Aim
The purpose of this experiment is to find out how a system in equilibrium responds to a change in concentration of components in the mixture.
Introduction
Iron(III) ions and thiocyanate ions react in solution to produce thiocyanatoiron(III), a complex ion, according to the equation:
Fe3+(aq) + SCN–(aq) → Fe(SCN)2+(aq)
Pale yellow + colourless → blood-red
The colour produced by the complex ion can indicate the position of equilibrium.
Requirements
- safety glasses
- 4 test-tubes and test-tube rack
- 2 teat-pipettes
- distilled water
- potassium thiocyanate solution, 0.5 M KSCN
- iron(III) chloride solution, 0.5 M FeCl3
- ammonium chloride, NH4Cl
- spatula
- glass stirring rod
Procedure
- Mix together one drop of 0.5 M iron(III) chloride solution and one drop of 0.5 M potassium thiocyanate solution in a test-tube and add about 5 cm3 of distilled water to form a pale orange-brown solution.
- Divide this solution into four equal parts in four test-tubes.
- Add one drop of 0.5 M iron(III) chloride to one test-tube. Add one drop of 0.5 M potassium thiocyanate to a second.
- Compare the colours of these solutions with the original samples. Record your observations.
- Add a spatula-full of solid ammonium chloride to a third test-tube and stir well. Compare the colour of this solution with the remaining tube and note your observation. Ammonium chloride removes iron(III) ions from the equilibrium by forming complex ions such as FeCl4–. A possible reaction is:
Fe3+(aq) + 4Cl–(aq) → FeCl4–(aq)
Interpretation of results
Having made three observations, suggest a cause for each colour change (in terms of the concentrations of the coloured species) and then suggest what can be inferred about a shift in the position of equilibrium. If a pattern has emerged, then you can make a prediction based on the results of the experiment.
| Observation | Explanation | |
| [Fe3+] increased | ||
| [SCN–] increased | ||
| [Fe3+] decreased |
Questions
- How would the position of equilibrium be affected by increasing the concentration of FeSCN2+?
- For each imposed change show how the shift in equilibrium position conforms to Le Chatelier’s principle.
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Consider the exothermic equilibrium system below. The [CuCl4]–2(aq) ion is light green while the [CuBr4]2(aq) ion is dark brown. Originally the equilibrium below was a dark green. [CuCl4]–2(aq) + 4 Br–(aq) ⇋ [CuBr4]–2(aq) + 4 Cl–(aq) Predict the color of the solution after the system has re-established equilibrium. Adding a few drops of Ag+(aq) solution (Ag+ reacts with Cl– ions). Answer: COLOR OF SOLUTION? Adding a few drops of colorless HCl to the solution. Answer: COLOR OF SOLUTION?
Consider the following system under equilibrium:
\[ \underbrace{\ce{Fe^{3+}(aq)}}_{\text{colorless}} + \underbrace{ \ce{SCN^{-}(aq)}}_{\text{colorless}} \rightleftharpoons \underbrace{\ce{FeSCN^{2+}(aq)}}_{\text{red}} \nonumber \]
If more \(Fe^{3+}\) is added to the reaction, what will happen?
According to Le Chatelier's Principle, the system will react to minimize the stress. Since Fe3+ is on the reactant side of this reaction, the rate of the forward reaction will increase in order to "use up" the additional reactant. This will cause the equilibrium to shift to the right, producing more FeSCN2+. For this particular reaction, we will be able to see that this has happened, as the solution will become a darker red color.
There are a few different ways to state what happens here when more Fe3+ is added, all of which have the same meaning:
- equilibrium shifts to the right
- equilibrium shifts to the product side
- the forward reaction is favored
What changes does this cause in the concentrations of the reaction participants?
| \(\ce{SCN^{-}(aq)}\) | Equilibrium will shift to the right, which will use up the reactants. The concentration of \(\ce{SCN^{-}(aq)}\) will decrease \(\ce{[SCN]^{-}\: \downarrow}\) as the rate of the forward reaction increases. |
| \(\ce{FeSCN^{2+}}\) | When the forward reaction rate increases, more products are produced, and the concentration of \(\ce{FeSCN^{2+}}\) will increase. \(\ce{[FeSCN]^{2+}} \uparrow \) |
How about the value of Keq? Notice that the concentration of some reaction participants have increased, while others have decreased. Once equilibrium has re-established itself, the value of Keq will be unchanged.
The value of Keq does not change when changes in concentration cause a shift in equilibrium.
What if more FeSCN2+ is added?
Again, equilibrium will shift to use up the added substance. In this case, equilibrium will shift to favor the reverse reaction, since the reverse reaction will use up the additional FeSCN2+.
- equilibrium shifts to the left
- equilibrium shifts to the reactant side
- the reverse reaction is favored
How do the concentrations of reaction participants change?
| \(\ce{SCN^{-}(aq)}\) | \(\ce{[SCN]^{-}\: \uparrow}\) as the reverse reaction is favored |
| \(\ce{FeSCN^{2+}}\) | \(\ce{[FeSCN]^{2+}} \uparrow \) because this is the substance that was added |
Concentration can also be changed by removing a substance from the reaction. This is often accomplished by adding another substance that reacts (in a side reaction) with something already in the reaction.
Let's remove SCN- from the system (perhaps by adding some Pb2+ ions—the lead(II) ions will form a precipitate with SCN-, removing them from the solution). What will happen now? Equilibrium will shift to replace SCN-—the reverse reaction will be favored because that is the direction that produces more SCN-.
- equilibrium shifts to the left
- equilibrium shifts to the reactant side
- the reverse reaction is favored
How do the concentrations of reaction participants change?
| \(\ce{SCN^{-}}\) | \(\ce{[SCN]^{-}\: \uparrow}\) as the reverse reaction is favored (but also ↓ because it was removed) |
| \(\ce{FeSCN^{2+}}\) | \(\ce{[FeSCN]^{2+}} \uparrow \) because this is the substance that was added |