Solutions

Introduction

We learned in chapter 1 that a solution is a mixture of substances with a uniform composition.

Solution = A homogeneous mixture

Examples of solutions include salt water, vodka, vinegar, hydrochloric acid, etc.

In chemistry we often have reason to work with solutions. For example consider the experiment you carried out in lab, where you mixed together aqueous solutions of Pb(NO₃)₂ and KI to form the yellow precipitate PbI₂. If you had mixed together pure Pb(NO₃)₂ and KI (both are solids) and stirred there would have been a minimal reaction. The water is essential in order to bring lead and iodide ions in contact.

In class demo – solid Pb(NO₃)₂ & KI vs. solution Pb(NO₃)₂ & KI

Thus we see that solution chemistry is a very important concept.

In chemistry we will often work with solutions containing only two components:

Solute – The minority component

Solvent – The majority component

For example in salt water, NaCl is the solute and water is the solvent.

Systems where water is the solvent are called aqueous solutions. In this course we will be concerned exclusively with aqueous solutions.

Intermolecular Forces in Solutions

To understand solutions at an atomic level let us consider the dissolution of sodium chloride in water.

Ionic Solids

A crystal of sodium chloride contains Na⁺ and Cl^- ions held together by the Coulombic attraction between cations (Na⁺) and anions (Cl^-).

Water is a molecular substance, meaning the hydrogen and oxygen share electrons. However, they don’t share electrons equally. The bonding electrons spend more time in the vicinity of the oxygen atom than near either of the hydrogens. This gives the oxygen end of the molecule a partial negative charge and the hydrogen end of the molecule a partial positive charge.

This unequal charge distribution makes water a polar molecule, and gives water its ability to dissolve compounds. When an ionic solid such as NaCl dissolves in water, the positive ends of the water molecule (hydrogen ends) are attracted to the negatively charged anions and the negative ends of the water molecule (oxygen end) are attracted to the positively charged cations.

The interaction between polar solvent molecules and ions allows a large part of the Coulombic attraction present in an ionic crystal to be maintained in solution. This allows the crystal to dissolve and prevents the ions from recombining to form a crystal.

Molecular Solids

Water can also dissolve molecular substances, we can separate these into molecular substances which dissociate into ions and those which do not.

Water is not the only example of a polar molecule, another example is ethanol, CH₃CH₂OH. You can think of this molecule as a water molecule where one of the hydrogens has been replaced with a C₂H₅ unit. As in water the oxygen ion has a partial negative charge and the hydrogen bound to water has a partial positive charge.

When ethanol dissolves in water the positive part of the water molecule (the hydrogen ends) are attracted to the negative part of the ethanol molecule (the oxygen end) and vice-versa. This is similar to what we saw earlier with NaCl, with one important difference. Ethanol molecules remain intact, they do not come apart to form ions.

Many molecules, for example the pure hydrocarbons (i.e. gasoline, propane, oil, etc.) are non-polar. In non-polar molecules there are no partially positive or negative ends to the molecule. In general non-polar molecules will not dissolve in polar solvents such as water. This is why water and oil do not mix.

There is another class of molecular compounds that we will learn about in this chapter, called acids. When a strong acid dissolves in water it does dissociate into ions. For example when HNO₃ (hydrogen nitrate) is dissolved in water the molecule comes apart to give H⁺ and NO₃^- ions.

Thus from a solution point of view strong acids behave very much like soluble ionic substances. As we will see later the distinguishing characteristic of an acid is the fact that the cation is always H⁺ (also remember that polyatomic ions remain intact).

Molarity and Solution Stoichiometry

In chapter 3 we learned how to calculate the number of moles/atoms /molecules from the mass of a substance. We then used this knowledge to calculate quantities such as limiting reactant and theoretical yield, which are very important when planning and interpreting chemical reactions.

If we try to apply the same concepts to solution reactions we are going to run into a serious problem. If we measure the mass of a solution we will be weighing the water as well as the compound of interest. How then can we define the quantity of the solute (which is what we are really interested in after all) in a solution?

To overcome this difficulty chemists came up with the concept of molarity. The molarity of a solution is defined as:

Molarity = Moles Solute

Liters of Solution

It is important to note that molarity is moles solute per liters of solution (solute + solvent) and not per liters of solvent.

When we work with solutions in laboratory they will always be defined in terms of their molarity (represented as M, i.e. 1 M CuSO₄ or 0.1 M CuSO₄). In calculations of quantities such as limiting reagents, theoretical yields, etc. the following analogies exist to reactions of pure substances.

Reactions of Pure Substances

Grams A [MW A] ® Moles A [Bal. equation] ® Moles B [MW B] ® Grams B

The quantities used to make conversions are given in parentheses.

Solution Reactions

Liters A [Molarity A] ® Moles A [Bal. equation] ® Moles B [Molarity B] ® Grams B

Molarity A Eq’n MW B

Just remember that whenever you need to know how many moles of a substance, which is in solution, use molarity for the conversion. Whereas, if you need to know how many moles there are of a solid or a gas use molecular weight.

Example

How much 0.5 M KI solution must be added to 10 mL of 0.1 M Pb(NO₃)₂ solution in order to completely precipitate the lead?

As always we need to begin with a balanced chemical reaction.

Pb(NO₃)₂ (aq) + 2KI (aq) ® PbI₂ (s) + 2KNO₃ (aq)

Next we use the chemical equation and molarity to calculate the amount of KI solution needed to fully react with the lead nitrate.

10 mL Pb(NO₃)₂ ´ [1 L/1000 mL] ´ [0.1 mol Pb(NO₃)₂/1 L] ´ [1 mol KI/1 mol Pb(NO₃)₂] ´ [1 L/0.5 mol KI]

= 4 ´ 10^-3 L = 4 mL