A 30 mL solution of TI* (0.45 M) is titrated with 0.60 M Co^3+. The anode is a p

Question

A 30 mL solution of TI* (0.45 M) is titrated with 0.60 M Co^3+. The anode is a platinum wire and the cathode is a saturated silver/silver chloride (Ag/AgCl) reference electrode; both are inserted into the titration beaker. Determine the cell potential at the titration volumes listed below: 21.65 mL of titrant added 45.00 mL of titrant added 49.72 mL of titrant added 53.09 mL of titrant added A 20 mL solution of Ag^3+ (52 mM) can be reduced by a basic sulfite (SO_3^2minu ) solution. A titration is performed with a saturated calomel reference electrode as the anode and a platinum wire as a cathode. Determine the cell potential at various titration volumes, below, if the titrant is 0.1 M SO_3^2 and both solutions are buffered to pH 9.5. 3.49 mL of titrant added 8.37 mL of titrant added 10.40 mL of titrant added 15.92mL of titrant added

Explanation / Answer

1) Titration

a) 21.65 ml of 0.6 M Co3+ added

moles of Co3+ added = 0.6 M x 21.65 ml = 12.99 mmol

moles of Tl+ present = 0.45 M x 30 ml = 13.5 mmol

[Tl+] remaining = 0.45 M x (30 - 21.65)ml/51.65 ml = 0.0099 M

[Tl3+] formed = 0.6 M x 21.65 ml/51.65 ml = 0.251 M

Using Nernst equation,

E = [Eo - 0.0592/n log([Tl+]/[Tl3+])] - 0.222

    = [0.77 - 0.0592/2 log(0.0099/0.251)] - 0.222

    = 0.589 V

b) 45 ml of 0.6 M Co3+ added

This is half equivalence point

[Co3+] = [Co2+]

E = 1.82 - 0.222 = 1.598 V

c) 49.72 ml of 0.6 M Co3+ added

moles of Co3+ = 0.6 M x 49.72 ml = 29.832 mmol

moles of Co2+ formed = 13.5 mmol

[Co3+] remaining = 16.332 mmol/79.72 ml = 0.205 M

[Co2+] formed = 0.45 M x 30 ml/79.72 ml = 0.169 M

E = (1.82 - 0.0592/2 log(0.169/0.205)) - 0.222

   = 1.600 V

d) 53.09 ml of 0.6 M Co3+ added

moles of Co3+ added = 0.6 M x 53.09 ml = 31.854 mmol

moles of Co2+ formed = 0.45 M x 30 ml = 13.5 mmol

[Co2+] formed = 13.5 mmol/83.09 ml = 0.162 M

[Co3+] remaining = 18.354 mmol/83.09 ml = 0.221 M

E = (1.82 - 0.0592/2 log(0.162/0.221)) - 0.222

   = 1.602 V

2. Titration

a) 3.49 ml of 0.1 M SO3^2- added

moles of SO3^2- added = 0.1 M x 3.49 ml = 0.349 mmol

moles of Ag3+ present = 0.052 M x 20 ml = 1.04 mmol

[Ag+] formed = 0.349 mmol/23.49 ml = 0.015 M

[Ag3+] remained = (1.04 - 0.349) mmol/23.49 ml = 0.029 M

E = (1.9 - 0.0592/2 log(0.015/0.029)) - 0.241

   = 1.667 V

b) 8.37 ml of 0.1 M SO3^2- added

moles of SO3^2- added = 0.1 M x 8.37 ml = 0.837 mmol

moles of Ag3+ present = 0.052 M x 20 ml = 1.04 mmol

[Ag+] formed = 0.837 mmol/28.37 ml = 0.029 M

[Ag3+] remained = (1.04 - 0.837) mmol/28.37 ml = 0.007 M

E = (1.9 - 0.0592/2 log(0.029/0.007)) - 0.241

   = 1.641 V

c) 10.40 ml of 0.1 M SO3^2- added

This is equivalence point

[SO3^2-] added = [Ag3+] present

E = (1.9 - 0.93)/2 = 0.485 V

d) 15.92 ml of 0.1 M SO3^2- added

moles of SO3^2- added = 0.1 M x 15.92 ml = 1.592 mmol

moles of SO4^2- formed = 1.04 mmol

[SO3^2-] remains = (1.592 - 1.04) mmol/35.92 ml = 0.015 M

[SO4^2-] formed = 1.04 mmol/35.92 ml = 0.03 M

E = (-0.93 - 0.0592/2 log(0.015/0.03)) - 0.241

   = -1.162 V

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