Electrolytes and Dissociation

As detailed in the previous section , when an ionic compound like NaCl dissolves in water, the ionic-bonded lattice structure is broken down, and the ions go their separate ways in the water, what we might call "free ions." We can represent this process with a chemical equation; we will learn later in this section the ins and outs of writing equations like this one.

NaCl (s) → Na⁺ (aq) + Cl^- (aq)

Note that in the formula for NaCl, we do not write the charges on the sodium and chloride ions, but we do once they are dispersed throughout an aqueous solution. This is because the charges become much more important once the ions are moving freely from each other. Solutions that contain free ions are called electrolyte solutions, and they are very important in chemistry, especially biological chemistry. The reason is related to the conduction of electrical current.

Conductivity of Solutions

The conductivity of solutions can be tested by using an ohmmeter such as the one shown here. The meter has two probes; in the picture at right, the probes are separated from each other, so they are not in electrical contact. When there is little or no electrical conductivity between the probes, as in this case the needle on the meter points to the left of the scale.

A battery in the ohmmeter passes electric current through the probes, if it is able. The amount of current that flows is a measure of the conductivity between the probes.

When there is very good electrical conductivity between the probes (as when they are touching) the needle on the meter points to the right of the scale.

When the probes are placed in water, as shown here, you can see that the meter needle is still far to the left. Pure water does not conduct electricity very well at all. This may surprise you, as you may have heard that water is a very good electrical conductor. However, as we will soon see, water is only conductive when it contains certain types of substances.

For example, when the probes are placed in a solution containing sodium chloride, the meter needle moves quite a ways to the right showing that the solution conducts electricity very well.

We will soon learn the reasons for this difference; it has to do with the way breaks apart into ions in solution.

When the probes are placed in a solution containing sugar, the meter needle does not appear to move away from the left end of the scale, showing that the solution does not conduct electricity any better than pure water.

In this exhibit, we have tried to make a fair comparison by using the same concentration of NaCl and sugar in each case.

When the probes are placed in a solution containing vinegar (acetic acid), the meter needle moves somewhat to the right showing that the solution conducts electricity somewhat.

So we have seen three different solutions with three different outcomes. Sodium chloride gave a very conductive solution, acetic acid moderately conductive, and sugar non-conductive.

Strong, Weak, and Non-Electrolytes

We refer to these three cases in terms of the "electrolyte" properties of the substances. Substances like sodium chloride - which produce highly conductive solutions - are called strong electrolytes. A substance is a strong electrolyte if it dissolves in water to produce a large number of free ions. This is because electrical current is the movement of charges; in order for current to flow through water, there must be charged particles (ions) able to move around.

In the case of NaCl, those ions are sodium and chloride. Other ionic compounds can also be electrolytes, but only if they are fairly soluble in water.

Thus, the solubility rules we saw in the previous section can tell us whether an ionic compound will be a strong electrolyte or not. Insoluble ionic compounds are non-electrolytes.

The sugar solution, however, does not conduct electricity. Sugar is a molecular compound; whereas salt dissolves in water by breaking up a lattice of ionic bonds to release free ions, sugar dissolves by breaking up a lattice of intermolecular forces to release individual sugar molecules (each with a formula of C₆H₁₂O₆). These molecules don't break up any further in solution, they just float around. Thus, sugar is a non-electrolyte.

What about the acetic acid solution? Acetic acid (HC₂H₃O₂) is a molecular compound, so it can dissolve into water as free neutral molecules, the way sugar does. However, once in the water, those molecules can react to release H⁺ and C₂H₃O₂^- ions, as shown at right. (We will discuss this type of reaction in much more detail in Lessons 5-7).

While this reaction does take place, it is highly incomplete. As the image shows, only a small percentage of acetic acid molecules react to form ions. This is why the solution of acetic acid only faintly conducts electricity, making it a weak electrolyte.

Dissociation Equations

When compounds release ions in solution (like sodium chloride and acetic acid do) we refer to that process as a dissociation. We can write chemical equations for dissociation processes. You have already seen the equation for the dissolution/dissociation of NaCl, above. The equation for the dissociation of acetic acid is shown below.

HC₂H₃O₂ (aq) → H⁺ (aq) + C₂H₃O₂^- (aq)

We should clarify here the difference between the terms "dissolution" and "dissociation," since they sound so similar and are partially overlapping. "Dissolution" processes refer to things dissolving in water. For molecular substances like sugar and acetic acid, this looks pretty simple.

C₆H₁₂O₆ (s) → C₆H₁₂O₆ (aq)

HC₂H₃O₂ (l) → HC₂H₃O₂ (aq)

These dissolution processes are not dissociations. "Dissociation" refers to a process that releases free ions. In the case of acetic acid, the dissolution process is followed by the dissociation process shown above. In the case of sugar, there is no dissociation - no free ions are formed, just neutral molecules distributed throughout the water.

For sodium chloride, however, the dissolution process is also a dissociation, because the solid NaCl dissolves into the water and releases free ions in one fell swoop.

NaCl (s) → Na⁺ (aq) + Cl^- (aq)

Writing Dissociation Equations

In this course, you will learn to write dissociation equations for three types of electrolytes: soluble ionic compounds, acids, and bases. Lessons 5-7 will cover acids and bases and their dissociations in great detail. In this section we will focus on soluble ionic compounds. Given any ionic compound, by the end of this section you should be able to identify whether it is an electrolyte or not, and if it is to write a dissociation equation for it. Note: Don't forget that ionic compounds are compounds containing a metal and a non-metal. The guidelines below will lead you astray if you don't keep that important fact in mind.

First, you must determine whether the compound is soluble in water. To do this, use the solubility rules presented in the previous section. For example, if you are given the compound Mg(OH)₂, you should look at the rule for OH^- and observe that compounds of that ion are mostly insoluble, and Mg²⁺ is not an exception. So Mg(OH)₂ would be classified as a non-electrolyte, which we can represent in one of the following ways (NR stands for "no reaction").

Mg(OH)₂ (s) → NR

Mg(OH)₂ (s) ⥇

Second, if the compound is soluble, you need to correctly identify the ions released when it dissolves. This means remembering, for example that aluminum ions are always Al³⁺, and sulfide ions are always S^2-. For variable cations, it means being able to determine charges, like the fact that VCl₃ contains the V³⁺ ion. Finally, it means knowing your polyatomic ions. When an ionic compound with polyatomic ions dissolves, the ions separate from each other, but the polyatomic ions stay intact. This can be seen in the examples shown below.

KOH (aq) → K⁺ (aq) + OH^- (aq)

MgSO₄ (s) → Mg²⁺ (aq) + SO₄^2- (aq)

Note: There are eight polyatomic ions whose names and formulas you must have memorized for this class: sulfate (SO₄^2-), nitrate (NO₃^-), phosphate (PO₄^3-), carbonate (CO₃^2-), chlorate (ClO₃^-), hydroxide (OH^-), acetate (C₂H₃O₂^-), and ammonium (NH₄⁺).

Finally, you must be mindful of the number of ions present in the formula. Like normal equations, dissociation equations must be balanced, so if there are three chloride ions in your compound, there must be three in the products. However, you have to be careful with the subscripts. When the compound Na₃PO₄ dissolves, not only do the phosphate ions move away from the sodium ions, the sodium ions move away from each other. So instead of a subscript (which implies connection via a bond) we write a coefficient of 3 on the sodium ions to keep things balanced when representing the dissociation of the compound.

Na₃PO₄ (s) → 3 Na⁺ (aq) + PO₄^3- (aq)

For practice, try writing dissociation equations for the ionic compounds shown below.

(NH₄)₂S

AlBr₃

Fe(OH)₂

KNO₃

BaCO₃

CaCl₂

You can check your answers below.

(NH₄)₂S (s) → 2 NH₄⁺ (aq) + S^2- (aq)

AlBr₃ (s) → Al³⁺ (aq) + 3 Br^- (aq)

Fe(OH)₂ (s) ⥇

KNO₃ (s) → K⁺ (aq) + NO₃^- (aq)

BaCO₃ (s) ⥇

CaCl₂ (s) → Ca²⁺ (aq) + 2 Cl^- (aq)

Google Sites

Report abuse