Chemistry Equilibrium constant

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2010-03-25 1:22 am

Measure 0.1 M iron (III) sulphate solution and 0.1 M potassium thiocyanate solution using measuring cylinders.Pour each combination of iron (III) sulphate and potassium thiocyanate into a thin-walled test tube which is clean and dry

Volume of Fe2(SO4)3 used-------2 cm^-3
Volume of KSCN used-------------6 cm^-3
Volume of water used-----12 cm^-3

Absorbance of colormeter:0.108
Equilibrium concentration of Fe(NCS)----0.008M

Calculate the equilibrium constant for the reaction between iron(III)inos
and thiosulphate ions.

回答 (1)

Heyman

2010-03-25 4:46 am

✔ 最佳答案

Initial conc. of [Fe]3+ = 0.1 x 2 x 2/(12+6+2) M = 0.02 M
(1 mol iron (III) sulphate gives 2 mol Fe3+)
Initial conc. of [SCN]- = 0.1 x 6/(12+6+2) M = 0.03 M
(The above calculations assume there is no significant volume change)

[Fe]3+ + [SCN]- → [Fe(SCN)]2+

As 1 mol [Fe]3+ forms complex with 1 mol [SCN]-, [SCN]- is in excess.
Now, as 1 mol [Fe]3+ gives 1 mol [Fe(SCN)]2+,
0.008 M [Fe]3+ has reacted.
So equilibrium conc. of [Fe]3+ = 0.02 - 0.008 = 0.012 M
Equilibrium conc. of [SCN]- = 0.03 - 0.008 = 0.022 M

Kc for the reaction = [Fe(SCN)2+] / ([Fe3+][SCN-])
= 0.008 / (0.012 x 0.022) = 30.3

2010-03-27 16:06:59 補充：
個 Kc 漏左個unit ─ mol dm^(-3)

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