Question

What is the equilibrium constant of zinc tetrahydrate? [Zn(H2O)4]2+

Please explain how you determined/calculated the value. If you found it from a certain source, please share the source.

Answer 1

The formation of zinc tetrahydrate, i.e. [Zn(H₂O)₄]²⁺ from Zn²⁺ and H₂O can be written as follows.

Zn²⁺(aq) + 4H₂O(l) [Zn(H₂O)₄]²⁺(s)

The equilibrium constant for the formation of zinc tetrahydrate is taken from the internet.

"https://books.google.co.in/books?id=j3OSKTCuO00C&pg=PA693&lpg=PA693&dq=equilibrium+constant+for+Zn2%2B+%2B+4CN-+%3D+Zn(CN)4%5E2-&source=bl&ots=W92El-Asz2&sig=ACfU3U10tr3d93qTpFSjEJNVGRZfaQUZ3A&hl=en&sa=X&ved=2ahUKEwiYg7PEvPvjAhW56nMBHRk6C9AQ6AEwBHoECAkQAQ#v=onepage&q=equilibrium%20constant%20for%20Zn2%2B%20%2B%204CN-%20%3D%20Zn(CN)4%5E2-&f=false"

[Zn(CN)₄]^2- Zn²⁺(aq) + 4CN^-(aq); K = 1.3×10^-17

http://www.docbrown.info/page07/transition10Zn.htm

[Zn(H₂O)₄]²⁺(aq) + 4CN^-(aq) [Zn(CN)₄]^2-(aq) + 4H₂O(l); K = 5×10¹⁶

The addition of the above two equilibria gives the following equilibrium.

[Zn(H₂O)₄]²⁺(aq) Zn²⁺(aq) + 4H₂O(l); resulting K = (1.3×10^-17)×(5×10¹⁶) = 0.65

Hence, the equilibrium constant for the formation of zinc Tetra hydrate can be written as follows.

K = 1/0.65 = 1.5385