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1 Solutions nt/Solutions_30SE.ppt

2 Definitions A solution is a homogeneous mixture of a solute dissolved in a solvent. The solvent is generally in excess an aqueous solution has water as solvent. Example The solution NaCl(aq) is sodium chloride NaCl(s) dissolved in water H 2 O(l) The solute is NaCl(s) and the solvent is H 2 O(l). Air is an example of a solution with one “ solvent ” (nitrogen) and many “ solutes ” (oxygen, helium, argon, carbon dioxide, etc.)

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4 Definitions Solutions can be classified as saturated or unsaturated. A saturated solution contains the maximum quantity of solute that dissolves at that temperature. A saturated solution represents equilibrium: rate of dissolving equals to rate of crystallization An unsaturated solution contains less than the maximum amount of solute that can dissolve at a particular temperature A SUPERSATURATED SOLUTIONS contain more solute than is possible to be dissolved A SUPERSATURATED SOLUTIONS contain more solute than is possible to be dissolved got-

5 Supersaturated solutions are unstable. The supersaturation is only temporary, and usually accomplished in one of two ways: 1. Warm the solvent so that it will dissolve more, then cool the solution 2. Evaporate some of the solvent carefully so that the solute does not solidify and come out of solution.

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7 Electrolyte and Non-electrolyte n Electrolyte: a substance that conducts electricity when dissolved in water. –Acids, bases and soluble ionic solutions are electrolytes. n Non-electrolyte: a substance that does not conduct electricity when dissolved in water. –Molecular compounds and insoluble ionic compounds are non-electrolytes. tr/IB/mustafa/AQUEOUS%2520SOLUTIONS.ppt+AQUEOUS+SOLUTIONS+ppt&cd=1&hl =en&ct=clnk&gl=uk

8 n Electrolytes n Some solutes can dissociate into ions. n Electric charge can be carried.

9 Types of solutes Na + Cl - Strong Electrolyte - 100% dissociation, all ions in solution high conductivity UEOUS%2520SOLUTIONS.ppt+AQUEOUS+SOLUTIONS+ppt&cd=1&hl=en&ct=clnk&gl=uk

10 Types of solutes CH 3 COOH CH 3 COO - H+H+ Weak Electrolyte - partial dissociation, molecules and ions in solution slight conductivity QUEOUS%2520SOLUTIONS.ppt+AQUEOUS+SOLUTIONS+ppt&cd=1&hl=en&ct=clnk&gl=uk

11 Types of solutes sugar Non- electrolyte - No dissociation, all molecules in solution no conductivity QUEOUS%2520SOLUTIONS.ppt+AQUEOUS+SOLUTIONS+ppt&cd=1&hl=en&ct=clnk&gl=uk

12 r/IB/mustafa/AQUEOUS%2520SOLUTIONS.ppt+AQUEOUS+SOLUTIONS+ppt&cd=1&hl=en&ct =clnk&gl=uk Types of Electrolytes n Weak electrolyte partially dissociates. –Fair conductor of electricity. n Non-electrolyte does not dissociate. –Poor conductor of electricity. Strong electrolyte dissociates completely. –Good electrical conduction.

13 mustafa/AQUEOUS%2520SOLUTIONS.ppt+AQUEOUS+SOLUTIONS+ppt&cd=1&hl=en&ct=clnk&gl =uk mustafa/AQUEOUS%2520SOLUTIONS.ppt+AQUEOUS+SOLUTIONS+ppt&cd=1&hl=en&ct=clnk&gl =uk Representation of Electrolytes using Chemical Equations MgCl 2 (s)+H 2 O → Mg 2+ (aq) + 2 Cl - (aq) A strong electrolyte: A weak electrolyte: CH 3 COOH(aq) ← CH 3 COO - (aq) +H + (aq) → CH 3 OH(aq) A non-electrolyte:

14 r/IB/mustafa/AQUEOUS%2520SOLUTIONS.ppt+AQUEOUS+SOLUTIONS+ppt&cd=1&hl=en&ct =clnk&gl=uk r/IB/mustafa/AQUEOUS%2520SOLUTIONS.ppt+AQUEOUS+SOLUTIONS+ppt&cd=1&hl=en&ct =clnk&gl=uk Strong Electrolytes Strong acids: HNO 3, H 2 SO 4, HCl, HClO 4 Strong bases: MOH (M = Na, K, Cs, Rb etc) Salts: All salts dissolving in water are completely ionized. Stoichiometry & concentration relationship NaCl (s) +H 2 O  Na + (aq) + Cl – (aq) Ca(OH) 2 (s) +H 2 O  Ca 2+ (aq) + 2 OH – (aq) AlCl 3 (s) +H 2 O  Al 3+ (aq) + 3 Cl – (aq) (NH 4 ) 2 SO 4 (s) +H 2 O  2 NH 4 + (aq) + SO 4 2– (aq)

15 3 Stages of Solution Process n Separation of Solute –must overcome IMF(Intermolecular forces) or ion-ion attractions in solute –requires energy, ENDOTHERMIC ( +  H) n Separation of Solvent –must overcome IMF of solvent particles –requires energy, ENDOTHERMIC (+  H) n Interaction of Solute & Solvent –attractive bonds form between solute particles and solvent particles –“Solvation” or “Hydration” (where water = solvent) –releases energy, EXOTHERMIC (-  H)

16 Dissolution at the molecular level? n Consider the dissolution of NaOH in H 2 O

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18 Factors Affecting Solubility 1. Nature of Solute / Solvent 1. Nature of Solute / Solvent - Like dissolves like (IMF) 2. Temperature - i) Solids/Liquids- Solubility increases with Temperature Increase K.E. increases motion and collision between solute / solvent. ii) Gas - Solubility decreases with Temperature Increase K.E. result in gas escaping to atmosphere. 3. Pressure Factor - i) Solids/Liquids - Very little effect Solids and Liquids are already close together, extra pressure will not increase solubility. ii) gas - Solubility increases with Pressure. Increase pressure squeezes gas solute into solvent.

19 k12.tr/IB/mustafa/A k12.tr/IB/mustafa/A Solubility curve Saturated Unsaturated Supersaturated

20 Solubility curve Any solution can be made saturated, unsaturated, or supersaturated by changing the temperature.

21 E.ppt E.ppt Solubilities of Solids vs Temperature Solubilities of several ionic solid as a function of temperature. MOST salts have greater solubility in hot water. A few salts have negative heat of solution, (exothermic process) and they become less soluble with increasing temperature.

22 The rate of solution The rate of solution is a measure of how fast a substance dissolves. Some of the factors determining the rate of solution are: n size of the particles -- When a solute dissolves, the action takes place only at the surface of each particle. When the total surface area of the solute particles is increased, the solute dissolves more rapidly. Breaking a solute into smaller pieces increases its surface area and hence its rate of solution. n (Sample problem: a cube with sides 1.0 cm long is cut in half, producing two pieces with dimensions of 1.0 cm x 1.0 cm x 0.50 cm. How much greater than the surface area of the original cube is the combined surface areas of the two pieces? n 2.0 cm 2

23 The rate of dissolution n stirring -- With liquid and solid solutes, stirring brings fresh portions of the solvent in contact with the solute, thereby increasing the rate of solution. n amount of solute already dissolved -- When there is little solute already in solution, dissolving takes place relatively rapidly. As the solution approaches the point where no solute can be dissolved, dissolving takes place more slowly. n temperature -- For solid, liquid and gaseous solutes, changing the temperature not only changes the amount of solute that will dissolve but also changes the rate at which the solute will dissolve.

24 Temperature & the Solubility of Gases The solubility of gases decreases at higher temperatures WHY???

25 The effect of partial pressure on solubility of gases Henry’s Law At pressure of few atmosphere or less, solubility of gas solute follows Henry Law which states that the amount of solute gas dissolved in solution is directly proportional to the amount of pressure above the solution. c = k P c = solubility of the gas (M) k = Henry’s Law Constant P = partial pressure of gas Henry’s Law Constants (25°C), k N M/mmHg O M/mmHg CO M/mmHg

26 How does Henry’s Law apply?? & Soft Drinks Henry’s Law & Soft Drinks n Soft drinks contain “carbonated water” – water with dissolved carbon dioxide gas. n The drinks are bottled with a CO 2 pressure greater than 1 atm. n When the bottle is opened, the pressure of CO 2 decreases and the solubility of CO 2 decreases, according to Henry’s Law. n Therefore, bubbles of CO 2 escape from solution.

27 Colligative Properties On adding a solute to a solvent, the properties of the solvent are modified. n Vapor pressure decreases n Melting point decreases n Boiling point increases n Osmosis is possible (osmotic pressure) These changes are called COLLIGATIVE PROPERTIES. They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

28 Vapor Pressure Lowering for a Solution n The diagram below shows how a phase diagram is affected by dissolving a solute in a solvent. Notice the changes in the freezing & boiling points.

29 Vapor Pressure Lowering n The presence of a non-volatile solute means that fewer solvent particles are at the solution’s surface, so less solvent evaporates!

30 Raoult’s Law Describes vapor pressure lowering mathematically n The lowering of the vapour pressure when a non-volatile solute is dissolved in a volatile solvent (A) can be described by Raoult’s Law: P A =  A P° A P A = vapour pressure of solvent A above solution X A = mole fraction of the solvent A in solution. P° A = vapour pressure of pure solvent A. only the solvent (A) contributes to the vapour pressure of the solution

31 Mixtures of Volatile Liquids Both liquids evaporate & contribute to the vapor pressure

32 Raoult’s Law: Mixing Two Volatile Liquids n Since BOTH liquids are volatile and contribute to the vapour, the total vapor pressure can be represented using Dalton’s Law: P T = P A + P B The vapor pressure from each component follows Raoult’s Law: P T =  A P° A +  B P° B Also,  A +  B = 1 (since there are 2 components) n Ideal solutions obtained if solute-solute, solute-solvent, n and solvent-solvent interactions are similar, n i.e. ΔH soln = 0.

33 Deviations from ideality occur if,, there are strong solute- solvent interactions as may be in H-bonding between solute and solvent. Such solutions are called nonideal solutions. ― Deviations from Raoult’s law ΔHsoln << 0 ⇒ negative deviation ΔHsoln >> 0 ⇒ positive deviation Benzene - Toluene mixture: –The vapor pressure from each component follows Raoult's Law. Recall that with only two components,  Bz +  Tol = 1 Benzene: when  Bz = 1, P Bz = P° Bz = 384 torr & when  Bz = 0, P Bz = 0 Toluene: when  Tol = 1, P Tol = P° Tol = 133 torr & when  Tol = 0, P Bz = 0

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35 Normal Boiling Process Normal Boiling Point: BP of 1atm When solute is added, BP > Normal BP Boiling point is elevated when solute inhibits solvent from escaping.

36 Boiling Point Elevation ΔT b = (T b -T b °) = i ·m ·k b Where, ΔT b = BP. Elevation T b = BP of solvent in solution T b ° = BP of pure solvent m = molality, k b = BP Constant Some Boiling Point Elevation and Freezing Point Depression Constants Normal bp (°C) K b Normal fp (°C) K f Solvent pure solvent (°C/m) pure solvent (°C/m) Water Benzene Camphor Chloroform (CH 3 Cl) (CH 3 Cl) Some Boiling Point Elevation and Freezing Point Depression Constants Normal bp (°C) K b Normal fp (°C) K f Solvent pure solvent (°C/m) pure solvent (°C/m) Water Benzene Camphor Chloroform (CH 3 Cl) (CH 3 Cl)

37 Boiling Point Elevation and Freezing Point Depression ∆T = i K m ∆T = i K m i = van’t Hoff factor = number of particles per molecule/formula unit. For covalent compounds, i = 1. For ionic compounds, i = the number of ions CompoundTheoretical Value of i glycol 1 NaCl2 CaCl 2 3 Ca 3 (PO 4 ) 2 5

38 When solution freezes the solid form is almost always pure. Solute particles does not fit into the crystal lattice of the solvent because of the differences in size. The solute essentially remains in solution and blocks other solvent from fitting into the crystal lattice during the freezing process. Freezing Point Depression n Normal Freezing Point: FP of 1atm n When solute is added, FP < Normal FP n FP is depressed when solute inhibits solvent from crystallizing.

39 Freezing Point Depression Phase Diagram and the lowering of the freezing point.  T f = i ·m ·k f Where,  T f = FP depression i = van’t Hoff Factor m = molality, k f = FP Constant Generally freezing point depression is used to determine the molar mass of an unknown substance. Derive an equation to find molar mass from the equation above.

40 Osmotic pressure n Osmosis is the spontaneous movement of water across a semi-permeable membrane from an area of low solute concentration to an area of high solute concentration n Osmotic Pressure - The Pressure that must be applied to stop osmosis  = i CRT where P = osmotic pressure i = van’t Hoff factor C = molarity R = ideal gas constant T = Kelvin temperature

41 Osmosis and Blood Cells (a) A cell placed in an isotonic solution. The net movement of water in and out of the cell is zero because the concentration of solutes inside and outside the cell is the same. (b) In a hypertonic solution, the concentration of solutes outside the cell is greater than that inside. There is a net flow of water out of the cell, causing the cell to dehydrate, shrink, and perhaps die. (c) In a hypotonic solution, the concentration of solutes outside of the cell is less than that inside. There is a net flow of water into the cell, causing the cell to swell and perhaps to burst.


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