Asked by Theia'a
Calculate the solubility (S) of Al(OH)3 based on [OH-] concentration (mol/L)?
Suppose the concentration of [OH-] is obvious and can be adjusted by the experimenter. Also, calculate the minimum molar solubility?
KdAl(OH)4-/Al3+=1.2x10-34 Ksp Al(OH)3=5x10-33
Suppose the concentration of [OH-] is obvious and can be adjusted by the experimenter. Also, calculate the minimum molar solubility?
KdAl(OH)4-/Al3+=1.2x10-34 Ksp Al(OH)3=5x10-33
Answers
Answered by
DrBob222
It's always better to give numbers instead of generalities. Say you know (OH)^- = 0.01 M, calculate solubility, S, of Al(OH)3. Then
.............Al(OH)3 ==> Al^3+ + 3OH^-
I.............solid..............0................0
C............solid..............S................3S
E.............solid..............S................3S
Ksp = (Al^3+)(OH)^3 = 5E-33
5E-33 = (S)(3S)^3
But you know (OH) = 3S = 0.01 M so
5E-33 = (S)(0.01)^3 = and solve for = solubility in moles/L.
.............Al(OH)3 ==> Al^3+ + 3OH^-
I.............solid..............0................0
C............solid..............S................3S
E.............solid..............S................3S
Ksp = (Al^3+)(OH)^3 = 5E-33
5E-33 = (S)(3S)^3
But you know (OH) = 3S = 0.01 M so
5E-33 = (S)(0.01)^3 = and solve for = solubility in moles/L.
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.