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I understand the theory, but am struggling to answer the following discussion questions:

Why isn't this represented by: $\ce{Fe^{3+} (aq) + 3 SCN- (aq) <=> Fe(SCN)3 (aq)}$
Why do we dilute the solution with water?

I don't understand what the question is asking, can someone please explain

I understand this.

I am unsure, can someone suggest a reason.

I understand this.

Constant volume of container?


  1. I thought ALL equilibrium reactions needed to be in a closed vessel - so if this experiment was conducted in an open test tube, wouldn't this be an error? How would this affect the accuracy of the results - would it be under/over estimated and why?

1 Answer 1


The second representation is more accurate because in water there can be no ``naked'' $\ce{Fe^3+}$ ion; as it goes into solution, iron combines with water to form an ion consisting of a $\ce{Fe}$ atom surrounded by six molecules of water. It is the oxygens that form bonds with the iron atom, in a roughly octahedral arrangement: two oxygen atoms in each coordinate direction, one on either side. The reaction consists of the replacement of one of these coordinated water molecules by the $\ce{SCN-}$ ion; now the iron has bonds to 5 oxygens and 1 nitrogen atom.

The addition of $\ce{H2O}$ to the formula doesn't mean we are diluting the solution with water. It just means we're letting the water that's present anyway play a role in the equation, to reflect reality.

It's not clear to me what they are asking in (ii). The colors change dramatically when you mix $\ce{[Fe(H2O)6]^3+}$ (pale yellow) with $\ce{SCN-}$ (colorless); the result is deep red. Perhaps some context will help. Read the discussion of the reaction in the text to see what they say about color intensity.

(iv) Again, what is the context? If the text is about the spectrospic measuring of the equilibrium constant, low concentrations ensure that the reation obeys the ideal law. In concentrated solutions the law doesn't work quite right unless you replace concentrations by activities, something that is only indirectly measurable (it's whatever makes the formula work).

(vi) Again, we need to know the context.

  1. No, not all equilibrium reactions need to be in a closed vessel. An aqueous reaction that only takes a short time (so that for example water won't evaporate appreciably, messing up the concentration) is just fine in a test tube.

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