QUESTION 9 10 MARKS You have a solution of 50.0 mL of 0.02 M AgNO3. To this solu

Question

QUESTION 9                  10 MARKS
You have a solution of 50.0 mL of 0.02 M AgNO3. To this solution 50.0 mL of 0.15 M
ligand solution was added. This ligand forms a strong complex with silver ions, AgL2,
and its overall formation constant log 2 = 10.5. What change in potential of a silver
indicator electrode would be recorded at 25 °C? Assume that AgL2 is the only
significant species of silver present.
E°(Ag) = 0.799 V. At 25 °C: (RT/nF)ln = (0.05916/n)log in V

old test paper question and im unsure how to answer

Explanation / Answer

Ag + 2L----------------> AgL2

Using Nerst Equation at 298K (25 °C)

E= E° - (RT/nF)ln[Products]/[Reactants] = E° - (RT/nF)ln[AgL2]/[Ag][L]2

E° = 0.799 V

n = 1

Initial amount of AgNO3 = 50.0 mL x 0.02M = 1 mmol (Limited Reagent)

Initial amount of Ligand = 50.0 mL x 0.15M = 7.5 mmol (Excess Reagent)

The conc. at equilbrium:

The stoichiometry of the reaction 1:2

1mmol of Ag ions will react with 2 mmol of L.

Ag+ ~ 0, L = 7.5-2 = 5.5mmol and AgL2 = 2mmol

Amount of solution = 100 mL

Conc of L = 5.5mmol/100mL = 0.055 M

Conc of AgL2 = 2 mmol/100 = 0.02 M

Using Nernst equation the potential is

E =0.799-0.05916 log ((0.02)/(0.055)2) = 0.751V

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