Because phosphoric acid is a weak acid, it is typically used only as an analyte. Phosphoric acid can be the weak acid in a weak acid-strong base titration. When the titration reaches the first equivalence point, the solution will contain the conjugate base H2PO4-. This will give the solution a pH greater than seven at that equivalence point .

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Phosphoric acid is added to lớn reduce the electrode potential for the Fe3+ → Fe2+ reaction by stabilising the ferric ion.


Because phosphoric acid is a weak acid, it is typically used only as an analyte. Phosphoric acid can be the weak acid in a weak acid-strong base titration. When the titration reaches the first equivalence point, the solution will contain the conjugate base H2PO4-. This will give the solution a pH greater than seven at that equivalence point .

Role -

Phosphoric acid is added khổng lồ reduce the electrode potential for the Fe3+ → Fe2+ reaction by stabilising the ferric ion.

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Because phosphoric acid is a weak acid, it is typically used only as an analyte. Phosphoric acid can be the weak acid in a weak acid-strong base titration. When the titration reaches the first equivalence point, the solution will contain the conjugate base H2PO4-. This will give the solution a pH greater than seven at that equivalence point .

Role -

Phosphoric acid is added khổng lồ reduce the electrode potential for the Fe3+ → Fe2+ reaction by stabilising the ferric ion.


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The equation for the reaction is

FeCl3 + H3PO4 → FePO4 + 3HCl

It is considered lớn be a double displacement reaction. Initially, the solution of FeCl3 would be light yellow. Upon mixing with phosphoric acid (H3PO4), the solution would become colorless, some/most of the iron phosphate (FePO4) would precipitate out of solution (solubility will be ~0.6 g/100 mL of water) & HCl gas may be released from solution.


The equation for the reaction is

FeCl3 + H3PO4 → FePO4 + 3HCl

It is considered to lớn be a double displacement reaction. Initially, the solution of FeCl3 would be light yellow. Upon mixing with phosphoric acid (H3PO4), the solution would become colorless, some/most of the iron phosphate (FePO4) would precipitate out of solution (solubility will be ~0.6 g/100 mL of water) và HCl gas may be released from solution.

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The equation for the reaction is

FeCl3 + H3PO4 → FePO4 + 3HCl

It is considered khổng lồ be a double displacement reaction. Initially, the solution of FeCl3 would be light yellow. Upon mixing with phosphoric acid (H3PO4), the solution would become colorless, some/most of the iron phosphate (FePO4) would precipitate out of solution (solubility will be ~0.6 g/100 mL of water) and HCl gas may be released from solution.


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Usually H2SO4 và H3PO4 are added the sulfuric is the main source of H3O+ the phosphoric acid is added lớn complex Fe+3. By lowering the concentration of ferric ion the change in redox potential is increased making the endpoint easier lớn see or measure electronically.


Usually H2SO4 và H3PO4 are added the sulfuric is the main source of H3O+ the phosphoric acid is added to lớn complex Fe+3. By lowering the concentration of ferric ion the change in redox potential is increased making the endpoint easier to see or measure electronically.

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Usually H2SO4 & H3PO4 are added the sulfuric is the main source of H3O+ the phosphoric acid is added khổng lồ complex Fe+3. By lowering the concentration of ferric ion the change in redox potential is increased making the endpoint easier to see or measure electronically.


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The oxidation state of Mn in KMnO4 is +7, which means that Mn does not have any d electrons left. Theoretically, Mn7+ complexes should not be colored because electronic transitions are not possible (since there are no d electrons left). However, KMnO4 has a deep purple color because of charge transfer from the ligand (O2–) to the metal center. This is called a ligand-to-metal charhmge transfer. Charge transfers are beyond Laporte và spin selection rules thus they are always allowed giving the complex very high molar extinction coefficients.


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The oxidation state of Mn in KMnO4 is +7, which means that Mn does not have any d electrons left. Theoretically, Mn7+ complexes should not be colored because electronic transitions are not possible (since there are no d electrons left). However, KMnO4 has a deep purple màu sắc because of charge transfer from the ligand (O2–) khổng lồ the metal center. This is called a ligand-to-metal charhmge transfer. Charge transfers are beyond Laporte và spin selection rules thus they are always allowed giving the complex very high molar extinction coefficients.

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The oxidation state of Mn in KMnO4 is +7, which means that Mn does not have any d electrons left. Theoretically, Mn7+ complexes should not be colored because electronic transitions are not possible (since there are no d electrons left). However, KMnO4 has a deep purple màu sắc because of charge transfer from the ligand (O2–) to lớn the metal center. This is called a ligand-to-metal charhmge transfer. Charge transfers are beyond Laporte & spin selection rules thus they are always allowed giving the complex very high molar extinction coefficients.


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H3PO4 is itself the chemical formula of Phosphoric Acid


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H3PO4 is itself the chemical formula of Phosphoric Acid

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H3PO4 is itself the chemical formula of Phosphoric Acid


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When FeCl3 dissolves in water, you get the aqueous Fe3+ ion và the aqueous Cl- ion. However, the aqueous Fe3+ ion is actually a complex ion of Fe3+ with 6 water molecules: Fe(H2O)6 (3+). This hexaaquairon(III) ion is actually a faint lilac color but may appear colorless. The reason that solutions containing Fe(H2O)6 (3+) are yellowish và even yellow-orange is because Fe(H2O)6 (3+) is acidic và can ionize according to lớn the following equilibria:

Fe(H2O)6 (3+) + H2O Fe(H2O)5(OH) (2+) + H3O+Fe(H2O)5(OH) (2+) + H2O Fe(H2O)4(OH)2 (1+) + H3O+Fe(H2O)4(OH)2 (1+) + H2O Fe(H2O)3(OH)3 (s) + H3O+

When water molecules complex with the Fe3+ ion, the H—O bond is weakened and is capable of ionizing just lượt thích any other acid. The interesting part is that Fe(H2O)5(OH) (2+) & Fe(H2O)4(OH)2 (1+) have yellow-orange colors in solution và Fe(H2O)3(OH)3 is a yellow-orange precipitate.

The Ka for Fe(H2O)6 (3+) is about 6.0 x 10^-3 and the Ka for the first ionization of H3PO4 is 7.1 x 10^-3.

This makes H3PO4 a stronger acid than Fe(H2O)6 (3+). So when you địa chỉ the H3PO4 lớn a solution of Fe(H2O)6 (3+), according lớn Le Châtelier’s principle, the ionization of Fe(H2O)6 (3+) will be “inhibited” because there is already H3O+ in the solution from the ionization of H3PO4

H3PO4 + H2O H2PO4 (1-) + H3O+ (1st ionization)

This shifts the equilibrium in favor of the Fe(H2O)6 (3+) ion which is faint lilac or nearly colorless.

I have had personal experience with this process. I initially made the Fe3+ solution & it was yellow-orange because of the reasons explained above. When I added a strong acid to the solution, it became virtually colorless, again because of the reasons above.


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When FeCl3 dissolves in water, you get the aqueous Fe3+ ion and the aqueous Cl- ion. However, the aqueous Fe3+ ion is actually a complex ion of Fe3+ with 6 water molecules: Fe(H2O)6 (3+). This hexaaquairon(III) ion is actually a faint lilac color but may appear colorless. The reason that solutions containing Fe(H2O)6 (3+) are yellowish và even yellow-orange is because Fe(H2O)6 (3+) is acidic & can ionize according lớn the following equilibria:

Fe(H2O)6 (3+) + H2O Fe(H2O)5(OH) (2+) + H3O+Fe(H2O)5(OH) (2+) + H2O Fe(H2O)4(OH)2 (1+) + H3O+Fe(H2O)4(OH)2 (1+) + H2O Fe(H2O)3(OH)3 (s) + H3O+

When water molecules complex with the Fe3+ ion, the H—O bond is weakened and is capable of ionizing just lượt thích any other acid. The interesting part is that Fe(H2O)5(OH) (2+) & Fe(H2O)4(OH)2 (1+) have yellow-orange colors in solution & Fe(H2O)3(OH)3 is a yellow-orange precipitate.

The Ka for Fe(H2O)6 (3+) is about 6.0 x 10^-3 and the Ka for the first ionization of H3PO4 is 7.1 x 10^-3.

This makes H3PO4 a stronger acid than Fe(H2O)6 (3+). So when you showroom the H3PO4 lớn a solution of Fe(H2O)6 (3+), according to lớn Le Châtelier’s principle, the ionization of Fe(H2O)6 (3+) will be “inhibited” because there is already H3O+ in the solution from the ionization of H3PO4

H3PO4 + H2O H2PO4 (1-) + H3O+ (1st ionization)

This shifts the equilibrium in favor of the Fe(H2O)6 (3+) ion which is faint lilac or nearly colorless.

I have had personal experience with this process. I initially made the Fe3+ solution and it was yellow-orange because of the reasons explained above. When I added a strong acid khổng lồ the solution, it became virtually colorless, again because of the reasons above.

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When FeCl3 dissolves in water, you get the aqueous Fe3+ ion and the aqueous Cl- ion. However, the aqueous Fe3+ ion is actually a complex ion of Fe3+ with 6 water molecules: Fe(H2O)6 (3+). This hexaaquairon(III) ion is actually a faint lilac màu sắc but may appear colorless. The reason that solutions containing Fe(H2O)6 (3+) are yellowish và even yellow-orange is because Fe(H2O)6 (3+) is acidic và can ionize according khổng lồ the following equilibria:

Fe(H2O)6 (3+) + H2O Fe(H2O)5(OH) (2+) + H3O+Fe(H2O)5(OH) (2+) + H2O Fe(H2O)4(OH)2 (1+) + H3O+Fe(H2O)4(OH)2 (1+) + H2O Fe(H2O)3(OH)3 (s) + H3O+

When water molecules complex with the Fe3+ ion, the H—O bond is weakened and is capable of ionizing just lượt thích any other acid. The interesting part is that Fe(H2O)5(OH) (2+) & Fe(H2O)4(OH)2 (1+) have yellow-orange colors in solution & Fe(H2O)3(OH)3 is a yellow-orange precipitate.

The Ka for Fe(H2O)6 (3+) is about 6.0 x 10^-3 và the Ka for the first ionization of H3PO4 is 7.1 x 10^-3.

This makes H3PO4 a stronger acid than Fe(H2O)6 (3+). So when you showroom the H3PO4 khổng lồ a solution of Fe(H2O)6 (3+), according to lớn Le Châtelier’s principle, the ionization of Fe(H2O)6 (3+) will be “inhibited” because there is already H3O+ in the solution from the ionization of H3PO4

H3PO4 + H2O H2PO4 (1-) + H3O+ (1st ionization)

This shifts the equilibrium in favor of the Fe(H2O)6 (3+) ion which is faint lilac or nearly colorless.

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I have had personal experience with this process. I initially made the Fe3+ solution & it was yellow-orange because of the reasons explained above. When I added a strong acid to the solution, it became virtually colorless, again because of the reasons above.