Procedure: Question: Part III - Prepare and Test Two Concentration Cells 13. Set

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Procedure:

Question:

Part III - Prepare and Test Two Concentration Cells 13. Set up and test a copper concentration cell Prepare 20 mL of 0.050 M CuSO4 solution by mixing 1 mL of 1.0 M CuSO4 solution with 19 mL of distilled water. Set up a concentration cell in two wells of the 24-well test plate by adding 5 mL of 0.050 M CuSO4 solution to one well and 5 mL of 1.0 M CuSO4 solution to an adjacent well. Use Cu metal electrodes in each well. Use a KNO3-soaked string as the salt bridge, as in Parts I and a. b. c. Clink >COLLECT to start data collection. d. Test and record the potential of the concentration cell in the same manner that vou tested the voltaic cells in Parts I and II 14. Set up a concentration cell to determine the solubility product constant, Kp, of Pbli 2. a. Prepare 10 mL of 0.050 M Pb(NO)2 solution by mixing 5 mL of 0.10 M Pb(NO3)2 solution with 5 mL of DI water. Mix 9 mL of 0.050 M KI solution with 3 mL of 0.050 M Pb(NO3)2 solution in a small beaker. In this reaction, most of the Pb-f and I will precipitate Pbl2 but a small amount of the ions will remain dissolved b. Lab 28 - 3 Set up the half cells in neighboring wells of the 24-well test plate. Place 5 mL of 0.050 M Pb(NO)2 solution in one half cell, and 5 mL of the Pbl2 mixture, from the small beaker, into an adjacent half cell. Use Pb electrodes in each half cell. Use a KNOj-soaked string as the salt bridge. c. d. Test and record the potential of the cell in the same manner that you tested the voltaic cells and the copper concentration cell 15. Discard the electrodes and the electrolyte solutions as directed. Rinse and clean the 24-well plate. CAUTION: Handle these solutions with care. If a spill occurs, ask your instructor how to clean up safeh

Explanation / Answer

a. moles of KI = 0.05 M x 0.009 L = 4.5 x 10^-4 mols

moles of Pb(NO3)2 = 0.005 M x 0.003 = 1.5 x 10^-4 mols

moles of PbI2 = 1.5 x 10^-4 mols

remaining moles of I- = 3 x 10^-4

[Pb2+] = 1.5 x 10^-4/0.012 = 0.0125 M

Nernst equation,

E = -0.0592/nlogQ

0.2761 = -0.0592/2 log([Pb2+]/0.0125)

[Pb2+] in solution equilibrium with PbI2 = 5.88 x 10^-12 M

b. Concentration of I- in solution

[I-] = 3 x 10^-4/0.012 = 0.025 M

c. Ksp = [ [Pb2+][I-]^2

           = (0.0125)(0.025)^2

           = 7.81 x 10^-6

d. The experimental value found here is lower than the Ksp value reported in the literature.

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