Precipitation Reactions

Bases

NaOH and NH₃

When in limited amounts, they form the same hydroxide precipitates

NaOH as the base

OH⁻

[Cu(H₂O)₆]²⁺ + 2OH⁻

[Cu(H₂O)₆]²⁺₍ₐq₎ + 2OH⁻ ₍ₐq₎ → Cu(H₂O)₄(OH)₂ ₍ₛ₎ + 2H₂O ₍ₗ₎

Blue solution → Blue ppt

Cu²⁺ + 2OH⁻

Cu²⁺ + 2OH⁻ → Cu(OH)₂ ₍ₛ₎

Blue solution → Blue ppt

[Mn(H₂O)₆]²⁺ + 2OH⁻

[Mn(H₂O)₆]²⁺ + 2OH⁻ → Mn(H₂O)₄(OH)₂ ₍ₛ₎ + 2H₂O ₍ₗ₎

Very pale pink solution → Pale brown ppt

Mn²⁺ + 2OH⁻

Mn²⁺ + 2OH⁻ → Mn(OH)₂ ₍ₛ₎

Very pale pink solution → Pale brown ppt

[Fe(H₂O)₆]²⁺ + 2OH⁻

[Fe(H₂O)₆]²⁺ + 2OH⁻ → Fe(H₂O)₄(OH)₂ ₍ₛ₎ + 2H₂O ₍ₗ₎

Pale green solution → Pale green ppt

Fe²⁺ + 2OH⁻

Fe²⁺ + 2OH⁻ →Fe(OH)₂ ₍ₛ₎

[Fe(H₂O)₆]³⁺ + 3OH⁻

[Fe(H₂O)₆]³⁺ + 3OH⁻ → Fe(H₂O)₃(OH)₃ ₍ₛ₎ + 3H₂O ₍ₗ₎

Yellow/brown solution → Brown ppt

Fe³⁺ + 3OH⁻

Fe³⁺ + 3OH⁻ → Fe(OH)₃ ₍ₛ₎

Yellow/brown solution → Brown ppt

[Cr(H₂O)₆]³⁺ + 3OH⁻

[Cr(H₂O)₆]³⁺ + 3OH⁻ → Cr(H₂O)₃(OH)₃ ₍ₛ₎ + 3H₂O ₍ₗ₎

Violet solution → Green ppt

Cr³⁺ + 3OH⁻

Cr³⁺ + 3OH⁻ → Cr(OH)₃ ₍ₛ₎

Green solution → Green ppt

NH₃ as the base

When ammonia is added in limited amounts, the same hydroxide precipitates form. The ammonia acts as a base, removes a proton from the aqueous complex, and becomes the ammonium ion (NH₄⁺)

[Mn(H₂O)₆]²⁺ + 2NH₃

[Mn(H₂O)₆]²⁺ + 2NH₃ → Mn(H₂O)₄(OH)₂ ₍ₛ₎ + 2NH₄⁺

[Fe(H₂O)₆]³⁺ + 3NH₃

[Fe(H₂O)₆]³⁺ + 3NH₃ → Fe(H₂O)₃(OH)₃ ₍ₛ₎ + 3NH₄⁺

Reacting excess NaOH to the Cr hydroxide

Cr(H₂O)₃(OH)₃ ₍ₛ₎ + 3OH⁻ → [Cr(OH)₆]³⁻ ₍ₐq₎ + 3H₂O

Green ppt → Green solution

Reacting acid with the Cr hydroxide

Cr(H₂O)₃(OH)₃ ₍ₛ₎ + 3H⁺ → [Cr(OH)₆]³⁺ ₍ₐq₎

What is Cr(H₂O)₃(OH)₃ ₍ₛ₎ (the Cr hydroxide) classed as?

Amphoteric

This is because it can react with alkali to give a solution and react with an acid to form the aqueous salt

Reaction with excess NH₃

Excess NH₃ ligand exchange reactions occur with Cu and Cr, and their hydroxide ppt dissolve in excess ammonia.

Neutral ligands

NH₃ and H₂O are neutral ligands, meaning they don't have an overall charge.

They are also similar in size, so swapping one for the other in a complex doesn't drastically affect the shape or stability.

Ligand exchange occurs without change of coordination number for Cr

Ligand exchange is when one ligand, like H₂O, is replaced by another (like NH₃)

Coordination number refers to the number of ligands directly bonded to the metal ion

So, for Cr, when H₂O ligands are replaced by NH₃, the number of ligands stay the same. Just the type of ligand changes.

Cr³⁺ reacting with excess NH₃

Cr(OH)₃(H₂O)₃ ₍ₛ₎ + 6NH₃ ₍ₐq₎ → [Cr(NH₃)₆]³⁺ ₍ₐq₎ + 3H₂O ₍ₗ₎+ 3OH⁻

Purple solution

Does Cu²⁺ react with excess NH₃?

The substitution may be incomplete

Cu(H₂O)₄(OH)₂ ₍ₛ₎ + 4NH₃ ₍ₐq₎ → [Cu(H₂O)₂(NH₃)₄]²⁺ + 2H₂O + 2OH⁻

Deep blue solution