Solutions

Topics so far

*Definition

*Parts, how solutions are formed and how to affect the rate of solvation

*Concentration

*Calculation of Concentration

*Solutions in Reactions

Double replacement reactions and solubility rules

Acid and Base Reactions

Solution Stoichiometry

*Acid and Base Solutions

Titration lab technique and calculations

Definitions and characteristics of acids and bases

Conjugate acid and base pairs

Behavior of acids and bases in aqueous solutions [H3O+], [OH-]

Strong vs Weak Acids

pH scale

Calculation of pH and pOH

Next and Last Solution Topic

Colligative Properties

Definition

The colligative properties we will work with are

*Lowering vapor pressure

*Depressing freezing-point

*Elevating boiling-point

Calculations for

Freezing-point depression

Boiling-point elevation

Ice Cream lab

______

Colligative Properties Introduction

The presence of solute molecules changes certain properties in the solution from what they are in the pure solvent. Solution properties that are dependent on concentration, are called colligative properties. Examples of colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. You worked with osmotic pressure in Biology.

When working with colligative properties we express solution concentration in terms of molality. As previously discussed we use molality because it is unaffected by changes in temperature. The magnitude of any colligative effect depends on the total concentration of solute particles in the solution, not on the number of moles of solute added to make up the solution. In short, you have to ask yourself “What are the concentrations of all solute particles in this solution?” For a strong electrolyte (think: ionic compounds, acids and bases), we can assume that all of thecompound has broken up into its component ions on dissolving.

For example, a 0.100 m solution of K3PO4 has four times the concentration of solute particles as a 0.100 m solution of a non-electrolyte such as sugar, because each formula unit of K3PO4 breaks up on dissolving into four ions:

three K+ ions and one PO43-ion

The number of solute particles formed is used in discussions and calculations of colligative properties and referred to as the i factor or the df dissociation factor.

Vapor Pressure

The Macroscopic View

When a solute is added to a solvent, the vapor pressure of the solvent (above the resulting solution) is lower than the vapor pressure above the pure solvent. Recall vapor pressure is the pressure exerted by a vapor that is in dynamic equilibrium with its liquid in a closed system. A solution that contains a nonvolatile solute (not easily vaporized) always has a lower vapor pressure than the pure solvent. How much lower, depends on the i or df of the solute.

In The Beaker View

.

Check your understanding

  1. What does it mean to say that a solute is volatile? Give an example.
  1. Which will have a lower vapor pressure, an aqueous solution of sugar or one made with calcium chloride? Explain your choice.
  1. A glass of iced tea containing excess sugar at the bottom is said to be ______.
  1. Categorize the following compounds.

Compound / Acid or Base or Ionic / Electrolyte Non-, Weak, Strong / If an electrolyte what are the ions formed / What is the df or i factor?
Ca (NO3)2 / Ionic / Electrolyte / Ca2+ and 2NO3- / df or I = 3
C6H12O6
HF
H2SO3
Ca(OH)2

Freezing Point Depression

Macroscopic View

When a substance freezes, the particles of the solid take on an orderly pattern. The presence of a solute disrupts the formation of the pattern. As a result more kinetic energy must be withdrawn from a solution than from the pure solvent to cause the solution to solidify. Therefore, the freezing point of a solution is lower than the freezing point of the pure solvent.

In The Beaker View

Calculations

The freezing points of solutions are all lower than that of the pure solvent. The freezing pointdepression is directly proportional to themolalityof the solute.

Freezing Point of the Solution = Freezing Point of the Solvent – Change in Freezing Point

The change in freezing point = K * df * m

Or

ΔTf=Tf(solvent)−Tf(solution)=Kf×m

WhereΔTf is the freezing point depression; the cahnge

Tf(solution) is the freezing point of the solution,

Tf(solvent) is the freezing point of the solvent,

Kf is the freezing point depression constant (given)

df is the dissociation constant andmis the molality.

Applications

Road salting takes advantage of this effect to lower the freezing point of the ice it is placed on. Lowering the freezing point allows the street ice to melt at lower temperatures. The maximum depression of the freezing point is about −18°C (0°F), so if the ambient temperature is lower,NaClwill be ineffective. Under these conditions,CaCl2can be used since it dissolves to make three ions instead of two forNaCl.

Check Your Understanding

  1. What is the molality of a solution prepared by dissolving 13.0 grams of barium nitrate in 450. grams of water?
  1. If I add 45 grams of sodium chloride to 500 grams of water, what is the freezing point of the resulting solution?

Constants you will need Kf(H2O) = 1.86 0C/m. Answer:Melting point = -5.730 C

Boiling Point Elevation

A solution has a higherboiling pointthan the pure solvent. This happens whenever a non-volatile solute, such as a salt, is added to a pure solvent, such as water.

Now it is your turn to provide the views. Consider your understanding of Freezing Point Depression and provide a particle diagram that illustrates what is happening “in the beaker” for Boiling Point Elevation.

Here are the captions you are illustrating:

Diagram A In this system liquid particle easily shift into the gas phase at the normal boiling point.

Diagram B In this system the solute particles prevent the liquid particles from escaping into the gas phase. There for the liquids in this system must posses ______.

Try this problem

What is the expected boiling point of water for a solution that contains 150 g of sodium chloride dissolved in 1.0 kg of water?