What is the K constant for the following reaction [Cu(H2O)6] aq + Cl- aq ⇌ [Cu(H2O)5Cl] aq + H2O (l) Given that [Cu(H2O)5Cl] aq + H2O (l) ⇌ [Cu(H2O)6] aq + Cl- aq. K = 0.23
Question
What is the K constant for the following reaction [Cu(H2O)6] aq + Cl- aq ⇌ [Cu(H2O)5Cl] aq + H2O (l) Given that [Cu(H2O)5Cl] aq + H2O (l) ⇌ [Cu(H2O)6] aq + Cl- aq. K = 0.23
Solution
The equilibrium constant (K) for a reaction is the reciprocal of the equilibrium constant for the reverse reaction.
Given that the equilibrium constant (K) for the reaction [Cu(H2O)5Cl] aq + H2O (l) ⇌ [Cu(H2O)6] aq + Cl- aq is 0.23, the equilibrium constant for the reverse reaction [Cu(H2O)6] aq + Cl- aq ⇌ [Cu(H2O)5Cl] aq + H2O (l) would be 1/K, or 1/0.23.
So, the K constant for the reaction [Cu(H2O)6] aq + Cl- aq ⇌ [Cu(H2O)5Cl] aq + H2O (l) is approximately 4.35.
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