A solution of 120 mL of 0.200 M KOH is mixed with a solution of 280 mL of 0.250 M NiSO4 Write the balanced chemical equation for the reaction that occurs. Answer as a balanced chemical equation.

Question

A solution of 120 mL of 0.200 M KOH is mixed with a solution of 280 mL of 0.250 M NiSO4 Write the balanced chemical equation for the reaction that occurs. Answer as a balanced chemical equation. Include the phases in your answer,

Answer 1

In this reaction, you will only have a double replacement if one or more of the products is insoluble. If you take Ni(OH)2 as being insoluble (it is actually partially soluble), then you have the aforementioned double replacement. Note that the question asks you to included phases, and this is the reason, i.e to show that one or more products is insoluble.

2KOH(aq) + NiSO₄(aq) ==> K₂SO₄(aq) + Ni(OH)₂(s)

Answer 2

KOH + NiSO4

First let's examine the net ionic equation of the reactants:

K⁺ + OH^- + Ni²⁺ + SO₄^2-

Notice how in the compound NiSO₄ is balanced by 2+ and 2- while the KOH is balanced only by +1 and -1. When these ions will exchange, there will have to be double the ions from KOH to balance charges from the NiSO₄. Therefore, to balance this equation, there will have to be a coefficent of 2 in front of KOH. Given this equation is an example of a exchange reaction, the products will have the ions from the compounds exchanged with each other. The balanced equation will look like this:

2KOH(aq) + NiSO4(aq) ⇒ K₂SO₄(aq) + Ni(OH)₂(s)