Stability Constants, Kstab (Cambridge (CIE) A Level Chemistry): Flashcards
Exam code: 9701
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Define stability constant (Kstab).
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Define stability constant (Kstab).
A stability constant (Kstab) is the equilibrium constant for the formation of a complex ion from its constituent central metal ion and ligands in solution. A larger value indicates a more stable complex.
Why is the concentration of water not included in the Kstab expression?
Water is present in excess as the solvent. Any water produced by the ligand substitution is negligible compared to the water already present, so its concentration is treated as constant and omitted.
True or False?
A larger Kstab value indicates a more stable complex.
True. The larger the Kstab, the further the equilibrium lies towards the products, meaning the complex formed is more stable.
For the reaction [Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O, the Kstab expression is: Kstab = .......... .
For the reaction [Cu(H2O)6]2+ + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4H2O, the Kstab expression is: Kstab = [[Cu(NH3)4(H2O)2]2+] / [[Cu(H2O)6]2+][NH3]4.
What is meant by stepwise stability constants in ligand substitution?
In stepwise substitution, each individual ligand replacement has its own equilibrium constant (K1, K2, etc.). The overall stability constant Kstab summarises all the stepwise constants combined.
True or False?
Stability constants are often expressed on a log10 scale to make comparison easier.
True. The log10 scale is used because Kstab values span many orders of magnitude, making them easier to compare.
In the Kstab expression, what does the position of equilibrium tell you if Kstab is very large?
A very large Kstab means the equilibrium lies far to the right, with products (the complex) strongly favoured over the constituent ions and ligands.
When a 0.15 mol dm-3 solution of [Cu(H2O)5Cl]+ reacts with 0.15 mol dm-3 Cl- to give 0.10 mol dm-3 Cu(H2O)4Cl2 at equilibrium, the Kstab = .......... dm3 mol-1.
When a 0.15 mol dm-3 solution of [Cu(H2O)5Cl]+ reacts with 0.15 mol dm-3 Cl- to give 0.10 mol dm-3 Cu(H2O)4Cl2 at equilibrium, the Kstab = 40 dm3 mol-1.
Define competing equilibrium.
A competing equilibrium in ligand exchange is the simultaneous existence of multiple substitution equilibria in solution, with the most stable complex (highest Kstab) being preferentially formed.
Why does ligand exchange occur in a transition metal complex?
Ligand exchange occurs because the incoming ligands form a more stable complex, with a larger Kstab, than the original complex. The equilibrium shifts to favour the more stable product.
True or False?
The complex [CoCl4]2- is more stable than [Co(NH3)6]2+ based on their stability constants.
False. The log10 Kstab of the NH3 complex (13.1) is greater than that of the Cl- complex (5.6), so the ammonia complex is more stable.
Define log10 Kstab.
log10 Kstab is the stability constant expressed on a base-10 logarithmic scale, making it easier to compare the relative stabilities of different complexes across a wide range of values.
When excess NH3 is added to [CoCl4]2-, the chloride ligands are replaced by ammonia because the ammonia complex has a .......... Kstab, shifting the equilibrium to the .......... .
When excess NH3 is added to [CoCl4]2-, the chloride ligands are replaced by ammonia because the ammonia complex has a larger Kstab, shifting the equilibrium to the right.
Given that log10 Kstab for [Ag(CN)2]- = 18.7 and for [Ag(NH3)2]+ = 6.2, which complex is more stable?
[Ag(CN)2]- is more stable because it has the larger log10 Kstab value (18.7 vs 6.2), meaning its equilibrium lies further to the right with a greater proportion of complex formed.
True or False?
A higher Kstab means the equilibrium lies further to the right, favouring the complex.
True. A higher Kstab value means the complex is more thermodynamically stable and the equilibrium position lies further towards the products.
Three silver complexes have Kstab values: [Ag(S2O3)2]3- = 2.9 × 1013, [Ag(CN)2]- = 5.3 × 1018, [Ag(NH3)2]+ = 1.6 × 107. Which has the lowest concentration at equilibrium if all ligands are present equally?
[Ag(NH3)2]+ has the lowest concentration because it has the smallest Kstab value and is the least stable complex.
Ligand exchange forms a new complex that is .......... than the original. The position of equilibrium shifts to the ...........
Ligand exchange forms a new complex that is more stable than the original. The position of equilibrium shifts to the right.
Define stability constant comparison.
Stability constant comparison is the use of Kstab values to determine which complex is more stable: the complex with the larger Kstab is preferentially formed when ligands compete for the same metal ion.
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