Calculate potentials after the addition of 10.00, 25.00, 50.00, and 50.10 mL of the reagent. Where necessary,Assume that [H+] = 1.00 throughout   50.00 mL of 0.05000 M U4+ with 0.02000 M MnO4

Q: How long would it take to titrate 1.00 mmol H2S with a current of 100.0 mA?

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Q: What is the potential after the addition of 25.00 mL of 0.05000 M Sn^ in 50.00 mL of 0.1000 M Fe".…

A: Given: Concentration of Sn2+ = 0.0500 M Volume of Sn2+ solution = 25.00 mL  Concentration of Fe3+ =…

Q: 1. Calculate the potential of a platinum indicator electrode dipping into a solution that is 0.1M in…

A: Since you have asked multiple questions, we will solve the first question for you. If you want any…

Q: Calculate the cell potential when 150.00 mL of 0.100 M Ag+ solution is titrated with 25.00 mL of…

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Q: Calculate the indicator electrode potential after adding 50.0 mL of 0.10 M Ce4+ to 5.0 mL of 0.10 M…

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Q: Complete the table. Calculate the cell potential when 20.0 mL of 0.05 M Fe2+ was titrated with 0.05…

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Q: A 25.0 mL solution of 0.0640 M Fe2* in 1 M HNO3 was titrated with 0.0500 M Pb4+ to give Fe3+ and…

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Q: A 60.0 mL solution of 0.00750M OSCI,3- in 1 M HCI was titrated with 0.0300 M IrCl,2- to give OsCl,2-…

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Q: At 25ºC, you conduct a titration of 20.00 mL of a 0.0400 M AgNo3 solution with a 0.050 M NaI soution…

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Q: acid dissociation constants for H3PO4 (Ka1 = 7.11 × 10 -3 , Ka2 = 6.32 × 10 -8 , Ka3 = 7.1 × 10…

A: The reaction is shown below: ........(i)

Q: A solution prepared by mixing 22.2 mL of 0.400 M NaCl and 22.2 mL of 0.400 M KI was titrated with…

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Q: Consider the following cell: Cu(s) | CuCl2(0.065 M) || FeCl2(0.035 M) | Fe(s) a) Using the standard…

A: a) Reduction half cell reaction ( Cathode): Fe2++2e-→Fe    Ecell=-0.45 Oxidation half cell reaction…

Q: 50.00 mL of 0.100 M Co³+ is titrated with 0.110 M MnO₂². Co³+ + eCo²+ Eº = 1.92 V MnO4 + e MnO4² Eº…

A: Known [Co3+] = 0.100 M = M1 ; Volume = 50 mL = V1 [MnO4-2] = 0.110 M = M2 , Volume = ? At equivalent…

Q: A solution prepared by mixing 22.2 mL of 0.400 M NaCl and 22.2 mL of 0.400 M KI was titrated with…

A: Answer is 7.48 x 10-16   M .

Q: 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. |…

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Q: Calculate the potential of the following cells. Assume a temperature of 25°C and that activities…

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Q: A titration of 50.0 mL of 0.10 M Sn2+ with 0.2 M Fe3+ in 1 M HCl to give Fe2+ and Sn4+ using Pt and…

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Q: What is the potential after the addition of 25.00 mL of 0.05000 M Sn2+ in 50.00 mL of 0.1000 M…

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Q: ionic strength of (a) 0.008 7 M KOH and (b) 0.000 2 M La(IO,)3

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Q: Calculate potentials after the addition of 10.00, 25.00,50.00, and 50.10 mL of the reagent. Where…

A: Given: Volume of reagent: 10.00, 25.00,50.00, and 50.10 mL   Cell:  50.00 mL of 0.1000 M V2+ with…

Q: Calculate potentials after the addition of 10.00, 25.00, 50.00, and 50.10 mL of the reagent. Where…

A: "Since you have asked multiple questions, we will solve the first question (4a) for you. If you want…

Q: A 50.00-mL of 0.0500 M Sn²+ was titrated against 0.100 M T1³+ in acidic medium. d. Calculate the…

A: 50.00 mL of 0.0500 M Sn2+ was titrated with 0.100 M of Tl3+ in acidic medium.Here we have to…

Q: Calculate potentials after the addition of 10.00, 25.00, 50.00, and 50.10 mL of the reagent. Where…

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Q: Calculate the potential at the equivalence point for titration of 0.100MFe^2+ with 0.100M…

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Q: A 30.0 mL solution of 0.0180 M Rh3+ in 1 M HClO4 was titrated with 0.0900 M Ce4+ to give Rh6+ and…

A: E =1.338 V Volume= 6 ml

Q: A soluion of 10.0 mL of 0.0500 mol/L AgNO3 is titrated with 0.0250 mol/L NaBr in the cell:…

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Q: Consider the titration of 60.00 mL of 0.100 M Co3+ with 0.130M MnO²- . What is the potential after…

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Q: Calculate potentials after the addition of 10.00, 25.00, 50.00, and 50.10 mL of the reagent. Where…

A: Given: 50.00 mL of 0.1000 M [Fe(CN)6]4- with 0.05000 M Tl3+ To Calculate: Potential after the…

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Potentiometric Titrations

Potentiometric titration is one of the important titration methods and is mainly carried out to characterize acid solutions. It follows a method similar to that of the normal titration technique, but the main difference is that there is no indicator used in the potentiometric titrations. Instead of using an indicator, electrodes are employed for monitoring the whole titration process. A scientist named Robert Behrend did the potentiometric titration for the first time, and it was done using the mercurous solution and potassium chloride.