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Adrian Hum

Why does adding excess NaOH prove amphoteric nature?

Dan Chadwick  Follow

According to IUPAC Goldbook:

A chemical species that behaves both as an acid and as a base is called amphoteric. This property depends upon the medium in which the species is investigated: $\ce{H2SO4}$ is an acid when studied in water, but becomes amphoteric in superacids.

Most common example for amphoteric compound is water, which can act either as an acid or base depending on the solute present:

$$\ce{HCl + H2O -> H3O+ + Cl-} \tag{acting as an base}$$$$\ce{NH3 + H2O -> NH4+ + OH-} \tag{acting as an acid}$$

OP describes the problem as: "This does not make sense to me because the definition of an amphoteric species is one that reacts with both acids and bases; yet, in this test, the species has only reacted with hydroxide ions; no acid - so there is no proof of amphoteric nature".

The problem is the question does not clarify which compound is the amphoteric. The starting compound, $\ce{[M(H2O)6]^3+}$, is not amphoteric for sure. I choose the most suitable amphoteric compound in this senario as the neutral precipitate, $\ce{M(H2O)3(OH)3}$ because it can shows the rest of amphoteric compounds in the series. For instance, according to the IUPAC description, this precipitate is an amphoteric compound because it will react with both acids and bases (as Mithoron pointed out in the comment) as follows:

$$\ce{M(H2O)3(OH)3 ->[H3O+] [M(H2O)4(OH)2]+ ->[H3O+] [M(H2O)5(OH)]^2+ ->[H3O+] [M(H2O)6]^3+}$$$$\ce{M(H2O)3(OH)3 ->[OH-] [M(H2O)2(OH)4]- ->[OH-] [M(H2O)(OH)5]^2- ->[OH-] [M(OH)6]^3-}$$

Note: Strictly speaking, according to the IUPAC description, all of these metal hydroxides except for $\ce{[M(H2O)6]^3+}$ and $\ce{[M(OH)6]^3-}$ are amphoteric compounds since each can go either way with an acid or a base.

For more examples of amphoteric compounds including lead(II) hydroxide, read here and here. For example, lead(II) hydroxide is an amphoteric compound, which shows following reactions with both acids and bases:$$\ce{Pb(OH)2 (s) + 2HCl (aq) -> PbCl2 (aq) + 2H2O}$$$$\ce{Pb(OH)2 (s) + 2NaOH (aq) -> Na2Pb(OH)4 (aq)}$$

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Gilbert Mendez  Follow
Here is a reference question from my exam board which explains why I have concerns over this: More
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Frank Gasparro  Follow
paste.pics/D2MURMore
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Harold Green  Follow
Apparently, the second reaction proves that lead hydroxide is amphoteric...?More
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Jean Hildesheim  Follow
Thank you for the response. I understand why the species is amphoteric. I just dont understand why if a species reacts with an excess of NaOH and dissolves, it is amphoteric. Could anyone help to clarify this?More
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Edward Blacksmith  Follow
@Matthew: I edited according to your concern. Hope you understand it better now.More
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