Question

When a strong base is added to the pale blue solution of cuso4, a precipitate forms and the solution above the precipitate is colorless what is the net ionic equation that describes this irreversible reaction?

Answer 1

Hello!

When a strong base reacts with CuSO₄ the following reaction happens:

Cu⁺²(aq) + OH⁻(aq) → Cu(OH)₂(s)

This is a reaction commonly used in titration experiments. One can follow this reaction through conductive measurements since the precipitate will decrease the conductivity of the solution as it doesn't conduct electrons as well as the ions in solution.

Have a nice day!

Answer 2

Answer: The net ionic equation is written below.

Explanation:

Strong base taken in context is NaOH.

Net ionic equation of any reaction does not include any spectator ions.

Spectator ions are defined as the ions which does not get involved in a chemical equation. They are found on both the sides of the chemical reaction when it is present in ionic form.

The chemical equation for the reaction of copper sulfate and sodium hydroxide is given as:

$CuSO_4(aq.)+2NaOH(aq.)\rightarrow Cu(OH)_2(s)+2Na_2SO_4(aq.)$

Ionic form of the above equation follows:

$Cu^{2+}(aq.)+SO_4^{2-}(aq.)+2Na^+(aq.)+2OH^-(aq.)\rightarrow Cu(OH)_2(s)+2Na^+(aq.)+SO_4^{2-}(aq.)$

As, sodium and sulfate ions are present on both the sides of the reaction. Thus, it will not be present in the net ionic equation and are spectator ions.

The net ionic equation for the above reaction follows:

$Cu^{2+}(aq.)+2OH^-(aq.)\rightarrow Cu(OH)_2(s)$

A white colored precipitate of copper (II) hydroxide is formed.

Hence, the net ionic equation is written above.