Composition of Solutions

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1 Chapter 11 Solutions The Composition of Solutions The Solution Process Factors that Affect Solubility Measuring Concentrations of Solutions Quantities for Reactions That Occur in Aqueous Solution Colligative Solutions Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Composition of Solutions A solution is: A homogeneous mixture with uniform composition throughout Composed of: Solute Substance being dissolved Usually present in the lesser amount Solvent Substance doing the dissolving Usually present in the greater amount Composition of Solutions What is the solute? What is the solvent?

2 Composition of Solutions Figure 11.5 Solubility How readily or completely a solute dissolves in a solvent A soluble ionic compound dissociates into ions when dissolved in water. Insoluble compounds remain mostly undissociated in their ionic, crystalline structure. The solubilities for ionic compounds do not follow a predictable pattern Must be recalled from experimental data Solubility rules are listed in Table Ions Rules Used to Predict the Solubility of Ionic Salts Rules Na +, K +, NH 4 + NO 3 - SO 4 2- Cl -, Br -, I - Ag + O 2-, OH - S 2- CrO 4 2- CO 3 2-, PO 4 3-, SO 3 2-, SiO 3 2- Most salts of sodium, potassium, and ammonium ions are soluble. All nitrates are soluble. Most sulfates are soluble. Exceptions: BaSO 4, SrSO 4, PbSO 4, CaSO 4, Hg 2 SO 4, & Ag 2 SO 4 Most chlorides, bromides, and iodides are soluble. Exceptions: AgX, Hg 2 X, PbX 2, & HgI 2 (X = Cl, Br, or I) Silver salts, except AgNO 3, are insoluble. Oxides and hydroxides are insoluble. Exceptions: NaOH, KOH, Ba(OH) 2, & Ca(OH) 2 (somewhat soluble) Sulfides are insoluble. Exceptions: Salts of Na +, K +, NH 4 +, and the alkaline earth metal ions Most chromates are insoluble. Exceptions: Salts of Na +, K +, NH 4+, Mg 2+, Ca 2+, Al 3+, and Ni 2+ Most carbonates, phosphates, sulfites, and silicates are insoluble. Exceptions: Na +, K +, and NH

3 Electrolytes A solution that conducts electricity Two kinds of electrolytes: Strong electrolyte A solute that dissociates completely into ions in aqueous solution Weak electrolyte A solute that dissociates partially into ions in aqueous solution Nonelectrolyte A solute that does not dissociate into ions in aqueous solution Electrolytes Practice Composition of Solutions What ions, atoms, or molecules are present when the following substances are mixed with water? Write a balanced equation to represent the process of dissolving each substance in water. ZnCl 2 (s) KNO 3 (s) C 5 H 11 OH(l)

4 Practice Solutions Composition of Solutions ZnCl 2 (s) ZnCl 2 (s) + H 2 O(l) Zn 2+ (aq) + 2 Cl - (aq) KNO 3 (s) KNO 3 (s) + H 2 O(l) K + (aq) + NO 3- (aq) C 5 H 11 OH(l) C 5 H 11 OH(l) + H 2 O(l) C 5 H 11 O - (aq) + H + (aq) Does not dissolve extensively Like Dissolves Like When the bonding in a solvent is similar to the bonding in a solute, then the solute will dissolve in the solvent. Polar covalent solvents dissolve polar covalent solutes. Nonpolar covalent solvents dissolve nonpolar covalent solutes. See Table 11.2 below: Solute Polar Solvent Nonpolar Ionic Soluble Insoluble Polar Soluble Insoluble Nonpolar Insoluble Soluble Solution Process Figure

5 Solution Process Ion-dipole force An attraction between an ion and a polar molecule When an ionic compound dissolves in water: Ionic bonds in the solute break. Hydrogen bonds between water molecules break. Ion-dipole forces form between ions and water molecules. Solution Process Heat of Solution The difference between the energy required for separation of solute and solvent and the energy released upon formation of ion-dipole interactions When the solvation process provides more energy than is needed to separate the pure solvent and solvent particles, the heat of solution is negative, and the overall dissolving process is exothermic

6 Endothermic Solution Process Figure 11.14B Solution Process for Nonpolar- Nonpolar Interactions Entropy A measure of the tendency for matter to become disordered or random in its distribution Matter spontaneously changes from a state of order to a state of randomness, unless energy changes oppose it. Solutions form because the makeup of the solution naturally tends toward disorganization

7 Entropy Figure Practice The Solution Process Iodine (I 2 ) dissolves in hexane (C 6 H 14 ) to form a solution. Describe the forces that must be broken for the solution to form. Practice Solutions The Solution Process Iodine (I 2 ) dissolves in hexane (C 6 H 14 ) to form a solution. Describe the forces that must be broken for the solution to form. London forces must be broken to allow I 2, the solute, to dissolve. London forces must also be broken to allow C 6 H 14, the solvent, to dissolve

8 Factors that Affect Solubility Three factors affect solubility: Structure Temperature Pressure Structure For a solid to be soluble in a given solvent, the energy released by the solvation process must compensate for all of the energy required to break up the forces at work both within the solute and the solvent. Entropy is also a factor The solubility (or miscibility) of liquids in other liquids depends largely on the polarity of the solute and solvent molecules. Temperature Affects the solubility of most substances in water Most ionic solids are more soluble in water at higher temperatures Gases are less soluble as temperature increases Figure

9 Pressure Affects gases but not liquids or solids The solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid An increase in pressure results in an increase in solubility of a gas Pressure Figure Measuring Concentrations of Solutions Concentration The relative amounts of the solutes and solvent that make up the solution Concentrat ion = amount of solute amount of solution (or solvent) Solubility The ratio that identifies the maximum amount of a solute that will dissolve in a particular solvent to form a stable solution under specified conditions A measure of maximum concentration Three types of concentrated solution: Saturated solution Unsaturated solution Supersaturated solution

10 Saturated Solution A solution that contains the maximum possible amount of solute When this type of solution forms, a dynamic equilibrium is established. Undissolved solute continues to dissolve, but at the same time, previously dissolved solute is deposited from solution. Supersaturated Solution A solution that contains more than the maximum possible amount of solute Unsaturated solution A solution that contains less than the maximum possible amount of solute Figure Table 11.3: Concentration Units Unit Percent by mass Percent by volume Mass/volume percent Parts per million Parts per billion Molarity (M) Molality (m) grams of solute x 100% grams of solution volume of solute x 100% volume of solution grams of solute x 100% volume of solution grams of solute x 10 6 grams of solution grams of solute x 10 9 grams of solution moles of solute liters of solution moles of solute kilograms of solvent Definition

11 Percent by Mass Is calculated by dividing the solute mass by the mass of the solution and multiplying it by 100% Percent by mass = mass (solute) x 100% mass (solution) To find the mass of the solution, we can add the mass of the solute to the mass of the solvent. Percent by Mass Example: What is the percent-by-mass concentration of acetic acid (CH 3 CO 2 H) in a vinegar solution that contains 2.70 g of acetic acid and g of water? Percent by mass = mass (solute) x 100% mass (solution) Percent by mass = 2.70 g x 100% (2.70 g g) Percent by mass = 2.15% Percent by Volume Is calculated by dividing the solute volume by the solution volume and multiplying by 100% Percent by volume = volume (solute) x 100% volume (solution) The volume of the solution must be measured. Volumes are not additive like masses

12 Percent by Volume Example: To prepare a solution, we dissolve 205 ml of ethanol in sufficient water to give a total volume of 235 ml. What is the percent by volume concentration of ethanol? Percent by volume = volume (solute) x 100% volume (solution) Percent by volume = 205 ml x 100% 235 ml Percent by volume = 87.2% Mass/Volume Percent Is calculated by dividing the solute mass by the volume of solution and multiplying by 100% Mass/Volume Percent = grams of solute x 100% volume of solution Example: The concentration of NaCl in a saline bag is 0.9%. Therefore, for every 100 ml of solution, the mass of NaCl present is 0.9 g. Parts per Million and per Billion Parts per million Is calculated by dividing the mass of solute by the mass of solution and multiplying by 1 million (10 6 ) Parts per million = mass (solute) x 10 6 mass (solution) Parts per billion Is calculated by dividing the mass of solute by the mass of solution and multiplying by 1 billion (10 9 ) Parts per billion = mass (solute) x 10 9 mass (solution)

13 Molarity The ratio of moles of solute to the volume of solution in liters Molarity (M) = moles (solute) liters (solution) Example A solution contains 117 g of potassium hydroxide (KOH) dissolved in sufficient water to give a total volume of 2.00 L. The molar mass of KOH is g/mol. What is the molarity of the aqueous potassium hydroxide solution? Molarity Example A solution contains 117 g of potassium hydroxide (KOH) dissolved in sufficient water to give a total volume of 2.00 L. The molar mass of KOH is g/mol. What is the molarity of the aqueous potassium hydroxide solution? Molarity (M) = moles (solute) liters (solution) Moles KOH = 117 g KOH x 1 mol KOH = 2.09 mol KOH g KOH Molarity (M) = 2.09 mol KOH = 1.04 M KOH 2.00 L Molality If the temperature of a solution increases, the molarity of the solution changes because the volume changes. Molality does not change with temperature because it divides moles of solute with the mass of solution. molality (m) = moles (solute) kilograms (solution)

14 Molality Example A solution contains 56.7 g of potassium hydroxide (KOH) dissolved in sufficient water to give a total mass of g. The molar mass of KOH is g/mol. What is the molality of the aqueous potassium hydroxide solution? molality (m) = moles (solute) kilograms (solution) moles (solute) = 56.7 g KOH x 1 mol KOH = 1.01 mol KOH g KOH kilograms (solvent) = g x 1 kg = kg 1000 g Molality Example A solution contains 56.7 g of potassium hydroxide (KOH) dissolved in sufficient water to give a total mass of g. The molar mass of KOH is g/mol. What is the molality of the aqueous potassium hydroxide solution? molality (m) = moles (solute) kilograms (solution) molality (m) = 1.01 mol KOH = 4.11 m KOH kg KOH Precipitation Reactions The cation from one reactant combines with the anion from another reactant to form an insoluble compound, called a precipitate. Example When a solution of lead(ii) nitrate is mixed with a solution of potassium chromate, a yellow precipitate forms according to the equation: Pb(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) 2 KNO 3 (aq) + PbCrO 4 (s) What volume of M lead(ii) nitrate is required to react with ml of M potassium chromate? What mass of PbCrO 4 solid forms?

15 Precipitation Reactions Pb(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) 2 KNO 3 (aq) + PbCrO 4 (s) What volume of M lead(ii) nitrate is required to react with ml of M potassium chromate? What mass of PbCrO 4 solid forms? To solve for volume of Pb(NO 3 ) 2 : M x V = moles 100.0mL K 2 CrO 4 x moles K 2 CrO 4 = moles K 2 CrO ml K 2 CrO moles K 2 CrO 4 x 1 mol Pb(NO 3 ) 2 = moles Pb(NO 3 ) 2 1 mol K 2 CrO moles Pb(NO 3 ) 2 x 1 L Pb(NO 3 ) 2 = L Pb(NO 3 ) mol Pb(NO 3 ) 2 Precipitation Reactions Pb(NO 3 ) 2 (aq) + K 2 CrO 4 (aq) 2 KNO 3 (aq) + PbCrO 4 (s) What volume of M lead(ii) nitrate is required to react with ml of M potassium chromate? What mass of PbCrO 4 solid forms? To solve for volume of Pb(NO 3 ) 2 : M x V = moles moles K 2 CrO 4 * 1 mol PbCrO 4 = moles PbCrO 4 1 mol K 2 CrO moles PbCrO 4 * g PbCrO 4 = 3.88 g PbCrO 4 1 mol PbCrO 4 Acid-Base Titrations Titration The process of determining the concentration of one substance in solution by reacting it with a solution of another substance that has a known concentration Equivalence point The point in a titration at which the moles of the acid equal the moles of the base End point The point in a titration where the indicator changes color Slightly different than the equivalence point, but a good estimation

16 Acid-Base Titrations Practice - Acid-Base Titrations We need ml of M HCl solution to completely titrate ml of Ba(OH) 2 solution of unknown concentration. What is the molarity of the barium hydroxide solution? Practice Solutions - Acid- Base Titrations We need ml of M HCl solution to completely titrate ml of Ba(OH) 2 solution of unknown concentration. What is the molarity of the barium hydroxide solution? First, write the balanced chemical equation: 2 HCl(aq) + Ba(OH) 2 (aq) BaCl 2 (aq) + 2 H 2 O(l) Next, find the moles of HCl: ml HCl x moles HCl 1000 ml HCl = x 10-3 moles HCl

17 Practice Solutions - Acid-Base Titrations HCl(aq) + Ba(OH) 2 (aq) BaCl 2 (aq) + H 2 O(l) Next, we need to find the volume (in L) of Ba(OH) 2 : ml Ba(OH) 1liter Ba(OH) = 2 2 x L Ba(OH) ml Ba(OH) 2 Finally, divide the moles of Ba(OH) 2 by the volume of Ba(OH) 2 : x 10 moles Ba(OH) Molarity [Ba(OH) = ] = M Ba(OH) L Ba(OH) 2 Acid-Base Titrations Volume of Reactant M reactant Moles of Reactant mole ratio Moles of Product M product Volume of Product Colligative Properties A property that does not depend on the identity of a solute in solution Vary only with the number of solute particles present in a specific quantity of solvent 4 colligative properties: Osmotic pressure Vapor pressure lowering Boiling point elevation Freezing point depression

18 Osmotic Pressure Osmosis A process in which solvent molecules diffuse through a barrier that does not allow the passage of solute particles The barrier is called a semipermeable membrane. A membrane that allows the passage of some substances but not others Osmotic Pressure Pressure that can be exerted on the solution to prevent osmosis Osmotic Pressure Figure Vapor Pressure Lowering Solutes come in 2 forms: Volatile Solutes that readily form a gas Nonvolatile Solutes that DO NOT readily form a gas Generally, the addition of a solute lowers the vapor pressure of a solution when compared to the pure solvent

19 Vapor Pressure Lowering Phase Diagram Boiling Point Elevation The addition of solute affects the boiling point because it affects the vapor pressure. The boiling point is raised with the addition of solute in comparison to the pure solvent. An equation that gives the increase in boiling point: T b = K b m Where T b is the increase in temperature from the pure solvent s boiling point, K b is the boiling point constant, which is characteristic of a particular solvent, and m is the molality (moles of solute per kg of solution)

20 Freezing Point Depression The freezing point is lowered with the addition of solute in comparison to the pure solvent. An equation that gives the decrease in freezing point: T f = K f m Where T f is the decrease in temperature from the pure solvent s freezing point, K f is the freezing point constant, which is characteristic of a particular solvent, and m is the molality (moles of solute per kg of solution) Practice Freezing Point Depression What is the change in freezing point for a 1.5 m solution of sucrose in water? Practice Solutions Freezing Point Depression What is the change in freezing point for a 1.5 m solution of sucrose in water? K f (water) = C/m T f = K f m T f = C * 1.5 m = C m

21 Colligative Properties and Strong Electrolytes Strong electrolytes dissociate most of the time into their constituent ions. Therefore, the number of particles (in this case ions) increases with the number of ions. Colligative properties are proportional to number of particles in solution. Example: MgCl 2 (s) Mg 2+ (aq) + 2 Cl - (aq) In this case the number of particles increases to 3 particles. Therefore, we would multiply the colligative property amount by 3. Practice Strong vs. Weak Electrolytes Which of the following aqueous solutions is expected to have the lowest freezing point? 0.5 m CH 3 CH 2 OH 0.5 m Ca(NO 3 ) m KBr Practice Solutions Strong vs. Weak Electrolytes Which of the following aqueous solutions is expected to have the lowest freezing point? 0.5 m CH 3 CH 2 OH 0.5 m Ca(NO 3 ) m KBr Because the molalities are the same, we only need to concentrate on the number of particles generated by each solute. CH 3 CH 2 OH will not dissociate into ions, Ca(NO 3 ) 2 dissociates into 3 ions, and KBr dissociates into 2 ions. Therefore, the largest number of particles is generated by Ca(NO 3 ) 2 and it will have the lowest freezing point

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