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The solubility product of nickel (II) hydroxide at 25°C is 1.6 x 10-16.

Calculate the molar solubility of nickel (II) hydroxide in 0.100 M sodium hydroxide.

Ni(OH)2 <=> Ni2+ + 2OH–
I = Ni2+ is 0 & 2OH– is 0.1
Change = Ni2+ is X and 2OH- is 2x
Equili = Ni2+ is X & 2OH- is (0.1+2x)

0.10/ Ksp = 0.10/1.6 x 10-16

To calculate the molar solubility of nickel (II) hydroxide in 0.100 M sodium hydroxide, you will need to use the solubility product constant (Ksp) and set up an equilibrium expression.

First, write the balanced equation for the dissociation of nickel (II) hydroxide:

Ni(OH)2 <=> Ni2+ + 2OH–

Next, set up an ICE table (Initial, Change, Equilibrium) to keep track of the concentrations:

I = Initial concentration
C = Change in concentration
E = Equilibrium concentration

I:
Ni2+ = 0 (since it is not present initially)
OH– = 0.100 M (as given)

C:
Ni2+ = X (concentration of Ni2+ at equilibrium)
OH– = 2X (concentration of OH- at equilibrium, as there are 2 moles of OH- produced for every mole of Ni2+)

E:
Ni2+ = X
OH– = 0.100 M + 2X

Now, write the equilibrium expression using the concentrations:

Ksp = [Ni2+] * [OH–]^2

Substitute the equilibrium concentrations into the equilibrium expression:

Ksp = X * (0.100 M + 2X)^2

Finally, you can solve for X, which represents the molar solubility of Ni(OH)2 in the 0.100 M NaOH solution, by rearranging the equation and solving for X:

Ksp / (0.100 M) = (0.100 M + 2X)^2

Divide Ksp by 0.100 M on both sides:

Ksp / (0.100 M) = (0.100 M + 2X)^2 / (0.100 M)

Take the square root of both sides:

sqrt(Ksp / (0.100 M)) = 0.100 M + 2X

Subtract 0.100 M from both sides:

sqrt(Ksp / (0.100 M)) - 0.100 M = 2X

Divide by 2 to solve for X:

X = (sqrt(Ksp / (0.100 M)) - 0.100 M) / 2

Plug in the given value for Ksp (1.6 x 10^-16) and the concentration of NaOH (0.100 M) to calculate the molar solubility of Ni(OH)2.