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 Solutions (homogeneous mixtures) are a part of our everyday life. Do you use mouthwash, toothpaste, cough syrup?  My swimming pool is one very large.

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Presentation on theme: " Solutions (homogeneous mixtures) are a part of our everyday life. Do you use mouthwash, toothpaste, cough syrup?  My swimming pool is one very large."— Presentation transcript:

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2  Solutions (homogeneous mixtures) are a part of our everyday life. Do you use mouthwash, toothpaste, cough syrup?  My swimming pool is one very large solution. The presence of algae makes my pool green and slimy. I then need to add an algicide which is another solution. Bromine is in solution as well as other chemicals (total alkalinity).

3  Solubility is of fundamental importance in a large number of practical applications ranging from ore processing to the use of medicines and the transport of pollutants.  Solubility is often said to be one of the “characteristic properties of a substance”.  used to describe a substance  indicates its polarity  helps distinguish it from another substance  becomes a guide for the applications of that substance  You will often see in a compounds description: “is insoluble in water or alcohol but soluble in concentrated sulfuric acid”

4  Solubility of a substance is useful when separating mixtures. The synthesis of chemical compounds, by the milligram in a lab or by the ton in industry makes use of the relative solubility of the end product.  Cooks, chemists, farmers, pharmacists and gardeners need to know which compounds are soluble and which are insoluble.  You will separate a mixture or five substances for first lab using the property of solubility as an important property.

5  What properties do solutions have?  How do we describe the concentrations?  Why do some substances dissolve while others do not?

6  Solution – a homogeneous mixture that is uniform throughout. The same substances will be in the same relative amounts. Solutions are not pure substances because they can have variable composition (example salt water).  Solvent – any substance that has other substances dissolved in it. The substance that is present in the largest amount (by volume, mass or number of moles) is usually the solvent.

7  Solute – a substance dissolved into the solvent.  Concentrated – a solution with a high level of solute ( > 1.0 M)  Dilute – a solution with a low level of solute (< 1.0 M)  A solution can be a gas, solid (alloy) or a liquid.  Aqueous – a liquid solution in which water is the liquid (aq)  Miscible – liquids which will dissolve in each other  Immiscible – liquids which do not dissolve in each other.

8  The ability of a solvent to dissolve a solute depends on the forces of attraction between the particles. There is always some attraction between the solvent and the solute.  Solubility – the mass of the solute that dissolves in a given quantity of solvent, at a certain temperature. May be g/mL or molar solubility (moles of solute in 1 L of solution).  Saturated – no more solute will dissolve in a solution. Excess solute can be seen.

9  Unsaturated – a solution which is not yet saturated. More solute can dissolve.  Sparingly soluble or slightly soluble – the solubility is between insoluble (less than 0.1g per 100 mL) and soluable (1g per 100 mL).  The solubility of gasses do not use the same terms as that for solids and liquids. An example is the amount of dissolved oxygen in water g/100 mL is the solubility of oxygen at 20 o C in fresh water. This is very soluble.

10  Substances of “like” polarity tend to be more soluble in each other than substances that differ in polarity.  Remember – polarity is a measure of how electrons are shared between bonding elements. It also is dependent on shapes of molecules determined by VESPR. “Like dissolves like.” Polar solvents dissolve other polar molecules and ionic compounds. Nonpolar molecules dissolve other nonpolar molecules. Alcohols which have characteristics of both tend to dissolve both polar and nonpolar but will not dissolve ionic solids.

11  Step 1 – the forces between the particles in the solid must be broken. Either between the ions or between the molecules. This requires energy.  Step 2 – Some of the intermolecular forces between the particles in the liquid must be broken. This also requires energy.  Step 3 – There is an attraction between the particles of the solid and the particles of the liquid. This step gives off energy.

12  Sugar (polar) in water (polar).  Iodine (nonpolar) in benzene (nonpolar)  Electronegativity: Electronegativity Difference Type of Bond 0.0 – 0.4nonpolarH-H – 2.0polarH-Cl0.9 > 2.0ionicNaCl2.1

13  Polar, nonpolar or ionic a) N and H 3.0 – 2.1 b) F and F c) Ca and O d) Al and Cl 3.0 – polar 0 nonpolar 2.5 ionic 1.5 polar

14  Not all substances dissolve in the same manner or amount.

15  1. The solubility of most substances increases as the temperature of the solvent increases. For some substances, the opposite is true. At 25 o C NaCl 36.2 g/100g H 2 O At 100 o C NaCl 39.2 g/100g H 2 O Sodium sulfate At 40 o C 50g/100g H 2 O At 100 o C 41g/100g H 2 O

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17  Gases have a higher solubility in cold water than in hot water.  …….So temperature affects the rate at which something will dissolve.

18  2. Pressure The solubility of a gas increases as the pressure of the gas above the solution increases. Soft drinks are bottled under high pressure. Henry’s Law At a given temperature the solubility of a gas in a liquid (S) is directly proportional to the pressure of the gas above the liquid (P). S 1 = S 2 P 1 = P 2 P then S

19  If the solubility of a gas in water is 0.77 g/L at 3.5 atm of pressure, what is the solubility in g/L at 1.0 atm of pressure? The temperature is constant at 25 o C.  If S 1 = S 2 then S 2 = S 1 x P 2 ___ ____ _______ P 1 P 2 P 1 = 0.77 g/l x 1.0 atm _________________ 3.5 atm = 0.22 g/L

20  3. Surface Area In order to dissolve, the solvent must come in contact with the solute. Smaller pieces increase the amount of surface of the solute interacting with the solvent. Term Used for SolubilitySolubility (g/L) Very SolubleGreater than 100 g SolubleBetween 10 g and 100 g Slightly SolubleBetween 1 g and 10 g InsolubleLess than 1 g (.1g /100 ml)

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22  When a solution of NaOH is added to a solution of CuSO 4 a blue solid precipitate forms. The mixture of solutions contains Na +, OH -, Cu 2+, and SO 4 2- ions.  When we consult a solubility table we find that the Cu(OH) 2 is insoluble, the precipitate formed must be Cu(OH) 2. An equation for this reaction can be written as: 2NaOH (aq) + CuSO 4 (aq) Na 2 SO 4 (aq) + Cu(OH) 2 (s)

23  In solution these compounds exist as dissociated ions as shown below: 2Na + (aq) + 2OH - (aq) + Cu 2+ (aq) + SO 4 2- (aq) Na + (aq) + SO 4 2- (aq) + Cu(OH) 2 (s)  Notice that the Na + (aq) and SO 4 2- (aq) remain unchanged by the reaction. These ions are called spectator ions because they do not react. So if we show only the ions changed by the reaction, the equation becomes: Cu 2+ (aq) + 2OH - (aq) Cu(OH) 2 (s) This is called the net ionic equation.

24  Precipitation Reactions and Solubility Rules A solid dissolves in water because the water molecules around the solid’s surface collide with the particles of the solid. The attraction between polar water and a partially charged particle or ion enables the water molecules to pull these particles away from the crystal, a process called dissociation. This is followed by solvation. Predicting Precipitates We can predict the formation of a precipitate if we know the solubility of the compounds. Patterns in solubility have lead to the creation of solubility rules. Solubility rules are based on what happens when salts are placed into water, they either dissolve or they do not dissolve.

25  Precipitation occurs because two SOLUBLE substances containing solvated ions are mixed and a pair of ions form a substance that is INSOLUBLE. The insoluble substance is the precipitate (solid) that is formed.  When you mix two solutions containing ionic compounds you can predict the products based on DOUBLE REPLACEMENT reaction.  If one of the predicted products is not soluble, it will precipitate out and a reaction occurs.  If both predicted products are soluble, no reaction occurs (all ions remain solvated).

26  Simple Rules for the Solubility of Ionic Compounds in Water 1. Most nitrate (NO 3 - ) salts are soluble. 2. Most salts containing the alkali metals cations (Li +, Na +, K +, Rb +, Cs + ) and the ammonium (NH 4 + ) cation are soluble. 3. Most chloride, bromide, and iodide salts are soluble. Except those containing Ag +, Pb 2+, and Hg + 4. Most sulfate salts are soluble. Except those containing Ba 2+, Pb 2+, Hg 2+, Sr 2+ and Ca Most hydroxide salts are insoluble except those of the alkali metal cations and a few alkaline earth metals: Ba 2+, Sr 2+, and Ca Most sulfides (S 2- ), carbonate (CO 3 2- ), chromate (CrO 4 2- ), and phosphate (PO 4 3- ) salts are insoluble.

27 Example: Will a precipitation reaction occur when Ba(NO 3 ) 2 (aq) and Na 2 SO 4 (aq) are mixed? Possible products from a double displacement reaction: BaSO 4 (s) and NaNO 3 (s)  From the solubility table BaSO 4 is not soluble so it would form a precipitate. Ba(NO 3 ) 2 (aq) + Na 2 SO 4 (aq) BaSO 4 (s) + 2 NaNO 3 (aq) We can also write a complete (total) ionic equation: Ba 2+ (aq) + 2NO 3 - (aq) + 2Na + (aq) + SO 4 2- (aq) 2Na + (aq) + 2NO 3 - (aq) + BaSO 4 (s) or finally we can write a net ionic equation which shows only the ions which precipitate from solution. Ba 2+ (aq) + SO 4 2- (aq) BaSO 4 (s)

28  Example: Will a precipitation reaction occur if 150 mL of 1.0M NaI (aq) is added to 200 mL of 0.5M AgNO 3 (aq)? The solubility table shows that AgI is insoluble. AgNO 3 (aq) + NaI (aq) AgI (s) + NaNO 3 (aq) Net ionic equation Ag + (aq) + I - (aq) AgI (s)

29  Molarity (M) is the most common method for describing the concentration of a solution, and is defined as the number of moles of solute present per litre of solution. M = n/V (V in litres) Example: Calculate the molarity of 75.0 g of KBr (s) dissolved in 500 mL of water. M= mol solute/ V (in litres)

30  Mol KBr = 75.0 g / 119 g/mol = mol V solution = 500 mL / 1000 mL = 0.5 L M = 0.630mol/0.5 L = 1.26 M KBr (aq)  Example: What are the concentrations of Ca 2+ (aq) and NO 3 - (aq) in 1.5 M Ca(NO 3 ) 2 (aq)? Ca(NO 3 ) 2 Ca 2+ (aq) + 2NO 3 1- (aq) 1.5 M 1.5 M M

31  Solutions prepared by dissolving a precise mass of solute in a precise volume of solvent are known as standard solutions. Volumetric flasks are used to prepare standard solutions.  Dilutions It is often necessary to dilute a certain solution. Dilution is the process by which a solution is made less concentrated by the addition of more solvent. When a solution has been diluted its solute particles are spread out more. When solutions are diluted, the amount of solute particles remains the same. Since n=MV (concentration x volume) we can use the following equation: M 1 V 1 = M 2 V 2

32  Example: mL of 3.0 M NaOH (aq) is diluted to 400 mL. What is the concentration of the diluted solution? Initail # moles = final # moles M 1 V 1 = M 2 V 2 (3.0 M x 0.2 L) = ( M 2 x 0.4L) M 2 = 1.5 M NaOH (aq)  Example: What volume of a stock solution of 6.0 M HCl (aq) is needed to make 900 mL of a 1.3 M solution? M 1 V 1 = M 2 V 2 (6.0 M x V 1 ) = (1.3 M x 0.9 L) V 1 = 0.2 L or 200 mL

33  Molality (m) is the number of moles of solute per kilogram of solvent.  M= n/kg  Example: A solution contains 15.5 g of NH 2 CONH 2 in 74.3 g H 2 O. Calculate the molality.  Molality = mol solute/kg solvent

34  Mol NH 2 CONH 2 = 15.5 g / 60.0 g/mol (molar mass) = mol  Kg H 2 O = 74.3 g x 1 kg 1000 g = Kg  m = mol solute Kg solvent = mol Kg = 3.48 m

35  Mole fraction (no units) is the number of moles of one compound in a mixture divided by the total number of moles of all components in the mixture. X a = n a / (n a + n b )  Example: What are the mole fractions of an ethanol and water solution prepared by dissolving 3.45 g C 2 H 5 OH in 21.1 g H 2 O? Mol ethanol = 3.45 g / 46.0 g/mol = mol Mol of water = 21.0 g / 18.0 g/mol = 1.18 mol Total # moles = mol of ethanol + mol of water = mol mol = mol

36  Mol fraction of ethanol = / = Mol fraction of water = 1.18 / =  Mol fraction ethanol + water = 1

37  % by mass mass of solute x 100 (mass of solute + mass of solvent)  Example: A 5 % NaCl solution would consist of 5.0 g of NaCl (s) and 95 g of water. ( I chose 100 g because it was easier to take 5%). 1 mL of H 2 O = 1 g H 2 O. So 5.0 g NaCl (s) in 95 mL of H 2 O (l).  Example: 24 g of NaCl (s) is dissolved in 152 g H 2 O (l). Calculate the % by mass of this solution. Mass % NaCl = 24 g x 100 (24 g + 152g) = 24/176 x 100 = 14 %

38  Properties of solutions which depend on the number of solute particles but not on their nature.  They are dependent on the number of solute particles dissolved in a given volume of solvent. They include: vapour pressure lowering, boiling point elevation, freezing point depression and osmotic pressure.

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40  The presence of a solute lowers the freezing point of a solution relative to that of the pure solvent. For example, pure water freezes at 0°C (32°F); if one dissolves 10 grams of sodium chloride (table salt) in 100 grams of water, the freezing point goes down to −5.9°C (21.4°F). If one uses sucrose (table sugar) instead of sodium chloride, 10 grams in 100 grams of water gives a solution with a freezing point of −0.56°C (31°F).

41  The reason that the salt solution has a lower freezing point than the sugar solution is that there are more particles in 10 grams of sodium chloride than in 10 grams of sucrose. Since sucrose, C 12 H 22 O 11 has a molecular weight of grams per mole and sodium chloride has a molecular weight of grams per mole, 1 gram of sodium chloride has almost six times as many sodium chloride units as there are sucrose units in a gram of sucrose. In addition, each sodium chloride unit comes apart into two ions (a sodium cation and a chloride anion ). The freezing point depression of a solution containing a dissolved substance, such as salt dissolved in water, is a colligative property.

42  One can calculate the change in freezing point (Δ T f ) relative to the pure solvent using the equation: Δ T f = K f m where K f is the freezing point depression constant for the solvent (1.86°C·kg/mol for water) (can be looked up on a table),m is molality (the number of moles of solute in solution per kilogram) of solvent for ALL particles. Because the presence of a solute lowers the freezing point, the Department of Highways puts salt on our roads after a snowfall, to keep the melted snow from refreezing. Also, the antifreeze used in automobile heating and cooling systems is a solution of water and ethylene glycol (or propylene glycol); this solution has a lower freezing point than either pure water or pure ethylene glycol.

43  ΔT f = K f x m What is the freezing point of these solutions? 1.4 mol of Na 2 SO 4 in 1750 of H 2 O m = 1.4 mol 1.75 kg = 0.8 m 3 particles are formed when Na 2 SO 4 ionizes 3 x 0.8m = 2.4 m Δ T f = K f x m = 1.86 o C/m x 2.4 m = 4.46 o C 0 o C – 4.5 o C = o C

44  What is the freezing point of 0.60 mol of MgSO 4 in 1300 g of H 2 O? m = 0.60 mol 1.3 kg = 0.46 m 2 particles x 0.46 m = 0.92 m ΔT f = 1.86 o C x 0.92 m = 1.71 o C 0 o C – 1.71 o C = o C

45  The boiling point of a solution is higher than that of the pure solvent. The actual boiling point elevation is the difference in temperature between the boiling points of a solution and the pure substance. Accordingly, the use of a solution, rather than a pure liquid, in antifreeze serves to keep the mixture from boiling in a hot automobile engine. As with freezing point depression, the effect depends on the number of solute particles present in a given amount of solvent, but not the identity of those particles. If 10 grams of sodium chloride are dissolved in 100 grams of water, the boiling point of the solution is 101.7°C (215.1°F; which is 1.7°C (3.1°F) higher than the boiling point of pure water).

46  The formula used to calculate the change in boiling point (Δ T b ) relative to the pure solvent is similar to that used for freezing point depression: Δ T b = K b m, where K b is the boiling point elevation constant for the solvent (0.52°C·kg/mol for water), and m has the same meaning as in the freezing point depression formula. Note that Δ T b represents an increase in the boiling point, whereas Δ T f represents a decrease in the freezing point.

47  Boiling Point Δ Tb = Kb x m What is the boiling point of a solution that contains 1.2 mol of sodium chloride in 800 g of water? m= 1.2 mol NaCl 0.8 kg = 1.5 m Total molality = 1.5 m x 2 = 3m Δ T b = K b x m = o C/m x 3 m = 1.54 o C 100 o C o C = o C

48  What is the boiling point of a solution that contains 1.25 mol of CaCl 2 in 1400 g of water? m= 1.25 mol 1.4 kg = 0.89 m CaCl 2 = 3 particles when it ionizes 3 x 0.89 m = 2.67 m Δ T b = K b x m = o C/m x 2.67 m = 1.37 o C 100 o C o C = o C

49  A solution of 7.50 g of a nonvolatile compound in g of water boils at o C. What is the molecular mass of the solute?  Molality of the solution: o C – 100 o C = 0.78 o C ( ΔT b ) If ΔT b = K b x m then m = Δ T b K b m = 0.78 o C o C/m m = 1.5 m

50  1.5 m x kg = mol Molecular mass of solute = mass of solute moles of solute = 7.5 g mol = 218 g/mol

51  The vapor pressure of a liquid is the equilibrium pressure of gas molecules from the liquid (evaporation) above the liquid itself in a closed system. A glass of water placed in an open room will evaporate completely (and thus never reach equilibrium); however, if a cover is placed on the glass, the space above the liquid will eventually contain a constant amount of water vapor. How much water vapor is present depends on the temperature, but not on the amount of liquid that is present at equilibrium.

52 If, instead of pure water, an aqueous solution is placed in the glass, the equilibrium pressure will be lower than it would be for pure water. A solution that contains a nonvolatile solute (it does not vaporize like sugar or salt) always has a lower vapour pressure than the pure solvent. In an aqueous solution of sodium chloride, there are sodium ions and chloride ions throughout the solution. The ions are surrounded by shells of water of solvation. This reduces the number of solvent molecules that have enough kinetic energy to escape as vapour. The result is a solution with a lower vapour pressure than the pure solvent.

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54  The decrease in the vapour pressure is proportional to the number of particles the solute makes in solution. A solute like sodium chloride which dissociates into several particles has a greater effect on the vapour pressure than the same concentration of a solute like glucose which does not dissociate.  NaCl – 2 particles when dissociated  Glucose – 1 molecule, no disassociation  CaCl 2 – 3 particles when dissociated

55  Raoult’s law states that the vapour pressure of the solvent over the solution is proportional to the fraction of solvent molecules in the solution, if two-thirds of the molecules are solvent molecules, the vapour pressure due to the solvent is approximately two-thirds of what it would be for pure solvent at that temperature.

56 If the solute has a vapour pressure of its own, then the total vapour pressure over the solution would be: vapour pressure = vapour pressure x mole fraction (solution) (solvent)

57  Osmosis is the process whereby a solvent passes through a semipermeable membrane from one solution to another (or from a pure solvent into a solution). A semipermeable membrane is a barrier through which some substances may pass (e.g., the solvent particles), and other species may not (e.g., the solute particles). Important examples of semipermeable membranes are the cell walls in cells of living things (plants and animals). Osmosis tends to drive solvent molecules through the semipermeable membrane from the low solute concentrations to the high solute concentrations; thus, a "complete” osmosis process would be one that ends with the solute concentrations being equal on both sides of the membrane. Osmotic pressure is the pressure that must be applied on the high concentration side to stop osmosis.

58  Osmosis is a very useful process. For example, meats can be preserved by turning them into jerky: The meat is soaked in a very concentrated salt solution, resulting in dehydration of the meat cells. Jerky does not spoil as quickly as fresh meat, since bacteria on the surface of the salty meat will fall victim to osmosis, and shrivel up and die. This process thus extends the life of the meat without the use of refrigeration. There are times when one wishes to prevent osmosis when two solutions (or a pure solvent and a solution) are on opposite sides of a semipermeable membrane.

59 . Osmosis can be prevented by applying pressure to the more concentrated solution equal to the osmotic pressure on the less concentrated side. This can be accomplished either physically, by applying force to one side of the system, or chemically, by modifying a solute concentration so that the two solute concentrations are equal. (If one applies a pressure greater than the osmotic pressure to the higher concentration solution, one can force solvent molecules from the concentrated solution to the dilute solution, or pure solvent. This process, known as reverse osmosis, is often used to purify water.)

60  A hospital patient receiving fluids intravenously receives an intravenous (IV) solution that is isotonic with (i.e., at the same solute concentration as) his or her cells. If the IV solution is too concentrated, osmosis will cause the cells to shrivel; too dilute a solution can cause the cells to burst. Similar problems would be experienced by freshwater fish swimming in salt water, or saltwater fish swimming in freshwater. The osmotic pressure, like other colligative properties, does not depend on the identity of the solute, but an electrolyte solute will contribute more particles per formula unit than a nonelectrolyte solute.


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