For [Fe(H2O)6]3+ + SCN− ⇌ [Fe(SCN)(H2O)5]2+ + H2O, the equilibrium concentrations are [Fe(H2O)6]3+ 0.0020 mol dm−3, SCN− 0.0050 mol dm−3 and [Fe(SCN)(H2O)5]2+ 0.0090 mol dm−3. What is Kstab?
- Kstab = [[Fe(SCN)(H2O)5]2+] ÷ ([[Fe(H2O)6]3+][SCN−]); water is left out.
- Bottom line: 0.0020 × 0.0050 = 1.0 × 10−5.
- Kstab = 0.0090 ÷ (1.0 × 10−5) = 900.
- Units: (mol dm−3) ÷ (mol dm−3)2 = dm3 mol−1.
Answer: Kstab = 9.0 × 102 dm3 mol−1