Chemistry 102: Acid Equilibrium: Net Equation

Question

When a strong base is added to a solution of CuSO₄, which is pale blue, a precipitate forms and the solution above the precipitate is colorless.

What is the net chemical equation that describes this reaction? 

Answer 1

For a strong base let’s choose NaOH (the most important part is that OH is present - the Na won’t really matter).

charge of Na = +1

charge of OH (hydroxide anion) = -1

charge of SO4 (sulfate anion) = -2

To cancel out with the SO4 charge, this means copper is +2

(Notice that we will need 2 of the ions that are +1 or -1 when they swap parters in this double replacement reaction because they will be matched with a +2 or -2)

Now let’s write the unbalanced reaction:

CuSO4 + NaOH = Na2SO4 + Cu(OH)2

Balance the equation by placing a 2 in front of NaOH

CuSO4 + 2NaOH = Na2SO4 + Cu(OH)2

To determine which product is the precipitate, look at a solubility chart. Follow the cation on one side and the anion on the other - the box where they meet should say soluble (S) or insoluble (I). From this we can see that Cu(OH)2 is insoluble and will be our solid precipitate that forms, while Na2SO4 remains soluble.

Lastly, we will write all aqueous solutions in the form of their ions, and cancel any ions that appear on both sides. This will give us the net chemical equation. (I’ve put the charges in parentheses but you would write them as a superscript).

Cu(2+) + SO4(2-) + 2Na(+) + 2OH(-) —> 2Na(+) + SO4(2-) + Cu(OH)2 (s)

The Na and SO4 ions appear on both sides and are spectators of this reaction, so they can be cancelled out. We are left with:

Cu(2+) + 2OH(-) —> Cu(OH)2 (s)