Ferric Hydroxide, Fe(OH)3, is an insoluble salt with a solubility of 10^-39.
1) What is the maximum concentration of ferric hydroxide that can dissolve?
2) Using Le Chatalier's Principle, explain why iron (III) solutions are more soluble in acidic solution.
5 answers
In this case, the OH- will predominate in the solution, adding acid will cause H+ ions to convert the OH- to water, LeChatalier predicts more will dissolve to undo that rduction in OH-
Thank you bobpursley..how can I determine the max concentration for Fe(OH)3 that will dissolve though?
Wouldn't it make sense that you need the pH of the solution to determine the solubility. You don't have a pH or the amount of the solute. And 10^-39 what
That's why I was confused, it just states that the Fe(OH)3 has a solubility product of 10^-39. Any ideas?
Fe(OH)3 <=> Fe^+3 + 3OH^-
Ksp = [Fe^+3][OH^-]^3
Ksp = (x)(3x)^3 = 27x^4
=> x = Solubility of Fe(OH)^3
---- = (Ksp/27)^1/4
---- = (1x10^-39/27)^1/4
---- = 7.8x10^11 Molar
[OH^-] = 3(7.8x10^-11)M = 2.4x10^-10M
=> pOH = -log[OH^-] = -log(2.4x10^-10) = 9.63
pH + pOH = 14 => pH = 14 - 9.63 = 4.37
Fe(OH)3 <=> Fe^+3 + 3OH^-
-------------------------
^
Adding H+ reacts with OH^- => HOH and leaves the product side of equilibrium needing OH^-. This causes more Fe(OH)3 to ionize to replace OH^- until all additional H^+ is converted to HOH => increasing solubility OF Fe(OH)3.
Ksp = [Fe^+3][OH^-]^3
Ksp = (x)(3x)^3 = 27x^4
=> x = Solubility of Fe(OH)^3
---- = (Ksp/27)^1/4
---- = (1x10^-39/27)^1/4
---- = 7.8x10^11 Molar
[OH^-] = 3(7.8x10^-11)M = 2.4x10^-10M
=> pOH = -log[OH^-] = -log(2.4x10^-10) = 9.63
pH + pOH = 14 => pH = 14 - 9.63 = 4.37
Fe(OH)3 <=> Fe^+3 + 3OH^-
-------------------------
^
Adding H+ reacts with OH^- => HOH and leaves the product side of equilibrium needing OH^-. This causes more Fe(OH)3 to ionize to replace OH^- until all additional H^+ is converted to HOH => increasing solubility OF Fe(OH)3.
2 answers