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.
Calculate the molar solubility of manganese II hydroxide in ph 12.85 solution.
2 answers
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
But I don't know how to solve for x the answer is 4.2 times 10^-11