The solubility product of nickel (II) hydroxide at 25°C is 1.6 x 10-16.
a) Calculate the molar solubility of nickel (II) hydroxide in pure water at 25°C.
b) Calculate the molar solubility of nickel (II) hydroxide in 0.100 M sodium hydroxide.
This is a fairly straight forward two problems. Tell us what you don't understand about a, then b. Perhaps we can help you through them.
To solve these problems, we need to understand the concept of solubility product (Ksp). The Ksp is the equilibrium constant for the dissolution of a sparingly soluble compound in water.
a) To calculate the molar solubility of nickel (II) hydroxide in pure water at 25°C, we need to set up an equilibrium expression for the dissolution of nickel (II) hydroxide:
Ni(OH)2(s) ⇌ Ni2+(aq) + 2OH-(aq)
The solubility product expression for this equilibrium is:
Ksp = [Ni2+][OH-]^2
Since nickel (II) hydroxide is an ionic compound, it dissociates completely in water. Therefore, at equilibrium, the concentration of Ni2+ ions will be equal to the concentration of OH- ions:
[Ni2+] = [OH-]
So, we can substitute [OH-] for [Ni2+] in the Ksp expression:
Ksp = [OH-][OH-]^2 = ([OH-])^3
Given that the solubility product (Ksp) of nickel (II) hydroxide is 1.6 x 10^-16, we can solve for the molar solubility ([OH-]).
1.6 x 10^-16 = ([OH-])^3
Now, take the cube root of both sides to solve for [OH-]:
[OH-] = (1.6 x 10^-16)^(1/3)
= 1.00 x 10^-5 M
Therefore, the molar solubility of nickel (II) hydroxide in pure water at 25°C is 1.00 x 10^-5 M.
b) To calculate the molar solubility of nickel (II) hydroxide in 0.100 M sodium hydroxide (NaOH), we need to consider the common ion effect. Sodium hydroxide is a strong base that significantly increases the concentration of hydroxide ions (OH-) in the solution.
Let's assume that x moles of nickel (II) hydroxide dissolve in 0.100 M NaOH. The concentration of nickel (II) ions ([Ni2+]) will be x M, and the concentration of hydroxide ions ([OH-]) will be (2x + 0.100) M (because NaOH already contributes 0.100 M OH-).
Using the same solubility product expression, we can substitute the concentrations into the Ksp expression:
Ksp = [Ni2+][OH-]^2
= (x)(2x + 0.100)^2
Given that the solubility product (Ksp) of nickel (II) hydroxide is 1.6 x 10^-16, we can solve for x:
1.6 x 10^-16 = (x)(2x + 0.100)^2
Solve this equation using algebraic techniques or numerical methods to find the value of x. This value represents the molar solubility of nickel (II) hydroxide in 0.100 M sodium hydroxide.