Topic 17 · page 13 of 20

Eight ions with ammonia

A few drops of ammonia give the same precipitates as sodium hydroxide, because ammonia is a base. In excess, ammonia is a ligand and five of the precipitates dissolve.

15 min read17.2217.23

Read first: Eight ions with sodium hydroxide

The observations 17.22

Ammonia solution does two jobs, one after the other.

Ion (colour of solution) A few drops of NH₃(aq) Excess NH₃(aq)
Cr³⁺ (green) green precipitate dissolves: purple solution
Mn²⁺ (very pale pink) off-white precipitate no change
Fe²⁺ (pale green) green precipitate no change
Fe³⁺ (yellow-brown) orange-brown precipitate no change
Co²⁺ (pink) blue precipitate dissolves: straw-coloured solution, darkens in air
Ni²⁺ (green) green precipitate dissolves: blue solution
Cu²⁺ (blue) blue precipitate dissolves: deep blue solution
Zn²⁺ (colourless) white precipitate dissolves: colourless solution

Ammonia as a base 17.2217.23

Each M²⁺ aqua ion loses two H⁺ and each M³⁺ ion loses three. The NH₃ takes the H⁺ and becomes NH₄⁺:

[M(HX2O)X6]X2++2 NHX3→[M(OH)X2(HX2O)X4]+2 NHX4X+ \begin{aligned} &\ce{[M(H2O)6]^2+ + 2NH3} \\ &\ce{-> [M(OH)2(H2O)4] + 2NH4+} \end{aligned}
[M(HX2O)X6]X3++3 NHX3→[M(OH)X3(HX2O)X3]+3 NHX4X+ \begin{aligned} &\ce{[M(H2O)6]^3+ + 3NH3} \\ &\ce{-> [M(OH)3(H2O)3] + 3NH4+} \end{aligned}

M is Mn, Fe, Co, Ni, Cu or Zn for the first equation and Cr or Fe for the second. The precipitates are the same as on the previous page. Ammonia solution is alkaline because of a small amount of NH₃ + H₂O ⇌ NH₄⁺ + OH⁻, so the OH⁻ route is also correct and the short ionic equation CuX2+(aq)+2 OHX−(aq)→Cu(OH)X2(s)\ce{Cu^2+(aq) + 2OH-(aq) -> Cu(OH)2(s)} is accepted.

Ammonia as a ligand 17.2217.23

In excess, NH₃ replaces water and hydroxide ligands. The precipitate dissolves because a charged complex ion forms, and charged ions are soluble.

Ion Complex formed Colour Coordination number
Cr³⁺ [Cr(NH₃)₆]³⁺ purple 6
Co²⁺ [Co(NH₃)₆]²⁺ straw 6
Ni²⁺ [Ni(NH₃)₆]²⁺ blue 6
Cu²⁺ [Cu(NH₃)₄(H₂O)₂]²⁺ deep blue 6
Zn²⁺ [Zn(NH₃)₄]²⁺ colourless 4

The equations, written from the precipitate:

[Cr(OH)X3(HX2O)X3]+6 NHX3→[Cr(NHX3)X6]X3++3 OHX−+3 HX2O \begin{aligned} &\ce{[Cr(OH)3(H2O)3] + 6NH3} \\ &\ce{-> [Cr(NH3)6]^3+ + 3OH- + 3H2O} \end{aligned}
[Co(OH)X2(HX2O)X4]+6 NHX3→[Co(NHX3)X6]X2++2 OHX−+4 HX2O \begin{aligned} &\ce{[Co(OH)2(H2O)4] + 6NH3} \\ &\ce{-> [Co(NH3)6]^2+ + 2OH- + 4H2O} \end{aligned}
[Ni(OH)X2(HX2O)X4]+6 NHX3→[Ni(NHX3)X6]X2++2 OHX−+4 HX2O \begin{aligned} &\ce{[Ni(OH)2(H2O)4] + 6NH3} \\ &\ce{-> [Ni(NH3)6]^2+ + 2OH- + 4H2O} \end{aligned}
[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}
[Zn(OH)X2(HX2O)X4]+4 NHX3→[Zn(NHX3)X4]X2++2 OHX−+4 HX2O \begin{aligned} &\ce{[Zn(OH)2(H2O)4] + 4NH3} \\ &\ce{-> [Zn(NH3)4]^2+ + 2OH- + 4H2O} \end{aligned}

Equations written straight from the aqua ion are also accepted, for example for nickel:

[Ni(HX2O)X6]X2++6 NHX3→[Ni(NHX3)X6]X2++6 HX2O \ce{[Ni(H2O)6]^2+ + 6NH3 -> [Ni(NH3)6]^2+ + 6H2O}

Copper keeps two water ligands: [Cu(HX2O)X6]X2++4 NHX3→[Cu(NHX3)X4(HX2O)X2]X2++4 HX2O\ce{[Cu(H2O)6]^2+ + 4NH3 -> [Cu(NH3)4(H2O)2]^2+ + 4H2O}.

The straw-coloured cobalt complex turns darker brown in air, because cobalt(II) is oxidised to cobalt(III).

Sodium hydroxide and ammonia together

Ion Dissolves in excess NaOH? Dissolves in excess NH₃?
Cr³⁺ yes (amphoteric) yes
Mn²⁺ no no
Fe²⁺ no no
Fe³⁺ no no
Co²⁺ no yes
Ni²⁺ no yes
Cu²⁺ no yes
Zn²⁺ yes (amphoteric) yes

Chromium and zinc dissolve in both. Cobalt, nickel and copper dissolve only in ammonia. Manganese and the two irons dissolve in neither.

Check yourself

  1. Give the observations when ammonia solution is added slowly, then in excess, to Ni²⁺(aq).
  2. Write the equation for the reaction of [Ni(OH)₂(H₂O)₄] with excess ammonia.
  3. Give the observations for Fe³⁺(aq) with a few drops of ammonia, and then in excess. Write the ionic equation for the first step.
  4. State which of the eight ions dissolve in excess NaOH, and which dissolve in excess NH₃.
  5. A solution gives a blue precipitate with a few drops of ammonia, which dissolves in excess to a deep blue solution. Name the ion, give the formula of the complex in the final solution and write the equation for its formation from the precipitate.
  6. Explain why the zinc hydroxide precipitate dissolves in excess ammonia solution.

Answers cover these until you have tried the questions

  1. A green precipitate forms, which dissolves in excess ammonia to form a blue solution.
  2. [Ni(OH)X2(HX2O)X4]+6 NHX3→[Ni(NHX3)X6]X2++2 OHX−+4 HX2O\ce{[Ni(OH)2(H2O)4] + 6NH3 -> [Ni(NH3)6]^2+ + 2OH- + 4H2O}
  3. An orange-brown precipitate forms, which does not dissolve in excess (no change). [Fe(HX2O)X6]X3++3 NHX3→[Fe(OH)X3(HX2O)X3]+3 NHX4X+\ce{[Fe(H2O)6]^3+ + 3NH3 -> [Fe(OH)3(H2O)3] + 3NH4+}
  4. Excess NaOH: Cr³⁺ and Zn²⁺. Excess NH₃: Cr³⁺, Co²⁺, Ni²⁺, Cu²⁺ and Zn²⁺.
  5. Cu²⁺. The complex is [Cu(NH₃)₄(H₂O)₂]²⁺. [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. In excess ammonia, NH₃ acts as a ligand and replaces the water and hydroxide ligands, forming the soluble complex ion [Zn(NH₃)₄]²⁺: [Zn(OH)X2(HX2O)X4]+4 NHX3→[Zn(NHX3)X4]X2++2 OHX−+4 HX2O\ce{[Zn(OH)2(H2O)4] + 4NH3 -> [Zn(NH3)4]^2+ + 2OH- + 4H2O} (a colourless solution).