Topic 17 · page 12 of 20

Eight ions with sodium hydroxide

Every one of the eight ions gives a hydroxide precipitate with sodium hydroxide. Learn the colours, which precipitates dissolve in excess, and which darken in air.

15 min read17.2217.23

Read first: Three reactions of aqua ions

The observations 17.22

Add aqueous sodium hydroxide a few drops at a time, then in excess, to a solution of each ion. Use the same words every time: say whether a precipitate forms (with its colour) and whether it dissolves to form a solution (with its colour).

Ion (colour of solution) A few drops of NaOH(aq) Excess NaOH(aq)
Cr³⁺ (green) green precipitate dissolves: dark green solution
Mn²⁺ (very pale pink) off-white precipitate, darkens to brown in air no change
Fe²⁺ (pale green) green precipitate, turns brown in air no change
Fe³⁺ (yellow-brown) orange-brown precipitate no change
Co²⁺ (pink) blue precipitate, turns pink on standing no change
Ni²⁺ (green) green precipitate no change
Cu²⁺ (blue) blue precipitate no change
Zn²⁺ (colourless) white precipitate dissolves: colourless solution

Two ions dissolve in excess alkali, chromium(III) and zinc. These are the two amphoteric hydroxides. For the other six, the precipitate does not change.

The precipitates 17.2217.23

The hydroxide ion removes H⁺ from water ligands until the complex is neutral. For a 2+ ion, two H⁺ are removed; for a 3+ ion, three.

Ion Precipitate (formula) Colour
Cr³⁺ [Cr(OH)₃(H₂O)₃], or Cr(OH)₃ green
Mn²⁺ [Mn(OH)₂(H₂O)₄], or Mn(OH)₂ off-white
Fe²⁺ [Fe(OH)₂(H₂O)₄], or Fe(OH)₂ green
Fe³⁺ [Fe(OH)₃(H₂O)₃], or Fe(OH)₃ orange-brown
Co²⁺ [Co(OH)₂(H₂O)₄], or Co(OH)₂ blue, then pink
Ni²⁺ [Ni(OH)₂(H₂O)₄], or Ni(OH)₂ green
Cu²⁺ [Cu(OH)₂(H₂O)₄], or Cu(OH)₂ blue
Zn²⁺ [Zn(OH)₂(H₂O)₄], or Zn(OH)₂ white

The equation is the same pattern for each M²⁺ ion (Mn, Fe, Co, Ni, Cu and Zn), with M replaced by the metal:

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

For the 3+ ions, Cr³⁺ and Fe³⁺:

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

The shorter ionic equations are accepted too, for example:

FeX2+(aq)+2 OHX−(aq)→Fe(OH)X2(s) \ce{Fe^2+(aq) + 2OH-(aq) -> Fe(OH)2(s)}
FeX3+(aq)+3 OHX−(aq)→Fe(OH)X3(s) \ce{Fe^3+(aq) + 3OH-(aq) -> Fe(OH)3(s)}

Worked exampleIron(II) and iron(III) with sodium hydroxide

  1. Fe²⁺ (pale green solution) gives a green precipitate of iron(II) hydroxide. It slowly turns brown from the surface, because oxygen in the air oxidises iron(II) to iron(III).
  2. Fe³⁺ (yellow-brown solution) gives an orange-brown precipitate of iron(III) hydroxide at once.
  3. The oxidation in air is: 4 Fe(OH)X2(s)+OX2(g)+2 HX2O(l)→4 Fe(OH)X3(s)\ce{4Fe(OH)2(s) + O2(g) + 2H2O(l) -> 4Fe(OH)3(s)}

Manganese(II) hydroxide darkens in air for the same reason: oxygen oxidises Mn(+2) to Mn(+3) and Mn(+4) compounds, which are brown. Cobalt(II) hydroxide first comes out blue and changes to pink on standing.

The two that dissolve 17.2217.23

Chromium(III) and zinc hydroxides are amphoteric. In excess hydroxide they become negative hydroxo ions and dissolve.

[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}
[Zn(OH)X2(HX2O)X4]+2 OHX−→[Zn(OH)X4]X2−+4 HX2O \begin{aligned} &\ce{[Zn(OH)2(H2O)4] + 2OH-} \\ &\ce{-> [Zn(OH)4]^2- + 4H2O} \end{aligned}
Ion Excess NaOH gives Colour of solution
Cr³⁺ [Cr(OH)₆]³⁻ dark green
Zn²⁺ [Zn(OH)₄]²⁻ colourless

Check yourself

  1. Give the observations when sodium hydroxide is added slowly, then in excess, to Cu²⁺(aq).
  2. Write the ionic equation for the reaction in question 1, including state symbols.
  3. A solution gives a green precipitate with sodium hydroxide that turns brown on standing. Name the ion and explain the colour change.
  4. Which two of the eight ions give a precipitate that dissolves in excess sodium hydroxide? Write the equation for one of them.
  5. Describe how you would tell apart solutions of Fe²⁺ and Fe³⁺ using sodium hydroxide.
  6. Cobalt(II) chloride solution is mixed with sodium hydroxide. State what you see and what happens on standing.

Answers cover these until you have tried the questions

  1. A blue precipitate forms. It does not dissolve in excess sodium hydroxide (no change).
  2. CuX2+(aq)+2 OHX−(aq)→Cu(OH)X2(s)\ce{Cu^2+(aq) + 2OH-(aq) -> Cu(OH)2(s)} (or [Cu(HX2O)X6]X2++2 OHX−→[Cu(OH)X2(HX2O)X4]+2 HX2O\ce{[Cu(H2O)6]^2+ + 2OH- -> [Cu(OH)2(H2O)4] + 2H2O}).
  3. Fe²⁺. The green precipitate is Fe(OH)₂. Oxygen in the air oxidises iron(II) to iron(III), forming brown Fe(OH)₃.
  4. Cr³⁺ and Zn²⁺. For example [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}, giving a colourless solution.
  5. Add sodium hydroxide to each. Fe²⁺ gives a green precipitate (which turns brown in air). Fe³⁺ gives an orange-brown precipitate.
  6. A blue precipitate forms. On standing it turns pink.