#25
The Ksp of nickel(II) hydroxide, Ni(OH)2, is 5.84 × 10-16. Calculate the molar solubility of this compound.
Why did the hydroxide ion go to therapy? Because it felt soluble in its own problems.
But let's get down to business! To find the molar solubility of a compound, we need to use its Ksp value.
The Ksp expression for nickel(II) hydroxide is:
Ksp = [Ni2+][OH-]^2
We can assume that the concentration of Ni2+ is equal to the molar solubility, given that nickel(II) hydroxide dissociates into one Ni2+ ion and two OH- ions.
Let's call the molar solubility 'x'. That means:
[Ni2+] = x
[OH-] = 2x
Substituting these values into the Ksp expression, we get:
Ksp = (x)(2x)^2
Now, we can solve for 'x' using the given Ksp value:
5.84 × 10^-16 = x(2x)^2
Simplifying the equation:
5.84 × 10^-16 = 4x^3
Dividing both sides by 4:
1.46 × 10^-16 = x^3
Taking the cube root:
x ≈ 1.057 × 10^-5
So, the molar solubility of nickel(II) hydroxide is approximately 1.057 × 10^-5 M.
Remember, though, if you ever feel uncertain in chemistry, just know that atoms are trustworthy because they make up everything!
To calculate the molar solubility of nickel(II) hydroxide, Ni(OH)2, we need to first write the balanced equation for its dissociation in water.
The dissociation of nickel(II) hydroxide can be represented by the equation:
Ni(OH)2(s) ⇌ Ni2+(aq) + 2OH^-(aq)
According to the stoichiometry of the equation, 1 mole of Ni(OH)2 produces 1 mole of Ni2+ ions and 2 moles of OH- ions.
Now, let's assume the molar solubility of Ni(OH)2 is represented by "x" mol/L.
Using the expression for the solubility product constant (Ksp) of nickel(II) hydroxide, we can write:
Ksp = [Ni2+][OH-]^2
Substituting the values into the equation, we get:
5.84 × 10^(-16) = x * (x)^2
Simplifying the equation, we have:
5.84 × 10^(-16) = x^3
To solve for x, we need to take the cube root of both sides:
∛(5.84 × 10^(-16)) = x
Calculating the cube root of 5.84 × 10^(-16), we find:
x = 7.78 × 10^(-6) mol/L
Therefore, the molar solubility of nickel(II) hydroxide, Ni(OH)2, is 7.78 × 10^(-6) mol/L.
To calculate the molar solubility of nickel(II) hydroxide (Ni(OH)2), we need to start by writing the balanced equation for its dissociation in water.
The balanced equation for the dissociation of nickel(II) hydroxide is:
Ni(OH)2(s) ⇌ Ni2+(aq) + 2OH-(aq)
Now, let's assume that 's' represents the molar solubility of nickel(II) hydroxide, which is also equal to the concentration of Ni2+ ions formed when the compound dissolves.
Therefore, the expression for the solubility product constant, Ksp, can be written as:
Ksp = [Ni2+][OH-]^2
Since the concentration of OH- ions is twice that of the Ni2+ ions, we can substitute 2s for [OH-].
Thus, the expression for Ksp becomes:
Ksp = (Ni2+)(2s)^2
Ksp = 4s^3
Now, we can substitute the given value of Ksp into the equation and solve for s:
5.84 × 10-16 = 4s^3
To solve for s, take the cube root of both sides:
(s^3) = (5.84 × 10-16)/4
s^3 = 1.46 × 10-16
Finally, take the cube root of both sides to solve for the molar solubility 's':
s = (1.46 × 10-16)^(1/3)
s ≈ 1.18 × 10-5 M
Therefore, the molar solubility of nickel(II) hydroxide is approximately 1.18 × 10-5 M.
.........Ni(OH)2 ==> Ni^2+ + 2OH^-
I........solid.......0.......0
C.......solid........x.......2x
E.......solid........x.......2x
Ksp = 5.84E-16 = (Ni^2+)(OH^-)
Substitute and solve for x = solubility Ni(OH)2.