V. Solutions and Solubility; Acid/Base Chemistry 12 Questions or 12% total

● Solutions and solubility

– Solution terminology and types

– Factors affecting solubility and dissolution rate

– Concentration terms and calculations

– Golligative properties and conductivity of solutions

– Ionic equilibrium in precipitation reactions and calculations involving Ksp

● Acid/Base chemistry

– Concepts and reactions

– Equilibrium and calculations

– Titrations and calculations

What is a solution? 

A solution is a homogeneous mixture of two or more substances.

Solute is the substance present in the smaller amount.

Solvent is the substance present in the larger amount.

Types of solutions:  The types of mixtures are “defined” by the phase of the solute(s) and the solvent:

A Solution may be:

Aqueous is solutions with water as the solvent.

Concentration is the amount of solute in solvent.

Saturated indicates the solution has maximum amount of solute dissolved at that temperature. 

Unsaturated indicates that more solute may be dissolved in the solvent at that temperature.

Supersaturated the solution contains more solute than a saturated solution of that solute in solvent at that temperature.  Supersaturated solutions are unstable and the “extra” solute is likely to precipitate when the solution is disturbed.

Factors influencing the dissolving process:  How do substances dissolve and what factors influence dissolving of a solute by a solvent?

1.  The general rule is “Like Dissolves Like” 

2. Intermolecular/Interparticle attractions hold particles together in liquids and solid samples.  These same interactions govern the dissolving process. There are three different attractions to consider:

A.      Solvent-solvent attractions that must break to allow solution formation.

B.      Solute-solute attractions that must break to allow solution formation.

C.      Solvent-solute attractions that must form to allow solution formation.

Each of these steps involves energy.  Energy is required for the breaking of attractions in the first two steps and energy is released in the formation of attractions.  The overall change in energy can be determined by considering the DH for each step and, by Hess’ Law, the DH for the overall process.  An overall exothermic change is one of two factors that drive physical or chemical changes to occur.

DH_solution = DH₁  + DH₂  +  DH₃

The second factor that drives reactions to occur is the tendency of nature to move toward disorder.  Mixing substances increases the disorder of the system and favors the formation of a mixture of the two materials. 

Some terms that are used to describe solutions:

Miscible describes two liquids that are completely soluble in all proportions.

Solvation/hydration is the process in which solvent molecules surround the solute molecules or ions in a cage-like pattern.  When the solvent is water, the process is called hydration.

The general rule “Like Dissolves Like” is quite useful.

Example: Predict the relative solubilities for bromine (Br₂) in benzene(C₆H₆) and in water.  [In the textbook, the problem provides the dipole moment (m = 0 for benzene indicating that it is non-polar; and  m = 1.87 for water indicating that it is polar.)]  Answer:  The non-polar molecules of bromine will be more soluble in the non-polar solvent benzene than in the polar solvent water.

Quick Review:  “Like Dissolves Like” General Behavior:

1.  Non-polar molecules are more soluble in non-polar solvents.

2.  Polar molecules are more soluble in polar solvents.

3.  Ionic compounds are more likely to dissolve in polar solvents.

The composition of a solution is given in units of concentration.  There are several ways to express the concentration of a solution by giving information about the proportion of solute to solvent. Concentration units one needs to know:

1.  Percent(%) by mass or weight  =        mass of solute           X 100%

                                                            Total mass of mixture   

2.  Mole Fraction (X) =               # moles of component A

                                                Total # moles in the mixture

3.  Molarity (M) =    # moles of solute

                             # Liters of solution

M₁V₁ = M₂V₂

4.  Molality (m)   =   # moles of solute

                             # Kilograms of solvent

5.  Volume percent  =        volume of solute           X 100%

                                               Total volume of mixture

6.        ppm (parts per million) = # grams of solute per 1,000,000 grams of solvent

7.        Normality (N) =     Equivalents solutes

Volume of solutions in liters

Normality: There is a relationship between normality and molarity. Normality can only be calculated when we deal with reactions, because normality is a function of equivalents.

The example below uses potassium hydroxide (KOH) to neutralize arsenic acid. By studying the reaction it is possible to determine the proton exchange number to determine the normality of the arsenic acid.

Look at the equation H₃AsO₄ + 2KOH  -->  K₂HAsO₄ + 2H₂O:

Equivalent weight = molar mass/(H+ per mole)

Equivalent = mass of compound / Equivalent weight

And Normality = (equivalents of X)/Liter

Normality = molarity x n (where n = the number of protons exchanged in a reaction).

You probably remember that when a hydrogen atom is ionized and loses its electron, you are left with only a proton. So a hydrogen ion is basically a proton.

Let's assume that we have a 0.25 M solution of H₃AsO₄ and want to determine the normality of it if it participates in the reaction

H₃AsO₄ + 2KOH  -->  K₂HAsO₄ + 2H₂O

When H₃AsO₄ is neutralized by KOH, H₃AsO₄ provides two protons to form 2H₂O.

Note that H₃AsO₄ has 3 H, but K₂HAsO₄ only has 1 H. That means that 2 protons were exchanged.

Again normality = molarity * n

where n = the number of protons exchanged in a reaction

Remember that normality of the solution is 0.25 mol H₃AsO₄ and there were two protons exchanged (2 equivalents/mole)

So, in short, while there is a relationship between the normality of a solution and the molarity of a solution, the normality can only be determined by examining reaction, determining the proton exchange and multiplying molarity by that number.

Normality is particularly useful in titrations calculations.

Where N = normality, V = volume, a = the substance on the left of the equation involved in proton exchange, and b=substance on the right of the equation involved in proton exchange:

N_aV_a = N_bV_b

Percentages are easy to calculate because they do not require information about the chemical nature of the substance. Percentages can be determined as percent by weight or percent by volume.