The Ksp of manganese(II) hydroxide, Mn(OH)2, is 2.00 × 10-13. Calculate the molar solubility of this compound.
...............Mn(OH)2 ==> Mn^2+ + 2OH^-
I..............solid.................0.............0
C............solid..................x............2x
E............solid...................x.............2x
Write the expression for Ksp for Mn(OH)2, plug in the E line above, and solve for x =-(Mn^2+)= [Mn(OH)2] and 2x = (OH^-)
Post your work if you get stuck.
Ksp=[Products]/[Reactants]
Mn(OH)2 -----------> Mn^+ + 2OH^-
..............Mn(OH)2 ==> Mn^2+ + 2OH^-
I..............solid.................0.............0
C............solid..................x............2x
E............solid...................x.............2x
Ksp=[x][2x]^2
Ksp=4x^3
Ksp/4=x^3
Solve for x.
To calculate the molar solubility of manganese(II) hydroxide (Mn(OH)2) using its solubility product constant (Ksp), we can follow these steps:
Step 1: Write the balanced chemical equation for the dissociation of Mn(OH)2.
Mn(OH)2(s) ⇌ Mn2+(aq) + 2OH-(aq)
Step 2: Write the expression for the Ksp of Mn(OH)2.
Ksp = [Mn2+][OH-]^2
Step 3: Since Mn(OH)2 dissociates into one Mn2+ ion and two OH- ions, the concentration of Mn2+ in the saturated solution will be equal to the molar solubility, let's assume it to be "x", and the concentration of OH- ions will be 2x.
Step 4: Substitute the concentrations of Mn2+ and OH- ions into the Ksp expression.
Ksp = x * (2x)^2
Ksp = 4x^3
Step 5: Substitute the given value for Ksp into the equation and solve for x.
2.00 × 10^-13 = 4x^3
Step 6: Solve the equation for x.
x^3 = (2.00 × 10^-13) / 4
x^3 = 0.5 × 10^-13
Taking the cube root of both sides:
x = ∛ (0.5 × 10^-13)
x ≈ 0.79 × 10^-5
Therefore, the molar solubility of manganese(II) hydroxide is approximately 0.79 × 10^-5 M.
To calculate the molar solubility of manganese(II) hydroxide (Mn(OH)2), we need to use the solubility product constant, Ksp. The Ksp expression for Mn(OH)2 is as follows:
Ksp = [Mn2+][OH-]^2
Where [Mn2+] represents the concentration of Mn2+ ions in the solution and [OH-] represents the concentration of hydroxide ions in the solution.
Since Mn(OH)2 dissociates into one Mn2+ ion and two OH- ions, we can assume that the concentration of Mn2+ ions is equal to 2 times the concentration of OH- ions:
[Mn2+] = 2[OH-]
Substituting this expression into the Ksp equation, we get:
Ksp = (2[OH-])[OH-]^2
Ksp = 2[OH-]^3
Now, let's solve for [OH-] by rearranging the equation:
[OH-]^3 = Ksp/2
Taking the cube root of both sides:
[OH-] = (Ksp/2)^(1/3)
Now we can substitute the given value of Ksp (2.00 × 10^-13) into the equation:
[OH-] = (2.00 × 10^-13 / 2)^(1/3)
[OH-] = (1.00 × 10^-13)^(1/3)
[OH-] = 1.00 × 10^-13^(1/3)
[OH-] = 1.00 × 10^-4
Since [OH-] represents the concentration of hydroxide ions, the molar solubility of Mn(OH)2 is 1.00 × 10^-4 M.