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Strong Electrolytes

Parts of this equation/concept include:

A -	Identifying Strong Electrolytes
B -	Calculating Molarity or Molality of Ions of Strong Electrolytes

Strong electrolytes come in two general categories: strong acids and soluble salts. It may help to remember that relatively few compounds are strong electrolytes. However, these compounds are also very common. Consequently, if the compound does not look familiar, it is probably not a strong electrolyte. The rules for strong electrolytes assume the solvent is water. Since there are few compounds that form ions in solvents other than water, you may assume that any substance dissolved in a solvent other than water is a nonelectrolyte.

Since there are only seven strong acids, the easiest way to identify strong acids is to learn the list. If it is not on the list, it is not a strong acid.

For a substance to fall in the category of “soluble salt” it must meet the criteria of both soluble and salt. Recall that “salt” is another name for an ionic compound, normally formed by a metal and an nonmetal. (Ammonium ion, NH₄+ , combined with another nonmetal or polyatomic ion is also a salt.) To determine whether the salt is soluble, you should learn the solubility rules. Since most salts are insoluble, it is easier to learn the few that are soluble, and if it is not one you have learned, it must be insoluble.

Strong acids: HCl, HBr, HI, HNO₃, HClO₄, HClO₃, H₂SO₄

Solubility rules from Table 5.4 (in the book):

1. All compounds containing alkali metal ions, ammonium (NH₄+), nitrate (NO₃–), or acetate (CH₃CO₂^– or C₂H₃O₂^–) ions are soluble.
2. All compounds containing halide ions except salt of Ag⁺, Cu⁺, Hg₂²⁺, or Pb²⁺ are soluble.
3. All compounds containing sulfate (SO₄^2–) ion except salts of Ba²⁺, Ca²⁺, Hg²⁺, Pb²⁺, and Sr²⁺ are soluble.
4. Ba(OH)₂, Ca(OH)₂, and Sr(OH)₂ are soluble.

When determining the ions formed by strong electrolytes in water, remember these tips:

• There are normally only two ions per compound, and you will probably recognize at least one of them.
• The cation is written first in the formula. For strong acids, the cation is H⁺.
• Polyatomic ions must not be broken up! (See Table 4.1 for list.)
• Subscripts become stoichiometric coefficients.
• Stoichiometric coefficients are not part of the formula.
• Ions without charges are wrong.

>> Example 1

Are the following compounds strong electrolytes? If so, what ions do they form in water?

a. NaNO3	b. CO2	c. HF	d. MgO	e. FeCl3

Solution:

1. NaNO₃ is a strong electrolyte. Ions: Na⁺ and NO₃^–
   Logic: It is a salt with the metal, sodium, and nonmetal, polyatomic anion, nitrate. Compounds containing either sodium or nitrate would be soluble (rule 1), so this certainly is. Sodium, as a group IA, alkali metal, always has a +1 charge; nitrate is a common polyatomic ion with a formula that should already be memorized.
   
2. CO₂ is not a strong electrolyte; therefore it does not form ions in water.
   Logic: It is a molecular compound of two nonmetals and not one of the strong acids.
   
3. HF is not a strong electrolyte.
   Logic: Since hydrogen is the first element, it is an acid. However, it is not on the list of strong acids; therefore it is not a strong electrolyte.
   
4. MgO is not a strong electrolyte.
   Logic: It is a salt of magnesium metal and oxide. However, neither element appears on the solubility rules. By default, that makes it insoluble. To be a strong electrolyte it must both be a salt and be soluble.
   
5. FeCl₃ is a strong electrolyte. In water it will make Fe³⁺ and 3 Cl^–.
   Logic: It is a salt of iron and chloride. According to solubility rule 2, chlorides (group VIIA or 17, the halogens) are normally soluble. The exceptions do NOT include iron, so this compound is soluble. Iron is a transition metal, so its charge depends on how it is combined. Chloride, however, always has a charge of –1. Since there are three and the entire compound has a net charge of zero, iron must have a charge of +3.

>> Example 2

Which of the following are strong electrolytes? For each strong electrolyte, what are the ions it makes in solution?

a. HNO3	b. PbBr2	c. K2CO3	d. Na	e. Ba(OH)2

Solution:

1. HNO₃ is a strong electrolyte. Ions = H⁺ and NO₃^–
   It is on the list of strong acids.
   
2. PbBr₂ is not a strong electrolyte.
   Bromide is a halide, which is usually soluble, but Pb²⁺ is one of the exceptions.
   
3. K₂CO₃ is a strong electrolyte. Ions = two ions of K⁺ and one ion of CO₃^2–
   Potassium is in group 1, is always soluble, and always has a +1 charge. Carbonate is a familiar polyatomic. If you forgot its charge, the two K⁺'s were a hint!
   
4. Na is not a strong electrolyte.
   It is not a strong acid. It is not a salt. It is sodium salts that are always soluble, this is sodium metal.
   
5. Ba(OH)₂ is a strong electrolyte. Ions = Ba²⁺ and two ions of OH^–
   Rule 4, it is one of the soluble hydroxides.

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First, you need to recognize that a solution is a strong electrolyte. Relatively few substances fit this category, but those that do are common.

There are no common examples of substances that are electrolytes in nonaqueous solutions. Therefore if the solvent is not water and if you are not given specific instructions otherwise, you mayassume that the substance is not an electrolyte.

In addition to determining if a substance forms ions, you must determine the ratio of moles of ions to moles of substance. This is the van't Hoff factor (i). One of the easiest ways to determine moles of ions (or particles) from molarity or molality is to use the van't Hoff factor, since

molarity of particles in solution = iM
molality of particles in solution = im

To predict the van't Hoff factor, you must know how the substance acts in the solvent. If the substance is not an ionic compound or an acid, i = 1. If the solvent is not water, i = 1. If the solvent is a weak electrolyte (e.g. a weak acid such as HF or a weak base such as NH₃), the van't Hoff factor cannot be predicted. If the substance is a strong electrolyte, you will normally assume complete ionization. i = 2 for all strong acids except sulfuric acid. Because it is a combination of a strong acid and a weak acid, you cannot predict its value of i. For soluble salts i = the number of ions in the overall compound.

Note that you are only predicting van't Hoff factors. The true values are somewhat less due to ion pairing. Nevertheless, the answer is better with the van't Hoff factor than without.

Therefore you should make the best prediction of a van't Hoff factor unless the problem asks you to find the true van't Hoff factor.

>> Example 3

Predict the van't Hoff factor for the following solutes in water

a. CH3OH	b. BaCl2	c. NiSO4	d. HNO3	e. (NH4)3PO4

Solution:

1. i = 1. This is a molecular compound and does not ionize in water.
2. i = 3. BaCl₂ Ba²⁺ + 2 Cl^–
3. i = 2. NiSO₄ Ni²⁺ + SO₄^2– (It is much easier if you know your polyatomic ions!)
4. i = 2. HNO₃ H⁺ + NO₃^– (Strong acids make H⁺ and the anion)
5. i = 4. (NH₄)₃PO₄ 3NH₄⁺ + PO₄^3–

>> Example 4

Predict the molarity of ions in the following solutions:

a. 0.10 M Na2CO3	b. 0.25 M AgNO3	c. 0.050 M HBr

Solution:

1. molarity of ions = iM = 3(0.10 M) = 0.30 M
2. molarity of ions = iM = 2(0.25 M) = 0.50 M
3. molarity of ions = iM = 2 (0.050 M) = 0.10 M

>> Example 5

Predict the molality of the following solutions:

a. 0.053 m LiF	b. 0.082 m C12H22O12	c. 0.33 m Cu(NO3)2

Solution:

1. molality of particles = im = 2(0.053 m) = 0.11 m (only two significant figures)
2. molality of particles = im = 1(0.082 m) = 0.082 m (it's a molecular compound)
3. molality of particles = im = 3(0.33 m) = 0.99 m

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