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Calculation of equilibrium constant

Average titre = 7.3 cm3

Number of moles of potassium thiocyanate = 0.020 ´ 7.3/1000

KCNS(aq) + AgNO3(aq) ⇒ AgCNS(s) + KNO3(aq)

Ag+(aq) º KCNS(aq)

\ number of moles of Ag+(aq) in 10 cm3 = 0.020 ´ 7.3/1000

\ number of moles of Ag+(aq) in 1000 cm3 = 0.020 ´ 7.3/1000 ´ 100 = 0.0146 mol dm-3

[Ag+(aq)]eqm = 0.0146 mol dm-3

As we started with the same concentration of Ag+ ions and Fe2+ ions, and they reacted in a 1:1 ratio, the equilibrium concentrations of these two ions must be the same.

[Fe2+(aq)]eqm = 0.0146 mol dm-3

The Fe2+ ions which have "disappeared" have been converted to Fe3+ ions:

[Fe2+(aq)]initial = 0.05 mol dm-3 (this is half of 0.1 mol dm-3 as it was diluted down by the added silver nitrate solution)

\ [Fe3+(aq)]eqm = [Fe2+(aq)]initial - [Fe2+(aq)]eqm = 0.05 - 0.0146 = 0.0354 mol dm-3

These concentration terms are now put into the Kc expression:

Kc = 0.0354/(0.0146 ´ 0.0146) = 166 mol-1 dm3 at room temperature

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