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Chapter 11 Solutions and their Properties. Solutions. A mixture is a combination of 2 or more substances. There are 3 types of mixtures, namely:

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slide1

Chapter 11

Solutions and their Properties

slide2

Solutions

  • A mixture is a combination of 2 or more substances.
  • There are 3 types of mixtures, namely:
  • Suspensions – are heterogeneous mixtures which means the individual substances are still visible. Mixing may momentarily result in the appearance of one phase but will soon separate upon standing.
  • Solutions – are homogeneous mixtures. There is only one phase visible. Solutions do not separate on standing. Particle size of solute are at the atomic or molecular level.
  • Colloidal Suspensions – have properties intermediate between suspensions and true solutions. Solute particle sizes are also intermediate.
slide3

Thermodynamics of Solution Formation

  • The change of entropy (∆S) for solution formation is positive because of the increase in randomness.
  • The change in enthalpy (∆H) however is dependent upon the 3 kinds of interactions between the solute and the solvent. These are:
    • Solute-Solute interaction – the energy absorbed to break solute bonds (lattice energy) of the crystal.
    • Solvent-Solvent interaction – the energy absorbed to
  • overcome the intermolecular forces of the solvent.
    • Solute-Solvent interaction – the energy released when the solvent molecules when they surround the solute molecule.
  • The overall energy absorbed or released depends on the magnitude of the 3 energies.
slide4

Types of Solutions

  • Gases form solutions in all proportions.
slide5

Concentration of Solutions

  • Concentration is a measure of the amount of solute dissolved in a solvent. A solution that has a relatively small amount of solute dissolved is known as a dilute solution while a solution with a relatively large amount of solute is called a concentrated solution.
  • Concentrations are also expressed more precisely by the following units:
slide6

Example Problem

A solution is made up of 10.0 ml ethyl alcohol (C2H5OH, d=0.785 g/ml) dissolved in 50.0 ml water. Calculate for the following concentrations:

1. % m/m, 2. X solute, 3.Molarity, 4. molality

mass solute = 10.0ml x 0.785g/ml = 7.85g

mass solvent = 50.0ml x 1.00g/ml = 50.0g

moles solute = 7.85g x 1 mol solute/46g = 0.171 mols

moles solvent = 50.0g x 1 mol solvent/18g = 2.78 mols

% m/m = 7.85g/57.85g x 100 = 13.7%

X = 0.171 / (0.171 + 2.78) = 0.0579

M = 0.171mol / 0.060L = 2.85M

m = 0.171mol / 0.050kg = 3.42m

slide7

Problem

  • 11.68 A 0.944 M solution of glucose, C6H12O6, in water has a density of 1.0624 g/ml at 20oC. What is the concentration of this solution in the following units?
            • Mole fraction
            • Mass percent
            • Molality
slide8

Some Factors Affecting Solubility

  • As more and more solute is added to a solution a state of equilibrium is reached where in the amount of solute going in and out of solution are equal. A saturated solution is formed.
  • Some substances form solutions where more solute is dissolved than in a saturated solution. Such solutions are very unstable and are called supersaturated solutions.
  • The amount of solute in a saturated solution depends on 2 factors:
    • (1) Temperature
    • (2) Pressure
slide9

Effect of Temperature on Solubility

  • The solubility of a substance is the amount of the substance in a saturated solution.
  • The effect of temperature on solubility depends on the nature and physical state of the solute.
  • For solid solutes, the trend is the higher the temperature the greater the solubility. Exceptions are Na2SO4 and a few others where solubility decreases and NaCl where solubility is virtually unaffected with temperature.
  • For liquids, there is no general trend.
  • For gases, solubility decreases with increase in temperature with few exceptions.
slide10

Effect of Pressure on Solubility

  • Pressure affects only the solubility of gases significantly. The solubility of liquids and solids are unaffected by pressure.
  • The higher the pressure above the gas the more gas is forced into solution thereby increasing solubility.
  • Henry’s Law allows the calculation of the solubility of gases given the partial pressure of the gas above the liquid:
  • Solubility [M] = k x P
  • where solubility: moles/L, k is Henry’s constant: moles/(L.atm), and P is the partial pressure of the gas: atm or mmHg depending on units of k.
slide11

Problems

11.70 Vinyl Chloride has a Henrys Law constant of 0.091 mol/(L.atm) at 25oC. What is the solubility of vinyl chloride in water in mol/L at 25oC and a pressure of 0.75 atm.

11.72 Fish generally need an O2 concentration in water of at least 4 mg/L for survival. What partial pressure of oxygen above the water in atmospheres at 0 oC is needed to obtain this concentration? The solubility of O2 in water at 0 oC and 1 atm partial pressure is 2.21 x 10-3 mol/L.

slide12

Colligative Properties

  • Colligative properties are properties of solutions that depend only on the number of solute particles but not on the chemical or physical nature of the solute.
  • HCl a strong acid, exhibits the same colligative properties as NaOH a strong base, since they both produce 2 particles when dissolved in water.
  • There are 4 colligative properties: vapor pressure, boiling point, freezing point and osmosis.
  • The presence of solute lowers the vapor pressure, elevates the boiling point, lowers the freezing point, and increases osmotic pressure of solutions.
slide13

Vapor Pressure with non-Volatile Solute

  • The presence of a non-volatile solute decreases vapor pressure since the presence of solute lowers the average kinetic energy of the solvent particles.
  • Raoult’s Law governs the magnitude of vapor pressure lowering according to the following equation:
  • Psoln = Psolventx Xsolvent
  • where Psoln is the vapor pressure of the solvent, Psolvent is the vapor pressure of the pure solvent, and X is the mole fraction of the solvent.
  • Raoult’s law only works for dilute solutions and the solute and solvent has similar intermolecular attractions.
slide14

Vapor Pressure with Volatile Solutes

  • Solutions with volatile solutes have vapor pressures higher than solutions with non-volatile solutes.
  • The vapor pressure of such a solution obeys Dalton’s Law of partial pressures which states that the total pressure of a mixture of gases is the sum of the partial pressures of each individual gas. For two gases:
  • Ptotal = PA + PB
  • For solutions and using Raoult’s Law PA = PoA * XA and PB = PoB * XB so:
  • Ptotal = PoA * XA + PoB * XB
  • where PoA andPoB are the vapor pressures of pure liquids A and B and XA and XB are the mole fractions of the two liquids.
slide15

Problems

11.13 What is the vapor pressure in mm Hg of a solution prepared by dissolving 5.00 g benzoic acid (C7H6O2) in 100.00 g ethyl alcohol (C2H6O) at 35oC? The vapor pressure of pure ethyl alcohol at 35oC is 100.5 mm Hg.

11.14 How many grams NaBr must be added to 250 g of water to lower the vapor pressure by 1.30 mm Hg at 40oC. Assuming complete dissociation? The vapor pressure of water at 40oC is 55.3 mm Hg.

slide16

Boiling Point Elevation Freezing Point Depression

  • The boiling points of solutions are always higher than the boiling point of the pure solvent.
  • The freezing points of solutions are always lower than the freezing point of the pure solvent.
  • The boiling point elevation and freezing point lowering can be calculated from the equations:
  • ∆Tb = i* Kb * m
  • ∆Tf = i* Kf * m
  • where ∆Tbis the increase in boiling point, ∆Tfis the decrease in freezing point, i is the van’t Hoff’s factor, Kb and Kf are the boiling and freezing point constants unique to the solvent and m is the molality of the solution.
slide17

Problem

11.36 The following phase diagram shows a very small part of the solid/liquid phase transition boundaries for two solutions of equal concentration. Substance A has i=1, and substance B has i=3.

P (atm)

2

1

0

  • Which line, red or blue, a solution of A, and which represents a solution of B?
  • What is the approximate melting point of the pure liquid solvent?
  • What is the approximate molal concentration of each solution, assuming that the solvent has a Kf = 3.0 oC/m?

10 15

T (oC)

slide18

Osmosis and Osmotic Pressure

  • Osmosis is the diffusion of water molecules across a semi-permiable membrane. The direction is always from a region of higher to a lower solution concentration.
  • The diffusion exerts pressure on the semi-permiable membrane. This pressure is known as osmotic pressure.
  • Osmotic pressure (∏) of solutions can be calculated using the following equation:
  • ∏ = iMRT
  • where i is the vant’s Hoff factor, M is the molarity of the solution, R is the ideal gas constant and T is the temperature in oK.
  • If the the value of R used is 0.0821 L-atm/(mole- oK) the unit of ∏ will be atm.
slide19

Problems

  • 11.96 What osmotic pressure in atmospheres would you expect for each of the following solution?
  • 5.00 g of NaCl in 350.0mL of aqueous solution at 50 oC.
  • (b) 6.33 g of sodium acetate, in 55.0 mL of aqueous solution at 10 oC.
  • 11.98 A solution of an unknown molecular substance in water at 300 oK gives rise to osmotic pressure of 4.85 atm. What is the molarity of the solution.
slide20

Some Uses of Colligative Properties

Deicing roads – the addition of soluble solute to ice lowers its melting point thereby melting it. Table salt (NaCl) or MgCl2 is used to deice roads. MgCl2 has the advantage of a higher van’t Hoff’s factor.

Reverse Osmosis – the addition of pressure on the salt solution side results in water diffusing through a semi-permiable membrane producing pure water.

Fractional distillation – used in refineries to separate different grades of hydrocarbons from oil.

Physiological effects – differences in salt concentrations inside and outside the cell causes the cell to shrink (crenate) or explode (lysis).