29.0 mLE =V29.9 mLE =V30.3 mLE =V49.0 mLE =V60.0 mLE =V68.0 mLE =V A solution prepared by mixing 60.0 mL of 0.360 M AGNO, and 60.0 mL of 0.360 M TINO, was titrated with 0.720 MNaBr in a cell containing a silver indicator electrode and a reference electrode with a constant potential of 0.175 V. Thereference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to thenegative terminal. The solubility constant of TIBr is Kp = 3.6 × 10-6 and the solubility constant of AgBr is= 5.0 x 10-13.KspWhich precipitate forms first?O TIBIO AgBrWhich of the expressions shows how the cell potential, E, depends on [Ag*]?E = 0.175 V – [0.799 V – 0.05916log ([Ag+])]1E = |0.799 V – 0.05916log- 0.175 V[Ag*)E = [0.799 V – 0.05916log ([Ag*]D] – 0.175 VE = 0.175 V - |0.799 V – 0.05916log[Ag** (ল)Calculate the first and second equivalence points of the titration.first equivalence point:mLsecond equivalence point:mLWhat is the cell potential after each of the given volumes of 0.720 M NaBr have been added?1.0 mLE =V19.0 mLE =V

(1) AgBr Precipitate will form first, as we have Ksp Value for AgBr is lower then…

A solution prepared by mixing 60.0 mL of 0.360 M AgNO, and 60.0 mL of 0.360 M TINO, was titrated with 0.720 M NaBr in a cell containing a silver indicator electrode and a reference electrode with a constant potential of 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TIB is Kp = 3.6 x 10-6 and the solubility constant of AgBr is Kp = 5.0 x 10-13. Which precipitate forms first? TIBr AgBr Which of the expressions shows how the cell potential E denends on [Agt1?

A solution prepared by mixing 60.0 mL of 0.360 M AgNO, and 60.0 mL of 0.360 M TINO, was titrated with 0.720 M NaBr in a cell containing a silver indicator electrode and a reference electrode with a constant potential of 0.175 V. The reference electrode is attached to the positive terminal of the potentiometer, and the silver electrode is attached to the negative terminal. The solubility constant of TIB is Kp = 3.6 x 10-6 and the solubility constant of AgBr is Kp = 5.0 x 10-13. Which precipitate forms first? TIBr AgBr Which of the expressions shows how the cell potential E denends on [Agt1?

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

See similar textbooks

Similar questions

The Na + concentration of a solution was determined by measurements performed with a sodium ion selective electrode. The electrode system has developed a potential for - 0.2831 V when immersed in 15.00 mL of the solution of unknown concentration. After adding 5.00 mL of NaCℓ 1.00 x 10-2 mol / L to the solution of unknown concentration, the potential went to -0.1546 V. Calculate the Na + concentration in the original solution.
The Sn in a 0.4352 g mineral specimen was reduced to the +2 state and titrated with 29.77 mL of 0.01735 M K2Cr2O7. Identify: (a) type of redox titrimetry used (b) appropriate indicator and its type (c) color at the end point (d) weight percent Sn (118.71 g/mol) in the sample
determine the electromotive force of the galvanic element Ni / NiSO4 // CoSO4 / Co if the concentration of nickel ions is 0.001M and the concentration of cobalt ions is 0.1M. The standard electrode potentials of nickel and cobalt are -0.23, respectively. and -0.28
Calculate the potential at the equivalence point for titration of 0.100MFe^2+ with 0.100M Ce^4+.Given the formal potential for the reduction E^0 of Fe^3+ to Fe^2+ is +0.77V, and the formal potential for the reduction of E^0 Ce^4+ to Ce^3+ is 1.45V.
Calculate the potentials for each of the following cells. Indicate whether the reaction will proceed spontaneously in the direction written. Pb|Pb* (0.1393M)||Ca* (0.0511M)|Cd a. Zn Zn* (0.0364M)|T7* (0.00906.M),TI" (0.062M)|Pt b. Pr|TiO*(0.079M),Tr* (0.00916M), H¨ (0.0147M)|vO*(0.0134M),V* (0.0784M),H* (0.0538M)|Hg c.
An electrochemical cell similar to the cell in Experiment 3 is constructed, but instead of using Cu electrodes and a Cu²+ solution, this cell uses Pb electrodes and a Pb²+ solution. Pb|Pb+²(1 M)||AsO(~ 1 M)|Pb₂ (AsO₂)₂(s) Pb(s) The measured potential, Ecell of this cell is -0.349 V. Calculate Ksp for Pb, (ASO₂)₂. Ksp = 1.479 x10¹2 Incorrect
22
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 Ce3+. Calculate the potential (vs. Ag | AgCl) at the equivalence point. At what volume of titrant will the potential be equal to the standard potential of the Ce half-reaction (ignoring the Ag | AgCl electrode)?
Calculate the cell potential when 50.00 mL of 0.100 M Ag+ solution is titrated with 50.00 mL of 0.150 M Cl- solution. Silver wire indicator electrode and Ag-AgCl electrode are used as reference electrodes. E0 (Ag/AgCl) = 0,197 V , E0 Ag+/Ag = 0,799 V, Kçç (AgCl) = 1,8 x10-8