The Ksp of manganese(II) hydroxide, Mn(OH)2, is 2.00 × 10-13. Calculate the solubility of this compound in g/L.
To calculate the solubility of manganese(II) hydroxide (Mn(OH)2) in g/L, we need to use the Ksp value. The Ksp expression for Mn(OH)2 is:
Ksp = [Mn2+][OH-]^2
Since Mn(OH)2 dissociates into two OH- ions, the concentration of OH- is twice that of Mn2+. Let's assume the molar solubility of Mn(OH)2 is x mol/L. Then:
[Mn2+] = x mol/L
[OH-] = 2x mol/L
Substituting these values into the Ksp expression:
Ksp = (x)(2x)^2
Simplifying:
Ksp = 4x^3
Rearranging the equation to solve for x:
x^3 = Ksp / 4
x = (Ksp / 4)^(1/3)
Substituting the given Ksp value:
x = (2.00 × 10^-13 / 4)^(1/3)
Calculating this value:
x = 5.62 × 10^-5 mol/L
To convert the solubility to g/L, we need to multiply by the molar mass of Mn(OH)2. The molar mass of Mn(OH)2 is:
Molar mass = (1 × atomic mass of Mn) + (2 × atomic mass of O) + (2 × atomic mass of H)
Molar mass = (1 × 54.938045) + (2 × 15.999) + (2 × 1.00784)
Molar mass = 54.938045 + 31.998 + 2.01568
Molar mass = 88.951725 g/mol
Now we can calculate the solubility in g/L:
Solubility = x mol/L × molar mass
Solubility = (5.62 × 10^-5 mol/L) × (88.951725 g/mol)
Calculating this value:
Solubility = 5.00 × 10^-3 g/L
Therefore, the solubility of manganese(II) hydroxide (Mn(OH)2) is 5.00 × 10^-3 g/L.
To calculate the solubility of manganese(II) hydroxide (Mn(OH)2) in g/L, we need to find the molar solubility first.
The solubility product constant (Ksp) for manganese(II) hydroxide is given as 2.00 × 10^-13. The Ksp expression for Mn(OH)2 is:
Ksp = [Mn2+][OH-]^2
In pure water, Mn(OH)2 dissociates completely into Mn2+ ions and OH- ions. Therefore, the concentration of Mn2+ ions and OH- ions is equal and can be represented as x.
Substituting these values into the Ksp expression:
Ksp = [x][x^2] = x^3
Now, we can solve for x. Taking the cube root of the Ksp value:
x = (Ksp)^(1/3)
x = (2.00 × 10^-13)^(1/3) ≈ 1.24 × 10^-4
The molar solubility of manganese(II) hydroxide is 1.24 × 10^-4 moles per liter (mol/L).
To calculate the solubility in g/L, we need to know the molar mass of Mn(OH)2, which can be calculated by summing the atomic masses of its constituent elements.
Molar mass of Mn(OH)2 = (1 × atomic mass of Mn) + (2 × atomic mass of O) + (2 × atomic mass of H) = (1 × 54.938) + (2 × 15.999) + (2 × 1.008) = 88.936 g/mol
Finally, we can convert the molar solubility to grams per liter (g/L):
Solubility (g/L) = (molar solubility in mol/L) × (molar mass of Mn(OH)2)
Solubility (g/L) = (1.24 × 10^-4 mol/L) × (88.936 g/mol) ≈ 1.10 × 10^-2 g/L
Therefore, the solubility of manganese(II) hydroxide in g/L is approximately 1.10 × 10^-2 g/L.
.......Mg(OH)2 ==> Mg^2+ + 2OH^-
I......solid........0.......0
C......solid........x......2x
E......solid........x......2x
Substitute the E line into the Ksp expression and solve for x = solubility in mols/L. Convert to grams/L.