# K Calculation Questions (1 Viewer)

##### Member
Theres been this part of calculations that I never understood. Say this question, a solution of iorn (II) hydroxide has a pH of 9.42. Calculate the solubility, in g/L of iorn (II) hydroxide in this solution.

You find [OH-] = 10^-4.58, since Fe(OH)2 --> Fe2+ + 2OH-, why can't we just half the concentration of OH- for the concentration/solubility of Fe(OH)2, since their ratio is 1:2. Ik we need to use the Ksp and solve for Fe+, then [Fe+] = [Fe(OH)2], but I don't get why the other way doesn't work if their ratios are the same.

#### Luukas.2

##### Well-Known Member
Theres been this part of calculations that I never understood. Say this question, a solution of iorn (II) hydroxide has a pH of 9.42. Calculate the solubility, in g/L of iorn (II) hydroxide in this solution.

You find [OH-] = 10^-4.58, since Fe(OH)2 --> Fe2+ + 2OH-, why can't we just half the concentration of OH- for the concentration/solubility of Fe(OH)2, since their ratio is 1:2. Ik we need to use the Ksp and solve for Fe+, then [Fe+] = [Fe(OH)2], but I don't get why the other way doesn't work if their ratios are the same.
If the solution were prepared by dissolving iron(II) hydroxide into neutral water and the pH was 9.42 at equilibrium, your method would work.

However, if the solution were prepared (say) by dissolving iron(II) chloride and then adding base to adjust the pH, then you will have no idea what the ratio of the species concentrations is except trough a Ksp calculation.

#### wizzkids

##### Active Member
Theres been this part of calculations that I never understood. Say this question, a solution of iorn (II) hydroxide has a pH of 9.42. Calculate the solubility, in g/L of iorn (II) hydroxide in this solution.
The short answer is: practically Zero solubility.
Iron(II) hydroxide is almost insoluble in water, it's solubility constant is 4.1 x 10-15. It only starts to become sparingly soluble once the pH starts to get up around 12.
Here is how you do these calculations. You can't just halve the concentration of [OH-] because their ratios are not the same. The values of [Fe2+] and [OH-] can change independently of each other if you are adding ions to the solution (the common ion effect).
We know that at 25oC the water ionization constant can be expressed as pH + pOH = 14. If pH is 9.42 then pOH is 4.58.
Write the expression for the solubility constant:
[Fe2+] [OH-]2 = 4.1 x 10-15
You substitute the value of pOH, 5.8 x 10-4 into the expression for the solubility constant.
Hence calculate the value of [Fe2+]
I think the answer comes to 3 x 10-8 mol/L which is stuff-all.

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#### Luukas.2

##### Well-Known Member
The short answer is: practically Zero solubility.
Iron(II) hydroxide is almost insoluble in water, it's solubility constant is 4.1 x 10-15. It only starts to become sparingly soluble once the pH starts to get up around 12.
This is backwards. Iron(II) hydroxide will not dissolve to any significant extent unless the hydroxide ion concentration is very low and so the solution needs to be acidic, not basic.

Since its Ksp = 4.87 x 10-17 at 298 K and as [Fe2+] = Ksp / [OH-]2 (in a saturated solution):
• at 298 K and pH = 12 ===> [OH-] = 10-2 mol L-1, [Fe2+] = 4.87 x 10-17 / (10-2)2 = 4.87 x 10-13 mol L-1
• at 298 K and pH = 9 ===> [OH-] = 10-5 mol L-1, [Fe2+] = 4.87 x 10-17 / (10-5)2 = 4.87 x 10-7 mol L-1
• at 298 K and pH = 6 ===> [OH-] = 10-8 mol L-1, [Fe2+] = 4.87 x 10-17 / (10-8)2 = 4.87 x 10-1 mol L-1

#### wizzkids

##### Active Member
This discussion lies totally outside high school syllabus. There are two ways that Iron(II) hydroxide can dissolve.
Iron(II) hydroxide has a region of solid stability between pH 8 and pH12.
The stability of the various hydroxides of iron can be conveniently shown on the Pourbaix Diagram.

I have circled the regions where Iron(II) hydroxide dissolves.
One occurs at very high pH iron(II) hydroxide becomes soluble as HFeO2- ions, i.e. hydrated iron oxide ions.
The other occurs at low pH iron(II) hydroxide becomes soluble as Fe2+ ions.

#### Luukas.2

##### Well-Known Member
@wizzkids is correct that some metal hydroxide increase their solubility at high pH - aluminium hydroxide, for example - which the HSC covers in the form of an example of amphotericism. It pops up in modules 6 and 8, though is generally covered very briefly. Aluminium hydroxide is amphoteric because it reacts with acid:

Al(OH)3 (s) + 3 H+ (aq) -----> Al3+ (aq) + 3 H2O (l)​

and it reacts with base:

Al(OH)3 (s) + OH- (aq) -----> [Al(OH)4]- (aq)​

Amphotericism is related to, but separate from, amphiprotism... and amphiprotic substance reacts with both acids and bases via BL proton transfer reactions, and so all amphiprotic substances are amphoteric. Substance like aluminium hydroxide and iron(II) hydroxide are amphoteric without being amphiprotic as the reactions involved are not BL proton transfer processes.

The formation of HFeO2- is an example of this same phenomenon, though a more complicated one than in the aluminium case, as the tetrahydroxyferrate(II) anion is undergoing subsequent rearrangement:

Fe(OH)2 (s) + 2 OH- (aq) <----- -----> [Fe(OH)4]2- (aq)

[Fe(OH)4]2- (aq) <----- -----> [FeO(OH)2]2- (aq) + H2O (l)

[FeO(OH)2]2- (aq) <----- -----> [FeO2]2- (aq) + H2O (l)

[FeO2]2- (aq) + H3O+ (aq) <----- -----> [HFeO2]- (aq) + H2O (l)​

Thus, the overall process is:

Fe(OH)2 (s) + OH- (aq) <----- -----> [HFeO2]- (aq) + H2O (l)​

However, this is unrelated to the calculation based on the Ksp of iron(II) hydroxide in a (relatively) low pH solution that is contemplated by this question. To take into account this high pH chemistry would require information on the equilibrium constant for the overall process above, at least.