Given the following information:

Ni(OH)2(s) Ni2+(aq) + 2 OH (aq)
Ksp=6.0 x 10 16
H2O(l) H+(aq) + OH (aq)
Kw=1.0 x 10 14

What is the equilibrium constant for the reaction,
Ni(OH)2(s) + 2 H+(aq) Ni2+(aq)+ 2 H2O(l)

Note you made a typo on Ksp for Nk(OH)2. It should be 6.0E-16.

Add eqn 1 to the reverse of 2x equn 2.
Keq for the sum = Ksp*(1/Kw^2)

To determine the equilibrium constant for the given reaction, we can use the equilibrium constant expressions for the individual reactions given.

The first reaction is:

Ni(OH)2(s) ⇌ Ni2+(aq) + 2OH-(aq)
Ksp = [Ni2+][OH-]^2

The second reaction is:

H2O(l) ⇌ H+(aq) + OH-(aq)
Kw = [H+][OH-]

In the final reaction:

Ni(OH)2(s) + 2 H+(aq) ⇌ Ni2+(aq) + 2 H2O(l)

We can combine the two equations above to obtain the equilibrium constant expression for the final reaction.

Since the second equation is the reverse of the dissociation of water, we can write:

Kw = [H+] [OH-] = 1.0 x 10^(-14)

Rearranging this equation, we get:

[OH-] = 1.0 x 10^(-14) / [H+]

Now we substitute this expression into the Ksp equation for the first reaction:

Ksp = [Ni2+][OH-]^2 = [Ni2+] * (1.0 x 10^(-14) / [H+])^2

Simplifying further:

Ksp = [Ni2+] * (1.0 x 10^(-14))^2 / [H+]^2
Ksp = [Ni2+] * (1.0 x 10^(-28)) / [H+]^2

Now we can relate the equilibrium constant for the final reaction (Kfinal) to the Ksp for the first reaction:

Kfinal = [Ni2+]/[H+]^2 = Ksp / (1.0 x 10^(-28))

Finally, substituting the given value of Ksp:

Kfinal = (6.0 x 10^16) / (1.0 x 10^(-28))
Kfinal = 6.0 x 10^44

Hence, the equilibrium constant for the reaction Ni(OH)2(s) + 2 H+(aq) Ni2+(aq) + 2 H2O(l) is 6.0 x 10^44.