The solubility of Fe(OH)2 is 3.00 10-3 g in 2.00 liters at 18°C. What is its Ksp at 18°C?
To find the Ksp (solubility product constant) of Fe(OH)2 at 18°C, we need to understand that Ksp is the product of the concentrations of the ions in a saturated solution at equilibrium.
First, let's write the dissociation equation for Fe(OH)2:
Fe(OH)2 ⇌ Fe2+ + 2OH-
From the equation, we can see that one mole of Fe(OH)2 dissociates into one mole of Fe2+ ions and two moles of OH- ions.
Now, let's determine the concentration of Fe2+ ions and OH- ions in the saturated solution.
From the given information, we know that the solubility of Fe(OH)2 is 3.00 × 10^-3 g in 2.00 liters of solution. We can convert this mass into moles by dividing by the molar mass of Fe(OH)2.
The molar mass of Fe(OH)2 can be calculated as follows:
Fe: 1 atom × 55.85 g/mol = 55.85 g/mol
(OH)2: 2 atoms × 16.00 g/mol + 2 atoms × 1.01 g/mol = 34.02 g/mol
Total molar mass: 55.85 g/mol + 34.02 g/mol = 89.87 g/mol
Converting the solubility into moles:
3.00 × 10^-3 g ÷ 89.87 g/mol = 3.34 × 10^-5 mol
Since one mole of Fe(OH)2 produces one mole of Fe2+ ions, the concentration of Fe2+ ions is also 3.34 × 10^-5 mol/L.
Each mole of Fe(OH)2 produces two moles of OH- ions, so the concentration of OH- ions is two times the concentration of Fe2+ ions, which is 2 × 3.34 × 10^-5 mol/L = 6.68 × 10^-5 mol/L.
Now, we have all the necessary information to calculate the Ksp.
Ksp = [Fe2+][OH-]^2
Substituting the concentrations we found:
Ksp = (3.34 × 10^-5 mol/L)(6.68 × 10^-5 mol/L)^2
Calculating the value of Ksp:
Ksp = 1.49 × 10^-14
Therefore, the Ksp of Fe(OH)2 at 18°C is 1.49 × 10^-14.