Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2015

Lecture Notes 11: 16 February

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Ionic reactions - dissolving and precipitates

Much of the chemistry around us involves the dissolution of ionic solids in water to give aqueous solutions and the precipitation of ions from aqueous solution to give precipitates (solids). So what I would like to do first is to look a little at the process of dissolving and the nature of aqueous solutions.

We will look at water later. The thing we need to know now is that the ions in water are not independent - they dissolve because they substitute interactions with water molecules for interactions with counter ions. And they stay in solution because they are insulated from each other by the water "shells" around each ion. A couple of corollaries.

• First, only so many ions will dissolve until we run out of water molecules to make the shells, then they will interact with each other and precipitate out.
• Second, the amount of a given ion which will dissolve in water depends on the competition between water and its counter ion for interaction with it. If the ion-water interactions are very strong, it will tend to be soluble, if the ion-ion interaction is very strong, then they will precipitate out at very low concentrations etc.

It is useful to remember some simple "rules" (really more like guide lines) to help in predicting reactions. For common compounds such as we see in general chemistry we can use the following rules:

1. Nitrates (NO₃^-) are all soluble.
2. Alkali metal (Li⁺, Na⁺, K⁺, Cs⁺, and Rb⁺) and ammonium (NH₄⁺) salts are all soluble, with the exception of a few Lithium salts.
3. Chloride, bromide, and iodide (Cl^-, Br^-, and I^-) salts are generally soluble, except for the salts of silver, lead(II) and mercury(I) (Ag⁺, Pb²⁺ and Hg₂²⁺).
4. Sulfates are soluble, except for the salts of barium {BaSO₄}, lead(II) {PbSO₄}, mercury(II) {HgSO₄}, and calcium {CaSO₄}.
5. Most hydroxides are only slightly soluble (but see rule 2).
6. Sulfides (S^2-), carbonates (CO₃^2-), phosphates (PO₄^3-), and chromates (CrO₄^2-) are only slightly soluble (but see rule 2).

Let's consider some chemical processes:

• Dissolve NaCl in water:

NaCl Na⁺_(aq) + Cl^-_(aq)

• Add NaCl solution to a AgNO₃ solution. First recall that each of the ions in solution is in an (aq)ueous complex, so we can write:

Na⁺_(aq) + Cl^-_(aq) + Ag⁺_(aq) + NO₃^-_(aq)  AgCl_(s) + Na⁺_(aq) + NO₃^-_(aq)

Notice that two ions don't change. So, if they don't prticipate, why show them. Instead we write a net ionic equation:

 Ag⁺ + Cl^- AgCl_(s)

Notice the ions that appeared on both sides are not shown (in mathematical terms they cancelled)

• Mix barium chloride and potassium sulfate:

Ba²⁺_(aq) + 2 Cl^-_(aq) + 2 K⁺_(aq) + SO₄^2-  BaSO_4(s) + 2 K⁺_(aq) + 2 Cl^-

Again, we want to write a net ionic equation showing only the ions which reacted:

Ba²⁺ + SO₄^2-  BaSO_4(s)

Notice that net ionic equations are very general expressions. Essentially they are saying that any time we have these species present they will react, regardless of what else happens to be there! (Sometimes folks are confused when they add ions which should react and they don't. This is usually a case where something else reacted first, so the ions of interest really weren't there!).

Since all ions in aqueous reactions are considered to be hydrated, we do not generally include _(aq) as part of the formula. But remember, it is assumed!

(Additional examples of solving net-ionic equations can be found in the Discussion Module.)

In these reactions we see a transfer of electrons from one atom or molecule to another. First let's look at some terms.

• Oxidation refers to taking electrons away from a substance. So to oxidize means to behave like oxygen normally does and "steal" electrons.
• Reduction refers to recieving electrons - it is the opposite of oxidation.
• Note that oxidation and reduction always go together. When oxygen oxidizes it is itself reduced (it gains electrons).
  • Consider the example of burning natural gas:

CH₄ + 2 O₂ CO₂ + 2 H₂O

Notice that the methane is oxidized by the oxygen. We say that the carbon and hydrogen are both oxidized to give the new covalent products, water and carbon dioxide, because the electrons are not evenly shared, they are pulled toward the oxygens in each bond.

Examples:

• A piece of clean copper is placed in a solution of silver nitrate. Assuming the copper goes to Cu(II) write a balanced net ionic equation. Notice that in this case you must balance the charge in order to get the final equation.

Ag⁺ + NO₃^- + Cu⁰ Ag⁰ + NO₃^- + Cu²⁺

Balancing: 2Ag⁺ + Cu⁰ 2Ag⁰ + Cu²⁺ 

• Consider the reaction of calcium metal with hydrochloric acid. This is similar to the reaction of sodium and water, and hydrogen gas is given off. Write a net ionic equation for this reaction:

H⁺ + Cl^- + Ca⁰ H_{2 (g)} + Ca²⁺ + Cl^-

Balancing: 2 H⁺ + Ca⁰ H_{2 (g)} + Ca²⁺

• A piece of clean iron is placed in a solution of copper(II) sulfate. Assuming iron goes to Fe(III) write a balanced equation. Again the problem is to balance the charge.

Cu²⁺ + SO₄^2- + Fe⁰ Cu⁰ + Fe³⁺ + SO₄^2-

Balancing: 3 Cu²⁺ + 2 Fe⁰ 3 Cu⁰ + 2 Fe³⁺

(Additional examples of solving net-ionic equations can be found in the Discussion Module.)

Balancing Redox Equations

There are two common methods for balancing redox reactions: the oxidation number method and the half-reaction method. The half-reaction method works very well for ionic reactions, it is relatively easy to give partial credit, and it is the only method I will use in this class. If you know how to do the other method you are welcome to do so, but be careful to make sure you show your work or I won't be able to give partial credit!

The Half-Reaction Method

In the half-reaction method what we do is first break an equation into two parts and then balance the parts individually. Presented stepwise:

Acid Solution:

Separate the reaction into two half-reactions.

Balance each half-reaction separately:

1. Balance atoms other than O & H by inspection.
2. Balance O by adding H₂O to the opposite side.
3. Balance H by addding H⁺ as appropriate.
4. Balance the charge by adding electrons (e^-) - add to same side as excess of positive charge, or opposite side if excess negative charge.
5. Balance the charges of the two half-reactions by multiplying appropriately.

Add two equations together

Cancel items appearing on both sides.

Example. Balance the following equation as it occurs in acid solution:

MnO₄^- + Cl^- Mn²⁺ + Cl₂

First break the equation into two half reactions, one for Mn and one for Cl

10 e^- + 16 H⁺ + 2 MnO₈^-+ 10 Cl^- 2 Mn²⁺ + 8 H₂O + 5 Cl₂ + 10 e^-

16 H⁺ + 2 MnO₄^- + 10 Cl^- 2 Mn²⁺ + 8 H₂O + 5 Cl₂

(Additional examples for balancing Redox equations in acidic solution can be found in the Discussion Module.

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© R A Paselk

Last modified 11 February 2015