The Ksp of iron(II) hydroxide, Fe(OH)2, is 4.87 × 10-17. Calculate the solubility of this compound in g/L.
To calculate the solubility of Fe(OH)2 in g/L, we need to determine the concentration of Fe2+ and OH- ions in the saturated solution.
The balanced equation for the dissociation of Fe(OH)2 is:
Fe(OH)2(s) ⇌ Fe2+(aq) + 2OH-(aq)
The equilibrium constant expression for this reaction is:
Ksp = [Fe2+][OH-]^2
Given the Ksp value of Fe(OH)2 (4.87 × 10^-17), we can set up the following equation:
4.87 × 10^-17 = [Fe2+][OH-]^2
Since Fe(OH)2 dissociates to Fe2+ and 2OH-, the concentrations of Fe2+ and OH- ions will be equal. Let's denote this concentration as x.
Therefore, the equation becomes:
4.87 × 10^-17 = x * (x)^2
Rearranging the equation:
x^3 = 4.87 × 10^-17
Taking the cube root of both sides:
x ≈ 9.76 × 10^-6
The concentration of Fe2+ and OH- ions is approximately 9.76 × 10^-6 M in the saturated solution.
To convert this to g/L, we need to multiply by the molar mass of Fe(OH)2:
Molar mass of Fe(OH)2 = (1 × atomic mass of Fe) + (2 × atomic mass of O) + (2 × atomic mass of H)
= (1 × 55.845) + (2 × 15.999) + (2 × 1.008)
≈ 89.855 g/mol
To convert the concentration to g/L:
Solubility (g/L) = (9.76 × 10^-6 mol/L) × (89.855 g/mol)
≈ 0.875 g/L
Therefore, the solubility of Fe(OH)2 is approximately 0.875 g/L.
To calculate the solubility of iron(II) hydroxide (Fe(OH)2) in grams per liter (g/L), we need to use the solubility product constant (Ksp) and the molar mass of Fe(OH)2.
The Ksp expression for Fe(OH)2 is written as:
Ksp = [Fe2+][OH-]^2
Since Fe(OH)2 dissociates into one Fe2+ ion and two OH- ions, the expression becomes:
Ksp = [Fe2+][OH-]^2
Let's assume that the solubility of Fe(OH)2 is "x." Therefore, the concentration of Fe2+ ions ([Fe2+]) will also be "x." The concentration of OH- ions ([OH-]) will be 2x because there are two OH- ions for every Fe2+ ion.
Substituting the values into the Ksp expression:
4.87 × 10-17 = x * (2x)^2
4.87 × 10-17 = 4x^3
x^3 = (4.87 × 10-17) / 4
Taking the cube root of both sides to solve for x:
x ≈ (4.87 × 10-17 / 4)^(1/3)
Calculating the value on the right side:
x ≈ 9.81 × 10-6
Therefore, the solubility of iron(II) hydroxide (Fe(OH)2) is approximately 9.81 × 10-6 moles per liter (mol/L).
To convert this solubility into grams per liter (g/L), we need to multiply the molar mass of Fe(OH)2 by the solubility in moles per liter and then multiply by 1,000 to convert moles to grams:
Solubility in g/L = (Molar mass of Fe(OH)2) * (Solubility in mol/L) * 1,000
The molar mass of Fe(OH)2 can be calculated as follows:
Fe: 55.85 g/mol
O: 16.00 g/mol (there are two O atoms in Fe(OH)2)
H: 1.01 g/mol (there are two H atoms in Fe(OH)2)
Molar mass of Fe(OH)2 = (55.85 g/mol) + (16.00 g/mol × 2) + (1.01 g/mol × 2) = 89.91 g/mol
Substituting the values into the formula:
Solubility in g/L = (89.91 g/mol) * (9.81 × 10-6 mol/L) * 1,000
Calculating the value on the right side:
Solubility in g/L ≈ 0.878 g/L
Therefore, the solubility of iron(II) hydroxide (Fe(OH)2) is approximately 0.878 grams per liter (g/L).
........Fe(OH)3 ==> Fe^3+ + 3OH^-
I.......solid........0........0
C.......solid........x........3x
E.......solid........x........3x
Ksp = (F^3+)(OH^-)^3
Substitute the E line into Ksp expression and solve for x, = [Fe(OH)3] in mols/L. That x molar mass = grams.