Topic 17 · page 11 of 20

Three reactions of aqua ions

Hydroxide ions and ammonia do three different things to a metal aqua ion. They remove H⁺ from water ligands (deprotonation), some hydroxides dissolve in excess alkali (amphoteric), and ammonia can replace ligands.

13 min read17.23

Read first: Ligands and complex ions

Deprotonation 17.23

In water a metal ion exists as a hexaaqua complex, [M(H₂O)₆]ⁿ⁺. The metal ion pulls electrons towards itself, so the O–H bonds in the water ligands are weakened. A base can remove an H⁺ ion from a ligand.

Each deprotonation lowers the charge of the complex by one. Remove as many H⁺ as the charge on the metal ion and the complex is neutral. A neutral complex is not soluble in water, so it comes out as a precipitate.

[Cu(HX2O)X6]X2++2 OHX−→[Cu(OH)X2(HX2O)X4]+2 HX2O \begin{aligned} &\ce{[Cu(H2O)6]^2+ + 2OH-} \\ &\ce{-> [Cu(OH)2(H2O)4] + 2H2O} \end{aligned}
[Fe(HX2O)X6]X3++3 OHX−→[Fe(OH)X3(HX2O)X3]+3 HX2O \begin{aligned} &\ce{[Fe(H2O)6]^3+ + 3OH-} \\ &\ce{-> [Fe(OH)3(H2O)3] + 3H2O} \end{aligned}

The number of OH⁻ in the precipitate equals the charge on the metal ion. You may also write the short form, CuX2+(aq)+2 OHX−(aq)→Cu(OH)X2(s)\ce{Cu^2+(aq) + 2OH-(aq) -> Cu(OH)2(s)}, if the question asks only for the precipitate.

A few drops of ammonia do the same job. Ammonia is a base: it takes the H⁺ and becomes NH₄⁺.

[Fe(HX2O)X6]X3++3 NHX3→[Fe(OH)X3(HX2O)X3]+3 NHX4X+ \begin{aligned} &\ce{[Fe(H2O)6]^3+ + 3NH3} \\ &\ce{-> [Fe(OH)3(H2O)3] + 3NH4+} \end{aligned}

Amphoteric behaviour 17.23

Some metal hydroxides react with excess hydroxide as well as with acid. Chromium(III) hydroxide and zinc hydroxide do.

With excess hydroxide, more H⁺ is removed from the water ligands and the complex becomes a negative ion, which dissolves. Here the hydroxide is acting as an acid, giving up H⁺:

[Cr(OH)X3(HX2O)X3]+3 OHX−→[Cr(OH)X6]X3−+3 HX2O \begin{aligned} &\ce{[Cr(OH)3(H2O)3] + 3OH-} \\ &\ce{-> [Cr(OH)6]^3- + 3H2O} \end{aligned}

With acid, the hydroxide ligands pick up H⁺ and become water again, so the precipitate dissolves and the aqua ion returns:

[Cr(OH)X3(HX2O)X3]+3 HX+→[Cr(HX2O)X6]X3+ \ce{[Cr(OH)3(H2O)3] + 3H+ -> [Cr(H2O)6]^3+}

Dissolving in both acid and alkali is what makes it amphoteric.

Ligand exchange 17.23

In excess ammonia, NH₃ acts as a ligand rather than as a base. Some precipitates dissolve because ammonia replaces the water and hydroxide ligands.

[Cu(OH)X2(HX2O)X4]+4 NHX3→[Cu(NHX3)X4(HX2O)X2]X2++2 OHX−+2 HX2O \begin{aligned} &\ce{[Cu(OH)2(H2O)4] + 4NH3} \\ &\ce{-> [Cu(NH3)4(H2O)2]^2+ + 2OH- + 2H2O} \end{aligned}
A flow chart. A hexaaqua ion on the left is deprotonated by NaOH or a few drops of NH3 to give a neutral hydroxide precipitate in the middle. From the precipitate, one arrow leads to a dissolved hydroxo complex with excess NaOH for amphoteric behaviour (Cr and Zn), and another to a dissolved ammine complex with excess NH3 for ligand exchange (Cr, Co, Ni, Cu, Zn).
The route for every precipitate. Excess NaOH shows amphoteric behaviour; excess NH3 shows ligand exchange.

Worked exampleChromium(III) with NaOH, then acid

  1. A few drops of NaOH: [Cr(HX2O)X6]X3++3 OHX−→[Cr(OH)X3(HX2O)X3]+3 HX2O\ce{[Cr(H2O)6]^3+ + 3OH- -> [Cr(OH)3(H2O)3] + 3H2O}. A green precipitate forms.
  2. Excess NaOH: [Cr(OH)X3(HX2O)X3]+3 OHX−→[Cr(OH)X6]X3−+3 HX2O\ce{[Cr(OH)3(H2O)3] + 3OH- -> [Cr(OH)6]^3- + 3H2O}. The precipitate dissolves to a dark green solution.
  3. Add acid to the precipitate: [Cr(OH)X3(HX2O)X3]+3 HX+→[Cr(HX2O)X6]X3+\ce{[Cr(OH)3(H2O)3] + 3H+ -> [Cr(H2O)6]^3+}. It dissolves again.

Check yourself

  1. Define deprotonation. Write an equation in which [Fe(H₂O)₆]²⁺ is deprotonated by hydroxide ions.
  2. Explain why the product of question 1 is a precipitate.
  3. What is meant by amphoteric? Write an equation to show that zinc hydroxide, [Zn(OH)₂(H₂O)₄], reacts with excess hydroxide.
  4. Write an equation to show that [Cr(OH)₃(H₂O)₃] reacts with acid.
  5. What is ligand exchange? Write an equation for the reaction of [Cu(OH)₂(H₂O)₄] with excess ammonia.
  6. A student writes that copper(II) hydroxide is amphoteric because it dissolves in excess ammonia. Explain the error.

Answers cover these until you have tried the questions

  1. Deprotonation is the removal of H⁺ from a species. [Fe(HX2O)X6]X2++2 OHX−→[Fe(OH)X2(HX2O)X4]+2 HX2O\ce{[Fe(H2O)6]^2+ + 2OH- -> [Fe(OH)2(H2O)4] + 2H2O}
  2. Two H⁺ ions have been removed, so the 2+ charge is cancelled by two OH⁻ ligands and the complex is neutral. A neutral complex is insoluble in water.
  3. An amphoteric species reacts with both acids and bases. [Zn(OH)X2(HX2O)X4]+2 OHX−→[Zn(OH)X4]X2−+4 HX2O\ce{[Zn(OH)2(H2O)4] + 2OH- -> [Zn(OH)4]^2- + 4H2O}
  4. [Cr(OH)X3(HX2O)X3]+3 HX+→[Cr(HX2O)X6]X3+\ce{[Cr(OH)3(H2O)3] + 3H+ -> [Cr(H2O)6]^3+}
  5. Ligand exchange is the replacement of one ligand by another. [Cu(OH)X2(HX2O)X4]+4 NHX3→[Cu(NHX3)X4(HX2O)X2]X2++2 OHX−+2 HX2O\ce{[Cu(OH)2(H2O)4] + 4NH3 -> [Cu(NH3)4(H2O)2]^2+ + 2OH- + 2H2O}
  6. It dissolves in excess ammonia because the NH₃ replaces ligands (ligand exchange). Amphoteric means a hydroxide dissolves in both acid and excess alkali (OH⁻), and copper(II) hydroxide does not dissolve in excess NaOH.