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A2 · Practice questions · Stability constants, Kstab

Rank the complexes.

Six original Cambridge-style questions on stability constants.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Expression, units, compare, entropy.

01
Recall
[3 marks]

Define the stability constant of a complex, and write the Kstab expression and units for [Co(H2O)6]2+ + 4Cl− ⇌ [CoCl4]2− + 6H2O.

  • The equilibrium constant for the formation of a complex ion in a solvent from its constituent ions or molecules. ✓
  • Kstab = [CoCl42−] / ([Co(H2O)62+][Cl−]4) (water omitted). ✓
  • Units: mol−4 dm12. ✓
02
Calculation
[3 marks]

At equilibrium, [Cu(NH3)4(H2O)22+] = 0.0500 mol dm−3, [NH3] = 0.200 mol dm−3 and [Cu2+] = 2.60 × 10−12 mol dm−3. Calculate Kstab and log Kstab.

  • Kstab = 0.0500 ÷ (2.60 × 10−12 × 0.2004). ✓
  • 0.2004 = 1.60 × 10−3, so Kstab = 0.0500 ÷ 4.16 × 10−15 = 1.20 × 1013 mol−4 dm12. ✓
  • log Kstab = 13.1. ✓
03
Calculation
[2 marks]

Kstab for [Ag(NH3)2]+ is 1.7 × 107 mol−2 dm6. In a solution at equilibrium [Ag(NH3)2+] = 0.0100 mol dm−3 and [NH3] = 0.100 mol dm−3. Calculate [Ag+].

  • [Ag+] = [complex] ÷ (Kstab × [NH3]2) = 0.0100 ÷ (1.7 × 107 × 0.0100). ✓
  • [Ag+] = 5.88 × 10−8 mol dm−3. ✓
04
Explain
[3 marks]

Explain, in terms of entropy, why [Cu(EDTA)]2− is much more stable than [Cu(NH3)4(H2O)2]2+.

  • [Cu(H2O)6]2+ + EDTA4− → [Cu(EDTA)]2− + 6H2O: 2 particles become 7. ✓
  • With NH3, 5 particles become 5, so there is little entropy change. ✓
  • ΔH is similar (similar bonds broken and made), so the large positive ΔS makes ΔG more negative and Kstab larger for EDTA. ✓
05
Analysis
[2 marks]

Using log Kstab values ([CuCl4]2− 5.6, [Cu(NH3)4(H2O)2]2+ 13.1, [Cu(EDTA)]2− 18.8), predict what is seen when EDTA4− is added to the deep blue ammine complex, and explain.

  • The deep blue colour fades to the paler blue of [Cu(EDTA)]2−. ✓
  • EDTA4− replaces the NH3 and H2O ligands because [Cu(EDTA)]2− has the larger Kstab (18.8 > 13.1). ✓
06
Calculation
[2 marks]

Calculate the equilibrium constant for [CuCl4]2− + 4NH3 ⇌ [Cu(NH3)4(H2O)2]2+ + 4Cl− using the log Kstab values 5.6 and 13.1.

  • K = Kstab(ammine) ÷ Kstab(chloro), so log K = 13.1 − 5.6 = 7.5. ✓
  • K = 107.5 = 3.16 × 107 (no units, since the powers cancel); the exchange goes almost to completion. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.