Asked by Henry
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.
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.
Answers
Answered by
bobpursley
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-
Answered by
Henry
Thank you bobpursley..how can I determine the max concentration for Fe(OH)3 that will dissolve though?
Answered by
DrBob222
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
Answered by
Scott
That's why I was confused, it just states that the Fe(OH)3 has a solubility product of 10^-39. Any ideas?
Answered by
DrRebel
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.
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