Reaction Fe 3 Scn Fescn 2 example essay topic
Lab 2 The solution in test-tube (B) is placed in five other test-tubes. I will now perform four different experiment and then compare with the original solution, and therefor I label the test-tubes a, b, c, d and e. In test-tube (a) I add some KSCN (crystals) and therefor raise the concentration of SCN- in the solution. The visible change is that the solution turn darker red. And since FeSCN 2+ is red, the conclusion is that the solution contains more FeSCN 2+. To test-tube (c) I add some FeNO 3 and the concentration of Fe 3+ is increased and the solution turns darker red the concentration of FeSCN 2+ is higher.
Since no more SCN- has been added the ratio Fe 3+ and SCN- must be in equilibrium. The solution now contains Fe 3+, SCN-, FeSCN 2+, NO 3- and K+. To test-tube (d) I add five drops of silver nitrate, AgNO 3, and the solution is turning brighter - the concentration of FeSCN 2+ must decrease. There seems to be a fight about the SCN-ions, both Ag+ and Fe 3+ want them, but the Ag+ions seems to win and form a salt, Ag SCN, that is hard diluted in water. The effect is, as described earlier, that the concentration of FeSCN- has decreased and the color of the solution is lighter red.
The reaction formula is Fe 3+ + SCN- FeSCN 2+, the reaction i to the left and the reaction is reversible. In the last test-tube I add some dinatriumhydrofosphate, Na 2 HPO 4, and the solution becomes colorless, there is no more FeSCN 2+. The concentration of Fe 3+ has decreased. In the reaction Fe 3+ + SCN- FeSCN 2+, the reaction is to the left.
CONCLUSION: In the reactions above I have changed the equilibrium by changing the concentration of different substances.