Chemistry 30

For each of the following,

• predict the products of the double displacement reaction and write a balanced equation.
• use a table of solubilities to predict whether or not a precipitate forms. Identify any precipitates in your balanced equation (part a); also identify products that remain in solution.
• if a precipitate occurs, write a balanced net ionic equation for the reaction.

CuSO_{4 (aq)} + NaOH_(aq)

Solution

Since a reaction does occur, we determine what the spectator ions are in the equation. It may help you to write out all aqueous species in their long form. Solids, however, remain together as a compound:

Cu²⁺_(aq) + SO₄^2- _(aq) + 2 Na⁺_(aq) + 2 OH^-_(aq) → Cu(OH)_{2 (s)} + 2 Na⁺_(aq)+ SO₄^2-_(aq)

Notice that Na₂SO_{4 (aq)} becomes 2 Na⁺_(aq)+ SO₄^2-_(aq)

The spectator ions are Na⁺_(aq)+ SO₄^2-_(aq). Remove them to get the net ionic equation:

Cu²⁺_(aq) + 2 OH^-_(aq) → Cu(OH)_{2 (s)}

CaS_(aq) + (NH₄)₃PO_{4 (aq)}

Remove spectator ions to produce the net ionic equation. With practice you will see that both S^2- and NH₄⁺ are spectator ions and you will not need to write the long form of the equation.

Long form:

3Ca²⁺_(aq)+ 3S^2-_(aq) + 6NH₄⁺_(aq) + 2PO₄^3-_(aq) →Ca₃(PO₄)_{2 (s)} + 6NH₄⁺_(aq) + 3 S^2-_(aq)

Net Ionic Equation

3Ca²⁺_(aq)+ 2PO₄^3-_(aq) →Ca₃(PO₄)_{2 (s)}

MgBr_{2 (aq)} + H₂SO_{4 (aq)}

Balanced Equation:

MgBr_{2 (aq)} + H₂SO_{4 (aq)} → MgSO_{4 (aq)} + 2 HBr_(aq)

A table of solubilities tells us that both MgSO₄ and HBr are soluble compounds. Thus no precipitate forms and there is NO REACTION. All ions remain in solution.

We may indicate there is no reaction as follows:

We wish to separate the cations from a mixture containing the following solutions:

Ra(NO₃)₂, Mg(NO₃)₂, and AgNO₃

In order to do so we are given the following separate solutions:

K₂SO₄, K₂S, and KOH

In what order should we add the separate solutions in order to remove the cations by selective precipitation? List the precipitates that form, in the proper order.

Begin by recognizing the spectator ions that we can ignore.

• In the cation solutions, nitrate, NO₃^-, is the spectator in all cases as it will never form a precipitate.
• In the separate solutions, the potassium ion, K⁺, is an alkali ion which likewise always forms soluble solutions.

Prepare a chart to help identify precipitates that will form. "ppt" indicates an insoluble precipitate; "sol" indicates a soluble compound that remains dissolved as ions.

Read your chart carefully! We are adding the negative anions one at a time, so we read the across the rows (not down the columns):

• SO₄^2- forms two precipitates, with Ag⁺ and Ra²⁺ so it cannot be used first.
• S^2- forms only one precipitate, with Ag⁺ to form Ag₂S_(s). It could be used first.
• OH^- forms two single precipitates, with Ag⁺ and Mg ²⁺ so it cannot be used first

The solution:

To summarize our solution:

1. First add K₂S to remove Ag⁺ , forming the preciptate Ag₂S_(s)
   

2. We could remove the next cation in either order:

   1. by adding K₂SO₄ to remove Ra²⁺, forming RaSO_4(s)

   2. or by adding KOH to remove Mg²⁺, forming Mg(OH)_{2 (s)}