Calculate the molar solubility of manganese II hydroxide in ph 12.85 solution.
The Ksp is 2.1 times 10 ^-13
But I don't know how to solve for x the answer is 4.2 times 10^-11
To calculate the molar solubility of manganese II hydroxide (Mn(OH)2) in a pH 12.85 solution, we need to consider the hydroxide ion (OH-) concentration.
Manganese II hydroxide dissociates in water as follows:
Mn(OH)2(s) ↔ Mn2+(aq) + 2OH-(aq)
In an alkaline solution like pH 12.85, the concentration of hydroxide ions (OH-) is related to the pH by the equation:
pOH = 14 - pH
Let's calculate the pOH first:
pOH = 14 - 12.85
pOH = 1.15
Now, we need to calculate the concentration of hydroxide ions (OH-) using the pOH value:
[OH-] = 10^(-pOH)
[OH-] = 10^(-1.15)
Using a scientific calculator, we find:
[OH-] ≈ 0.045 M
Since each mole of manganese II hydroxide (Mn(OH)2) dissociates into 2 moles of OH- ions, we can say that the concentration of Mn2+ ions will also be 2 times the concentration of OH- ions.
Therefore, the molar solubility of manganese II hydroxide in a pH 12.85 solution is approximately:
[ Mn(OH)2 ] = 2 × [ OH- ]
[ Mn(OH)2 ] = 2 × 0.045 M
[ Mn(OH)2 ] ≈ 0.09 M
So, the molar solubility of manganese II hydroxide in a pH 12.85 solution is approximately 0.09 M.
I would convert pH to pOH first, then convert that to OH^-
pH + pOH = pKw = 14
You know pH and pKw, solve for pOH, then
pOH = -log(OH^-)
Solve for OH^- and substitute in equations below. I obtained about 0.07M
...........Mn(OH)2 ==> Mn^2+ + 2OH^-
I..........solid........0.......0.07
C..........solid........x.......2x
E..........solid........x.......2x+0.07
Ksp = (Mn^2+)(OH^-)^2
Look up Ksp = (x)(2x+0.07)^2
Solve for x.
Be sure and run the numbers yourself, the above are estimates.
x = solubility Mn = solubility Mn(OH)2.