A saturated Ni(OH)_2_ can be prepared by dissolving 0.239 mg of Ni(OH)_2_ in water to make 500.0 mL of solution. What is the Ksp for Ni(OH)_2_?
M Ni(OH)2 = moles/L.
You have grams, convert to moles and you have L. That is the solubility of Ni(OH)2 which I will call x for simplicity.
.........Ni(OH)2 ==> Ni^2+ + 2OH^-
...........x.........x..........2x
Substitute into Ksp expression and solve for Ksp.
To find the solubility product constant (Ksp) for Ni(OH)2, we need to use the given information about the mass of the compound and the volume of the solution.
Step 1: Convert the mass of Ni(OH)2 to moles.
Calculate the number of moles of Ni(OH)2 using the molecular weight of Ni(OH)2:
Molar mass of Ni(OH)2 = atomic weight of Ni + (atomic weight of O × 2) + (atomic weight of H × 2)
Molar mass of Ni(OH)2 = 58.69 + (16.00 × 2) + (1.01 × 2)
Molar mass of Ni(OH)2 = 92.69 g/mol
Mass of Ni(OH)2 = 0.239 mg = 0.000239 g
Number of moles of Ni(OH)2 = mass / molar mass
Number of moles of Ni(OH)2 = 0.000239 g / 92.69 g/mol
Step 2: Calculate the molar concentration of Ni(OH)2.
Molar concentration (M) is calculated using the formula:
M = moles / volume (in L)
Volume of solution = 500.0 mL = 0.500 L
Molar concentration of Ni(OH)2 = moles / volume
Molar concentration of Ni(OH)2 = (0.000239 g / 92.69 g/mol) / 0.500 L
Step 3: Write the balanced chemical equation for the dissociation of Ni(OH)2.
Ni(OH)2 ↔ Ni^2+ + 2OH^-
Step 4: Use the stoichiometry of the balanced equation to determine the concentrations of Ni^2+ and OH^-.
Since 1 mole of Ni(OH)2 produces 1 mole of Ni^2+ ions and 2 moles of OH^- ions, the concentration of Ni^2+ can be considered equal to the molar concentration of Ni(OH)2.
The concentration of OH^- ions is twice the molar concentration of Ni(OH)2.
Step 5: Write the expression for the solubility product constant (Ksp) using the concentrations of Ni^2+ and OH^-.
Ksp = [Ni^2+] [OH^-]^2
Step 6: Substitute the concentrations into the Ksp expression.
Ksp = (molar concentration of Ni(OH)2) × (concentration of OH^-)^2
Ksp = (0.000239 g / 92.69 g/mol) / 0.500 L × [2 × (0.000239 g / 92.69 g/mol)]^2
Now, calculate the value of Ksp using a calculator.
To find the Ksp (solubility product constant) for Ni(OH)2, we need to first determine the concentration of Ni2+ and OH- ions in the saturated solution.
Given:
Mass of Ni(OH)2 = 0.239 mg
Volume of solution = 500.0 mL = 0.5000 L
First, let's convert the mass of Ni(OH)2 to moles:
Molar mass of Ni(OH)2 = (58.69 g/mol) + 2(1.01 g/mol) + 2(16.00 g/mol) = 92.69 g/mol
Number of moles of Ni(OH)2 = mass / molar mass
= 0.239 mg / 92.69 g/mol
= 2.576e-6 mol
Since Ni(OH)2 dissociates into Ni2+ and 2OH-, the concentration of Ni2+ and OH- ions is twice the concentration of Ni(OH)2.
Concentration of Ni2+ ions = 2 * (2.576e-6 mol / 0.5000 L) = 2 * 5.152e-6 M
Concentration of OH- ions = 2 * (2 * 5.152e-6 M) = 4 * 5.152e-6 M
The solubility product constant (Ksp) expression for Ni(OH)2 is:
Ksp = [Ni2+] * [OH-]^2
Substituting the concentrations we calculated:
Ksp = (2 * 5.152e-6) * (4 * 5.152e-6)^2
Calculating the value of Ksp:
Ksp = 2 * 5.152e-6 * (4 * 5.152e-6)^2
= 2 * 5.152e-6 * 4 * (5.152e-6)^2
= 2 * 5.152e-6 * 4 * 2.656384e-11
= 1.060736e-16
Therefore, the Ksp value for Ni(OH)2 is 1.060736e-16.