2Ce3+ + Sn4+The Eo for each part:Ce4+ + e- -------> Ce3+ Eo= 1.470VSn4+ + 2e- -------> Sn2+ Eo= 0.139V\What is the cell potential when 60mL Ce4+ is added?
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A redox titration: 160.00mL of 0.0500M Sn2+ solution was titrated with 0.200M Ce4+ according to the reaction below:
2Ce4+ + Sn2+ ------> 2Ce3+ + Sn4+
The Eo for each part:
Ce4+ + e- -------> Ce3+ Eo= 1.470V
Sn4+ + 2e- -------> Sn2+ Eo= 0.139V
\What is the cell potential when 60mL Ce4+ is added?
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- What cathode potential (versus SCE) would be required to lower the total Hg(II) concentration of the following solutions to 1.00 10-6 M (assume reaction product in each case is elemental Hg): (a) an aqueous solution of Hg2+? (b) a solution with an equilibrium SCN- concentration of 0.100 M? Hg2+ + 2SCN- Hg(SCN)2(aq) = Kf = 1.8 107 c) a solution with an equilibrium Br- concentration of 0.100 M? HgBr42++ 2e- Hg(l) + 4Br- E0= 0.223 VSulfide ion (S2- ) is formed in wastewater by the action of an aerobic bacteria on organic matter. Sulfide can be readily protonated to form volatile, toxic H2S. In addition to the toxicity and noxious odor, sulfide and H2S cause corrosion problems because they can be easily converted to sulfuric acid when conditions change to aerobic. One common method to determine sulfide is by coulometric titration with generated silver ion.At the generator electrode, the reaction is Ag Ag+ + e-. The titration reaction is S2- + 2Ag+ Ag2S(s). (a) A digital chloridometer was used to determine the mass of sulfide in a wastewater sample. The chloridometer reads out directly in ng Cl-.In chloride determinations, the same generator reaction is used,but the titration reaction is Cl- + Ag+ AgCI(s). Derive an equation that relates the desired quantity, mass S2- (ng), to the chloridometer readout in mass Cl- (ng). (b) A particular wastewater standard gave a reading of 1689.6 ng Cl-. What total charge in coulombs was required to generate the Ag+ needed to precipitate the sulfide in this standard? (c) The following results were obtained on 20.00-mL samples containing known amounts of sulfide.17 Each standard was analyzed in triplicate and the mass of chloride recorded. Convert each of the chloride results to mass S2- (ng). (d) Determine the average mass of S2- (ng), the standard deviation, and the %RSD) of each standard. (e) Prepare a plot ofthe average mass of S2- determined (ng) versus the actual mass (ng). Determine theslope, the intercept, the standard error, and the R2 value. Comment on the fit of the data to a linear model. (f) Determine the detection limit (ng) and in parts per million using a k factor of 2 (see Equation 1-12). (g) An unknown wastewater sample gave an average reading of 893.2 ng Cl. What is the mass of sulfide (ng)? If 20.00 mL of the wastewater sample was introduced into the titration vessel, what is the concentration of S2- n parts per million?A redox titration: 160.00mL of 0.0500M Sn2+ solution was titrated with 0.200M Ce4+ according to the reaction below:2Ce4+ + Sn2+ ------> 2Ce3+ + Sn4+The Eo for each part:Ce4+ + e- -------> Ce3+ Eo= 1.470VSn4+ + 2e- -------> Sn2+ Eo= 0.139Va. What is the VE (the equivalence volume ) (of Ce4+) needed to reach the endpoint?
- A redox titration: 160.00mL of 0.0500M Sn2+ solution was titrated with 0.200M Ce4+ according to the reaction below:2Ce4+ + Sn2+ ------> 2Ce3+ + Sn4+The Eo for each part:Ce4+ + e- -------> Ce3+ Eo= 1.470VSn4+ + 2e- -------> Sn2+ Eo= 0.139VCalculate the potential (voltage), when 10.00mL of Ce4+ is added. (SCE= +0.241 V)A redox titration: 160.00mL of 0.0500M Sn2+ solution was titrated with 0.200M Ce4+ according to the reaction below:2Ce4+ + Sn2+ ------> 2Ce3+ + Sn4+The Eo for each part:Ce4+ + e- -------> Ce3+ Eo= 1.470VSn4+ + 2e- -------> Sn2+ Eo= 0.139VWhat is the cell potential at the equivalence point?A soluion of 10.0 mL of 0.0500 mol/L AgNO3 is titrated with 0.0250 mol/L NaBr in the cell: VKE||onbekende oplossing||Ag(s) Calculate the potential of the cell after adding 0.1 mL, 10.0 mL, 20.0 mL and 30.0 ml of titrant. E\deg (Ag+, Ag = 0.799 V; E(VKE = 0.244 V; Ksp(ABr) = 5.0*10-13
- A 10.0 mL solution of 0.050 M AgNO3 was titrated with 0.0250 M NaBr in the cell: S.C.E || titration solution | Ag(s) Find the cell voltage for 30.0 mL of titrant. (A) +0.093 V (B) -0.039 V (C) -0.093 V (D) +0.039 V112. Solutions of sodium thiosulfate are used to dissolve unexposed AgBr (Kp and-white film. What mass of AgBr can dissolve in 1.00 L of 0.500 M Na,S,O;? Ag* reacts with S,O,- to form a complex ion: = 5.0 × 10-13) in the developing process for black- Ag*(aq) + 2S,O;²-(aq) = Ag(S,O;),-(aq) K = 2.9 X 1013The basic salt Cr(OH)3 (s) was allowed to partially dissolveinto 1.50 L of solution that was fixed at pH=12.53. How many grams of theoriginal salt was dissolved at equilibrium? Molar Mass for Cr(OH)3=103 g/mol. Cr(OH)3(s) <=> Cr3+(aq) +3OH-(aq) Ksp=6.30*10-31
- At 25 °C, a 10.0-mL solution of 0.0500 M AGNO3 was titrated with 0.0250 M NaBr in the cell: S.C.E. titration solution | Ag(s). Find the cell voltage for 0.1 mL and 30.0 mL of titrant if the Ksp of AgBr is 5.0 x 10-13. (ESCE = 0.244 V; Ag* + e -> Ag(s) E° = 0.799 V).The E0 for the reaction Pb3(AsO4)2 (s) + 6e- ⇆ 3Pb(s) + 2AsO43- is -0.475 V. Calculate the solubility product constant for Pb3(AsO4)2 ?Pb2 + + 2e- ⇆ Pb(s) E0=-0.126 VA 48.0048.00 mL aliquot from a 0.4500.450 L solution that contains 0.3700.370 g of MnSO4MnSO4 (MW=151.00MW=151.00 g/mol) required 35.835.8 mL of an EDTAEDTA solution to reach the end point in a titration. What mass, in milligrams, of CaCO3CaCO3 ( MW=100.09 MW=100.09 g/mol) will react with 1.011.01 mL of the EDTAEDTA solution?
