For the titration of 40.0 mL of 0.0100 M Sn^2+ by 0.0500 M Tl^3+ in 1 M HCl, usi

Question

For the titration of 40.0 mL of 0.0100 M Sn^2+ by 0.0500 M Tl^3+ in 1 M HCl, using Pt and Ag | AgCl electrodes:

(a) What is the balanced titration reaction?

(b) What are the two half-reactions that occur at the indicator electrode? Answer by adding charges, coefficients, and products to the following.

(c) What are the two Nernst equations for the cell voltage? The potential for the Ag | AgCl electrode is 0.197 V.

(d) What is the value of E at the following volumes of added Tl^3+ ?

At 1.00mL?

At 4.00 mL?

At 7.90 mL?

At 8.00 mL?

At 8.10 mL?

At 13.0 mL?

Explanation / Answer

(a) Balanced titration reaction : Sn2+ + Tl3+ ---> Sn4+ + Tl+

(b) half reactions are,

Sn4+ + 2e- ---> Sn2+ .....Eo = 0.139 V

Tl3+ + 2e- -----> Tl+ ......Eo = 0.77 V

(c) Nernst equations thus become,

E = {0.139 - 0.0592/nlog([Sn2+]/[Sn4+]} - 0.197 ......[A]

E = {0.77 - 0.059/2log([Tl+]/[Tl3+]} - 0.197 -----[B]

(d) Value of E after,

(i) 1 ml of Tl3+ is added

Using equation [A]

initial moles of Sn2+ = 0.01 x 0.04 = 4 x 10^-4 mols

moles of Tl3+ added = 0.001 x 0.05 = 5 x 10^-5 mols

moles Sn2+ remaining = 3.5 x 10^-4 mols

moles of Sn4+ formed = 5 x 10^-5 mols

molarity of Sn2+ = 3.5 x 10^-4/0.041 = 8.54 mM

molarity of Sn4+ = 5 x 10^-5/0.041 = 1.22 mM

Feed values,

E = 0.139 - 0.0592/2 log(1.22/8.54) - 0.197 = -0.033 V

(ii) 4 ml of Tl3+ added

Using equation [A]

initial moles of Sn2+ = 0.01 x 0.04 = 4 x 10^-4 mols

moles of Tl3+ added = 0.004 x 0.05 = 2 x 10^-4 mols

remaining Sn2+ = 2 x 10^-4 mols

log term = 0

E = 0.139 - 0.197 = -0.058 V

(iii) after 7.9 ml Tl3+ added

Using equation [A]

initial moles of Sn2+ = 0.01 x 0.04 = 4 x 10^-4 mols

moles of Tl3+ added = 0.0079 x 0.05 = 3.95 x 10^-4 mols

moles Sn2+ remaining = 5 x 10^-6 mols

moles of Sn4+ formed = 3.95 x 10^-4 mols

molarity of Sn2+ = 5 x 10^-6/(0.04+0.0079) = 0.1044 mM

molarity of Sn4+ = 3.95 x 10^-4/(0.04+0.0079) = 8.25 mM

Feed values,

E = 0.139 - 0.0592/2 log(8.25/0.1044) - 0.197 = -0.114 V

(iv) 8 ml of Tl3+ added

Using equation [A]

initial moles of Sn2+ = 0.01 x 0.04 = 4 x 10^-4 mols

moles of Tl3+ added = 0.008 x 0.05 = 4 x 10^-4 mols

this is equivalence point,

E = 0.139 V

(v) after 8.1 ml Tl3+ added

Using Equation [B]

E = {0.77 - 0.059/2log([Tl+]/[Tl3+]} - 0.197 -----[B]

excess molarity of Tl3+ = 0.0001 x 0.05/(0.04+0.0081) = 0.104 M

Molarity of Tl+ = 4 x 10^-4/(0.04+0.0081) = 8.32 M

E = {0.77 - 0.059/2log(8.32/0.104} - 0.197 = 0.517 V

(vi) after 13 ml Tl3+ added

molarity of Tl3+ remaining = 0.005 x 0.05/(0.04+0.013) = 4.717 M

molarity of Tl+ = 8.32 M

E = {0.77 - 0.059/2log(8.32/4.717} - 0.197 = 0.566 V

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