Question

KOH(aq) + CuSO4(aq) write the general and ionic equations for the reactions

Answer 1

The general equation for the reaction is:

`2KOH\left(aq\right)+CuSO_4\left(aq\right)\rightarrow K_2SO_4(aq)+Cu(OH)_2(s)`

The ionic equation for the reaction is:

`2K^+(aq)+2OH^-(aq)+Cu^(2+)(aq)+SO_4^(2-)\rightarrow2K^+(aq)+SO_4^(2-)+Cu(OH)_2(s)`

Step-by-step explanation

Given unbalanced equation: KOH(aq) + CuSO4(aq)

The general equation for the reaction is:

`2KOH\left(aq\right)+CuSO_4\left(aq\right)\rightarrow K_2SO_4(aq)+Cu(OH)_2(s)`

The reaction type is a double displacement reaction.

The ionic equation for the reaction is:

`2K^+(aq)+2OH^-(aq)+Cu^(2+)(aq)+SO_4^(2-)\rightarrow2K^+(aq)+SO_4^(2-)+Cu(OH)_2(s)`
`2K^+(aq)+2OH^-(aq)+Cu^(2+)(aq)+SO_4^(2-)\rightarrow2K^+(aq)+SO_4^(2-)+Cu(OH)_2(s)`

Cross out the spectators' ions on both sides of the complete ionic equation to get the net ionic equation as shown below:

`2OH^-(aq)+Cu^(2+)(aq)\rightarrow Cu(OH)_2(s)`