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

Titration

1) Tl+ with Co3+

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+] remained = 0.51 mmol/51.65 ml = 0.0099 M

[Tl3+] formed = 12.99 mmol/51.65 ml = 0.251 M

E = (1.252 - 0.0592/2 log(0.0099/0.251)) - 0.222 = 1.07 V

b) 45 ml of Co3+ added

moles of Co3+ added = 0.6 M x 45 ml = 27 mmol

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

This is half equivalence point

E = 1.82 - 0.222 = 1.598 V

c) 49.72 ml Co3+ added

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

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

[Co2+] formed = 13.5 mmol/79.72 ml = 0.17 M

[Co3+] remained = 16.332 mmol/79.72 ml = 0.20 M

E = (1.82 - 0.0592/2 log(0.17/0.2)) - 0.222 = 1.60 V

d) 53.09 ml of Co3+ added

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

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

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

[Co3+] remained = 18.354 mmol/83.09 ml = 0.22 M

E = (1.82 - 0.0592/2 log(0.162/0.22)) - 0.222 = 1.60 V

2) Ag3+ with SO3^2-

a) 3.49 ml 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

[Ag3+] remained = 0.691 mmol/23.49 ml = 0.029 M

[Ag+] formed = 0.349 mmol/23.40 ml = 0.015 M

E = (1.9 - 0.0592/2 log(0.015/0.029)) - 0.241 = 1.667 V

b) 8.37 ml 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

[Ag3+] remained = 0.203 mmol/28.37 ml = 0.007 M

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

E = (1.9 - 0.0592/2 log(0.029/0.007)) - 0.241 = 1.641 V

c) 10.40 ml of SO3^2- added

moles of SO3^2- added = 0.1 M x 10.40 ml = 1.04 mmol

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

Equivalence point

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

d) 15.92 ml of SO3^2- added

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

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

[SO3^2-] remained = 0.552 mmol/35.92 ml = 0.015 M

[SO4^2-] formed = 1.04 mmol/35.92 ml = 0.030 M

pOH = 14 - pH = 14 - 9.5 = 4.5

[OH-] = 3.16 x 10^-5 M

E = (-0.93 - 0.0592/2 log(0.015 x (3.16 x 10^-5)^2/0.03) - 0.241 = -0.895 V

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