How to receive copper hydroxide

How to receive copper hydroxide

Hydroxide of copper (II) – the substance of bright blue color not soluble in water. Has crystal or amorphous structure. This weak basis is applied when processing agricultural plants, in textile and chemical industry. Cu(OH) ₂ receive action on salt of copper of the strong bases (alkalis).


1. Receiving from sulfate of CuSO copper (II) ₄ - the white crystal powder soluble in water. In interaction with damp air or with water sulfate of copper forms crystalline hydrate (pentahydrate of sulfate of copper (II)) better known as copper vitriol CuSO ₄ • 5H₂O. Therefore when receiving hydroxide not pure sulfate of copper, but its crystalline hydrate actually participates. Add to this solution alkali (for example NaOH) and watch effect of reaction: CuSO ₄ + 5H₂O + 2NaOH = Na₂SO ₄ + Cu(OH) ₂↓ +5 NO.Pri addition of proportional amount of reagents solution becomes colourless, and the received hydroxide of copper drops out in the form of a blue deposit. Further this solution can participate in high-quality reaction to proteins.

2. Receiving from nitrate of Cu copper (II) (NO ₃) ₂ - colourless crystal substance. Reacts exchange with the strong bases. It is possible to carry out reaction of receiving hydroxide from salt, having added colourless crystals of nitrate of copper (II) to NaOH solution. As a result you receive colourless solution of nitrate of sodium and a deposit of hydroxide of copper (II) of blue color: Cu (NO ₃) ₂ + 2NaOH = Cu(OH) ₂↓ + 2NaNO ₃.

3. Receiving from chloride of CuCl copper (II) ₂ - under normal conditions represents powder of yellow or yellow-brown color. Let's well dissolve in water. Fill chloride of copper in a test tube and add equivalent amount of alkali. Yellow crystals disappear and the blue deposit is formed. If necessary to emit substance from solution, filter a deposit and dry up. Do not use high-thermal types of drying since at a temperature close to 100 °C, Cu(OH) ₂ decays on oxide of copper (II) and water: CuCl ₂ + 2NaOH = 2NaCl + Cu(OH) ₂↓.

4. Receiving from acetate of copper (II) (CH₃COO) ₂Cu - the substance of dark green color soluble in water. At dissolution, solution gets blue coloring. Add the calculated amount of alkali to solution of acetate of copper (II) and observe formation of hydroxide (amorphous blue deposit): (CH₃COO) of ₂Cu + 2NaOH = Cu(OH) ₂↓ + CH₃COONa. Since solutions of salts of copper (II) are painted in blue or blue color, reactions of decolouration of solutions with the subsequent loss of a color deposit look very effectively.

Author: «MirrorInfo» Dream Team