Consider the titration of 25.00 mL of 0.0100 M Sn2+ by 0.0500 M Tl3+ in 1.0 M HC
ID: 1029153 • Letter: C
Question
Consider the titration of 25.00 mL of 0.0100 M Sn2+ by 0.0500 M Tl3+ in 1.0 M HCl, using Pt and saturated calomel electrodes to find the end point. Read the textbook (Quantitative Chemical Analysis, 8th Ed. Daniel Harris, page 341-344). Sn2+ + Tl3+ Sn4+ + Tl+
Find the net cell voltage E (v) when 3.60 mL of Tl3+ is consumed.
Calculate the volume (mL) of Tl3+ used at the equivalence point.
Find the net cell voltage E (v) at the equivalence point.
Find the net cell voltage E (v) when 9.10 mL of Tl3+ is consumed.
Explanation / Answer
Titration,
Sn2+ + Tl3+ <==> Sn4+ + Tl+
when 3.60 ml Tl3+ added
initial Sn2+ = 0.01 M x 25 ml = 0.25 mmol
Tl3+ added = 0.05 M x 3.60 ml = 0.18 mmol
moles Sn4+ formed = 0.18 mmol
Sn2+ remained = 0.07 mmol
Ecell = Eo - 0.0592/n log[Sn2+/Sn4+] - 0.241
= 0.15 - 0.0592/2 log(0.07/0.18) - 0.241
= 0.162 V
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Volume Tl3+ added to reach equivalence point = 0.25 mmol/0.05 M = 5 ml
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at equivalence point
Ecell = 1/2(0.15 + 1.21) - 0.241 = 0.439 V
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after 9.10 ml Tl3+ added
initial Sn2+ = 0.01 M x 25 ml = 0.25 mmol
Tl3+ added = 0.05 M x 9.10 ml = 0.455 mmol
moles Tl+ formed = 0.25 mmol
moles Tl3+ remained = 0.205 mmol
Ecell = Eo - 0.0592/n log([Tl+]/[Tl3+]) - 0.241
= 1.21 - 0.0592/2 log(0.25/0.205) - 0.241
= 0.966 V