A LevelChemistryCIETopic Questions28. Chemistry of Transition Elements28.5 Stability Constants, Kstab

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First exams 2025

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Stability Constants, Kstab (CIE A Level Chemistry)

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1a4 marks

When chromium(III) sulfate dissolves in water, a green solution containing the [Cr(H2O)6]3+ ion forms.

i)
State the bond angles found in this complex ion.
 
[1]
 
ii)
Explain why the chromium(III) complex ion is coloured.
 
[3]

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1b5 marks

Ethylenediaminetetraacetate, EDTA4−, shown in Fig. 1.1. is a polydentate ligand.

edta-ligand

Fig. 1.1

When a solution of EDTA4− is added to a solution of [Cr(H2O)6]3+ ions, a new complex ion is formed.

[Cr(H2O)6]3+ + EDTA4– ⇌ [Cr(EDTA)]– + 6H2O

i)
Name the type of reaction occurring here.
 
[1]
 
ii)
Write an expression for the stability constant, Kstab, of [CrEDTA]– in this reaction.
 
[1]
 
iii)
The numerical value of the Kstab of [CrEDTA]– is 2.51 × 1023 in this reaction. 
 
Suggest what this indicates about the position and entropy of the equilibrium.
 
[3]

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1c
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Chromium (III) picolinate, shown in Fig. 1.2, is a neutral complex that can be prepared from the weak acid, picolinic acid.2-5

Fig. 1.2

Chromium(III) picolinate is used in tablets as a nutritional supplement for chromium. 

i)
Draw the structure of the ligand in chromium(III) picolinate.
 
[1]
 
ii)
A typical tablet of chromium(III) picolinate contains 200 μg of chromium.
 
Calculate the mass, in g, of chromium (III) picolinate in a typical tablet. Give your answer to three significant figures.
 
[2]

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1d3 marks

Compound L is a complex with the empirical formula CrN4H12Cl3 

The formula of compound L contains one chloride ion and a complex ion M, which has two stereoisomers.

Complete three-dimensional diagrams to show the shape of the stereoisomers of complex ion M.

 
 
 
 
 
Cr 
 
 
 
 
 
 
 
 
 
Cr 
 
 
 
 
 

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1a1 mark

Sketch the shape of a 3dxy orbital.

q7a-9701-y22-sp-4-cie-ial-chem

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1b3 marks
i)
Some transition elements and their compounds behave as catalysts. Explain why transition elements behave as catalysts.

[2]

ii)
Catalysis can be classified as heterogeneous or homogeneous.
Complete Table 1.1 by placing one tick (✓) in each row to indicate the type of catalysis in each reaction.

Table 1.1

  type of catalysis
reaction heterogeneous homogeneous
Fe in the Haber process    
Fe2+ in the I– / S2O82– reaction    
NO2 in the oxidation of SO2    

[1]

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1c2 marks

A solution containing a mixture of Sn2+(aq) and Sn4+(aq) is added to a solution containing a mixture of Fe2+(aq) and Fe3+(aq).

Table 2.2 lists electrode potentials for some electrode reactions of these ions.

Table 2.2

electrode reaction Eɵ / V
Fe2+ + 2e– ⇌ Fe –0.44
Fe3+ + 3e– ⇌ Fe –0.04
Fe3+ + e– ⇌ Fe2+ +0.77
Sn2+ + 2e– ⇌ Sn –0.14
Sn4+ + 2e– ⇌ Sn2+ +0.15

Eɵ data from the table can be used to predict the reaction that takes place when the two solutions are mixed.

i)
Write an equation for this reaction.

 [1]

ii)
Calculate E subscript cell superscript straight ɵ for this reaction.

[1]

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1d5 marks

Hexaaquairon(III) ions are pale violet. They form a colourless complex with fluoride ions, F–, as shown in equilibrium 1, and a deep-red complex with thiocyanate ions, SCN–, as shown in equilibrium 2.

equilibrium 1 [Fe(H2O)6]3+ + F– ⇌ [Fe(H2O)5F]2+ + H2O Kstab = 2.0 × 105 mol–1 dm3   

                          violet                          colourless

equilibrium 2 [Fe(H2O)6]3+ + SCN– ⇌ [Fe(H2O)5SCN]2+ + H2O Kstab = 1.0 × 103 mol–1 dm3

                           violet                                deep-red

The following two experiments are carried out.

Experiment 1: A few drops of KSCN(aq) are added to 5 cm3 of Fe3+(aq), followed by a few drops of KF(aq).

Experiment 2: A few drops of KF(aq) are added to 5 cm3 of Fe3+(aq), followed by a few drops of KSCN(aq).

i)

Predict and explain the sequence of colour changes you would observe in each of Experiment 1 and Experiment 2.

Experiment 1 ...................................................

Experiment 2 ....................................................

   [4]

ii)
Name the type of reaction occurring during the experiments in (d)(i).

[1]

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1e2 marks

Solutions of iron(III) salts are acidic due to the equilibrium shown.

 
[Fe(H2O)6]3+ (aq) ⇌ [Fe(H2O)5(OH)]2+ (aq) + H+ (aq)   Ka = 8.9 × 10–4 mol dm–3
 

Calculate the pH of a 0.25 mol dm–3 FeCl3 solution.
Show your working.

pH = ..........................................................

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2a1 mark

Define the term ligand.

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2b1 mark

[Cu(H2O)6]2+ reacts with ammonia to form a new complex. Write an expression for the overall stability constant, Kstab, for the formation of [Cu(H2O)2(NH3)4]2+

   Kstab = 

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2c
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Table 2.1 shows the stability constants for each stage in the replacement of four water molecules for the [Cu(H2O)6]2+ complex

Table 2.1

Kn value / dm3 mol-1
K1 1.78 x 104
K2 4.07 x 103
K3 9.55 x 102
K4 1.74 x 102

i)
Using Table 2.1, calculate the stability constant, Kstab, and state the units.



Kstab = ...................................... 

Units = .......................................

[2]

ii)
Explain how this value relates to the relative stabilities of the two complexes
[1]

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3a1 mark

State the meaning of the term stability constant, Kstab

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3b4 marks

Silver forms different complexes. The stability constant, Kstab, values at 298 K are shown in Table 3.1

Table 3.1

complex stability constant, Kstab
[Ag(CN)2]– 5.3 x 1018
[Ag(NH3)]+ 1.7 x 107
[Ag(S3O3)2]3– 2.9 x 1013

The following equilibrium exists between two complex ions of silver in the +1 oxidation state.

Ag+ + NH3 ⇌ [AgNH3]+

i)
Write the expression for the stability constant, Kstab, for this equilibrium

   Kstab = 


[1]

ii) 
When a 0.500 mol dm–3 solution of AgNO3 (aq) is mixed with 0.500 mol dm–3 NH3 (aq) the equilibrium mixture of [Ag(NH3)]+ (aq) was found to be 0.4998 mol dm–3

Calculate Kstab and state the units
Kstab = ...................................

Units = .................................
[3]

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3c
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Using Table 3.1, deduce the order of stability of the silver complexes

most stable .......................................................
  .......................................................
least stable .......................................................

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