Didn't I do this for you yesterday?
Convert pH to OH^-. I would do that by converting to pOH by using
pH + pOH = pKw = 14, then to OH^- by
pOH = -log (OH^-).
Let x = solubility Mn(OH)2.
........Mn(OH)2 => Mn^2+ + 2OH^-
...........x........x........2x
Ksp = (Mn^2+)(OH^-)^2
Substitute Ksp.
(Mn^+) = x
(OH^-) = x + (OH^-) from above.
Solve for x = Mn(OH)2 in moles/L. Convert to g/L by g = moles x molar mass. The others are done the same way. Post your work if you get stuck.
Calculate the solubility of Mn(OH)2 in grams per liter when buffered at each of the following.
a) pH 7.4
b) pH 9.2
c) pH 11.5
ALL IN GRAMS PER LITER.
PLEASE HELP ME AND EXPLAIN
2 answers
676t