CHAPTER 13 SOLUTIONS CHAPTER 13 Solution Topics • • • • • • Solution review Solution forming process Solution concentration definitions Solution Concentration calculations Solution stoichiometry Colligative Properties 13.2 SOLUTION REVIEW Solutions are homogenous mixtures. They consist of a larger component called the solvent and one or more smaller components called the solutes. Can be in the solid, liquid, or gaseous state. 13.2 Solution Examples •Margarine •Tap Water •Steel •18 Carat Gold •Air •Sterling Silver 13.2 Solution Examples Composition of air: Dry air contains roughly (by volume) 78% nitrogen, 21% oxygen, 0.93% argon, 0.038% carbon dioxide, and small amounts of other gas Air also contains a variable amount of water vapor, on average around 1% What is the solvent in air ? 13.2 Solution Examples Composition of air: Dry air contains roughly (by volume) 78% nitrogen, 21% oxygen, 0.93% argon, 0.038% carbon dioxide, and small amounts of other gases. Air also contains a variable amount of water vapor, on average around 1% What is the solvent in air ? Nitrogen, N2 13.2 Solution Examples Composition of air: Dry air contains roughly (by volume) 78% nitrogen, 21% oxygen, 0.93% argon, 0.038% carbon dioxide, and small amounts of other gases. Air also contains a variable amount of water vapor, on average around 1% What is the solvent in air ? Nitrogen, N2 What is a solute in air? 13.2 Solution Examples Composition of air: Dry air contains roughly (by volume) 78% nitrogen, 21% oxygen, 0.93% argon, 0.038% carbon dioxide, and small amounts of other gases. Air also contains a variable amount of water vapor, on average around 1% What is the solvent in air ? Nitrogen, N2 What is a solute in air? Oxygen, O2 13.2 Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ? 13.2 Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ? Gold 13.2 Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ? Gold What are the solutes? 13.2 Solution Examples Composition of 18 carat gold: 75% gold, 12.5% silver, 12.5% copper. What is the solvent in 18 ct gold ? Gold What are the solutes? Silver and Copper 13.2 Solution Examples Examples of solutions include: Salt water: what is the solvent in salt water ? 13.2 Solution Examples Examples of solutions include: Salt water: what is the solvent in salt water ? Water, H2O 13.2 Solution Examples Examples of solutions include: Salt water: what is the solvent in salt water ? Water, H2O What is a solute in sea water? 13.2 Solution Examples Examples of solutions include: Salt water: what is the solvent in salt water ? Water, H2O What is a solute in sea water? NaCl (salt) 13.2 Solution Properties Some general properties of solutions include: Solutions may be formed between solids, liquids or gases. They are homogenous in composition They do not settle under gravity They do not scatter light (Called the Tyndall Effect) Solute particles are too small to scatter light and therefore light will go right through a solution like is shown on the next slide. 13.3 Solution Properties Tyndall Effect Laser light reflected by a colloid. In a solution you would not see any red light. 13.3 Solution Properties Soluble substances are those that can dissolve in a given solvent. Insoluble or immiscible substances are those that cannot dissolve in a given solvent. Which of the following are soluble in water? NaCl, sugar, cooking oil, alcohol, gasoline, motor oil 13.3 Solution Properties Soluble substances are those that can dissolve in a given solvent. Insoluble or immiscible substances are those that cannot dissolve in a given solvent. Which of the following are soluble in water? NaCl, sugar, cooking oil, alcohol, gasoline, motor oil 13.3 Solution Properties Soluble substances are those that can dissolve in a given solvent. Insoluble or immiscible substances are those that cannot dissolve in a given solvent. Which of the following are soluble in water? NaCl, sugar, cooking oil, alcohol, gasoline, motor oil Which of the following are immiscible in cooking oil? NaCl, sugar, alcohol, gasoline, motor oil, water 13.3 Solution Properties Soluble substances are those that can dissolve in a given solvent. Insoluble or immiscible substances are those that cannot dissolve in a given solvent. Which of the following are soluble in water? NaCl, sugar, cooking oil, alcohol, gasoline, motor oil Which of the following are immiscible in cooking oil? NaCl, sugar, alcohol, gasoline, motor oil, water 13.3 Solution Properties The maximum amount of a given solute a solvent can dissolve is called the solubility. The solubility is dependent on the temperature and pressure. Solubility is often expressed in terms of grams of solute per 100 g of solvent but may have other units. When a solvent contains the minimum amount of a solute possible the solutions is said to be unsaturated. When a solvent contains the maximum amount of a solute possible the solutions is said to be saturated. When a solvent contains more than the maximum amount of a solute possible the solutions is said to be supersaturated. 13.3 Solution Properties Solutions form when a soluble solute(s) is dissolved in a solvent. In biological systems aqueous (solutions where water is the solvent) are of particular importance. The solubility of most liquids and solids in water increases with temperature. The effect of pressure on the solubility of liquid or solid solutes in water is negligible. 13.3 Solution Properties Solubility Curves 13.3 Solution Properties By forming a solution at a high temperature then slowly cooling it we can form supersaturated solutions that contain more solute than in a saturated solution. These kinds of solutions are very unstable and tend to separate out the excess solute with the slightest disturbance. http://www.youtube.com/watch?v=uy6eKm8IRdI&NR=1 http://www.youtube.com/watch?v=aC-KOYQsIvU&feature=related 13.3 Solution Properties The solubility of gases in water decreases with temperature. Are cold carbonated drinks bubblier than warm carbonated drinks? The solubility of many gases in water is directly proportional to the pressure being applied to the solution. i.e. double the pressure, double the solubility What happens when the cork is removed from a bottle of champagne? What is the origin of decompression sickness? Anyone heard of hyperbaric therapy? 13.3 Solution Properties When we place an ionic solid in water there will be attractive forces between the ions at the surface of the crystal and the water molecules. These attractive forces are called ion-dipole forces. Water molecules orient such that the positive end of the molecule is oriented towards the negative ions at the surface and vice versa. 13.3 Solution Properties How do solutions form? Why do some substances leave one phase and enter the solution and others don’t? How can we use chemistry to predict solubility's? Lets first look at the formation of a solution between an ionic solute and a polar solvent such as H2O. 13.3 Solution Properties Ionic compounds are composed of oppositely charged ions arranged in a repeating 3-d arrangement. They are held together by attractive forces between oppositely charged ions. 13.3 Solution Properties Ionic compounds are composed of oppositely charged ions arranged in a repeating 3-d arrangement. They are held together by attractive forces between oppositely charged ions Why is chloride ion larger than sodium ion? 13.3 Solution Properties red is the region where electrons are found most often and blue is where electrons are rarely found 13.3 Solution Properties If the attractive force between the surface ion and the solvent is greater than the forces between the ion and the solid then the ion will enter the solution phase. The ion that has left the solid H2O and becomes completed surrounded by water molecules. It has become solvated or hydrated. Strong K+ Ion dipole forces hold the ions to water. 13.3 Solution Properties Note the different orientation of water molecules around the oppositely charged ions. Positive pole of water directed to the negative ions and the negative pole directed to the positive ions 13.3 Solution Properties In a solution of an ionic compound a solvated ion will occasionally collide with the surface of the solid. Sometimes when this happens the ion will “stick” to the surface and become part of the solid phase again. This will happen more frequently the more concentrated the solution is. 13.3 Solution Properties When the rate of ions leaving the solid equals the rate of ions going back to the solid the system is at equilibrium and the solution is saturated. When a solution is at equilibrium with its solute macroscopically there will be no change occurring. However, at the molecular level lots is happening, just in equal and opposite directions. 13.3 Solution Properties Supersaturated solutions can form because there are no sites for solute ions to collide with. When we place a “seed” crystal in a supersaturated solution this provides the needed sites and the excess solute crystallizes very quickly. 13.3 Solution Properties In the you tube video we watched you can just see the tiny seed crystals on the persons finger. 13.3 Solution Properties Polar but non-ionic solutes dissolve in water via a similar mechanism as for ionic compounds. 13.3 Solution Properties A solute will be insoluble in a solvent if: 1. Forces between solute particles are greater than the forces between solute particles and the solvent. 13.3 Solution Properties A solute will be insoluble in a solvent if: Forces between the solvent particles and solute particles are stronger than forces between the solvent and the solute. e.g. The only attractive force between oil and water will is dispersion forces. These are weak compared to hydrogen bonds between water molecules. 13.3 Solution Properties In a polar solvent there will be attraction between the oppositely charged ends of the molecule. Hydrogen bonds are represented by dotted lines between the water molecules. A hydrogen bond is and intermolecular force between hydrogen of one molecule and O, N, or F of another molecule. Hydrogen must be directly attached to O, N, or F in at least one of the two hydrogen bonded molecules. 13.3 Solution Properties A good “rule of thumb” that works especially well for nonionic compounds is: “Like dissolves like” i.e. Polar solvents dissolve polar solutes well and non-polar solvents dissolve non-polar solutes well. 13.3 Solution Properties The rate of dissolution is dependent upon: 1. The surface area of the solute. i.e. how finely divided it is. Increasing rate 13.3 Solution Properties 2. How hot the solution is. i.e. the kinetic energy of solute and solvent. 3. The rate of stirring. Typically when we are preparing a solution in the lab we will both heat and stir. 13.3 Solution Properties When a solute dissolves in a solvent heat can be released or absorbed. When heat is absorbed the process is endothermic and the solution becomes cooler. This effect is used in instant cold packs for sporting injuries and first aid. 13.3 Solution Properties Endothermic Solution Solvent temperature 22.2° Solvent temperature 11.3° 13.3 Solution Properties Exothermic Solution More commonly dissolution is an exothermic process and heat is released when a solute is dissolved. Sometimes when we make a solution it will get so hot it boils!! 13.5 SOLUTION CONCENTRATION The ratio of the amount of solute to amount of solution, or solvent is defined by the concentration. Concentration = solute solution = solute solvent There are various combinations of units that are used in these rations. Ratio X 102 X 103 X 106 X 109 g solute g solution = % (w/w) g solute = % (w/v) mL solution mL solute mL solution = % (v/v) ppt (w/w) ppm (w/w) ppb (w/w) ppt (w/v) ppm (w/v) ppb (w/v) ppt (v/v) ppm (v/v) ppb (v/v) 13.5 SOLUTION CONCENTRATION 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 25.2 g NaCl 1. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 13.5 SOLUTION CONCENTRATION 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 25.2 g NaCl 33.6g H2O + 25.2 g NaCl 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 13.5 SOLUTION CONCENTRATION 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 25.2 g NaCl 100 33.6g H2O + 25.2 g NaCl 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 13.5 SOLUTION CONCENTRATION 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 25.2 g NaCl 100 33.6g H2O + 25.2 g NaCl = 43.0 % NaCl 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 13.5 SOLUTION CONCENTRATION 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 25.2 g NaCl 100 33.6g H2O + 25.2 g NaCl = 43.0 % NaCl 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 44.6 g NaCl 100 g solution 13.5 SOLUTION CONCENTRATION 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 25.2 g NaCl 100 33.6g H2O + 25.2 g NaCl = 43.0 % NaCl 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 44.6 g NaCl 333 g solution 100 g solution 13.5 SOLUTION CONCENTRATION 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 25.2 g NaCl 100 33.6g H2O + 25.2 g NaCl = 43.0 % NaCl 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 44.6 g NaCl 333 g solution 100 g solution 13.5 SOLUTION CONCENTRATION 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 25.2 g NaCl 100 33.6g H2O + 25.2 g NaCl = 43.0 % NaCl 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution 13.5 SOLUTION CONCENTRATION 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 25.2 g NaCl 100 33.6g H2O + 25.2 g NaCl = 43.0 % NaCl 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution Mass of water? 13.5 SOLUTION CONCENTRATION 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 25.2 g NaCl 100 33.6g H2O + 25.2 g NaCl = 43.0 % NaCl 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution Mass of water? 333 g solution – 149 g NaCl = 184 g H2O 13.5 SOLUTION CONCENTRATION 1. Find the % (w/w) when 25.2 g NaCl is combined with 33.6g H2O. 25.2 g NaCl 100 33.6g H2O + 25.2 g NaCl = 43.0 % NaCl 2. Find the mass of water and salt required to make 333 g of a 44.6 % (w/w) solution. 44.6 g NaCl 333 g solution = 149 g NaCl 100 g solution Mass of water? 333 g solution – 149 g NaCl = 184 g H2O 13.5 SOLUTION CONCENTRATION 3. How many grams of NaCl are required to dissolve in 88.2 g of water to make a 29.2% (w/w) solution. 4. A sugar solution is 35.2%(w/v) find the mass of sugar contained in a 432 mL sample of this sugar solution. 13.5 SOLUTION CONCENTRATION The solution concentration can also be defined using moles. The most common example is molarity (M). The molarity of a solution is defined as: “The number of moles of solute in 1 L of solution” and is given the formula: Molarity (M) = Moles solute Liters solution 13.5 SOLUTION CONCENTRATION 1. A student dissolves 25.8 g of NaCl in a 250 mL volumetric flask. Calculate the molarity of this solution. (picture of volumetric flask is on the next slide) 2. Find the mass of HCl required to form 2.00 L of a 0.500 M solution of HCl. 3. A student evaporates the water form a 333 mL sample of a 0.136 M solution of NaCl. What mass of salt remains? 4. Find the molarity of sodium ions in a solution containing 2.35g of in a one liter volumetric flask. 13.5 SOLUTION CONCENTRATION In the lab we would use a piece of glassware called a volumetric flask to prepare this solution. 13.5 SOLUTION CONCENTRATION 13.5 SOLUTION CONCENTRATION 13.7 SOLUTION CONCENTRATION Often we will want to make a dilute solution from a more concentrated one. To determine how to do this we use the formula : C1V1 = C2V2 Where: C1 = concentration of more concentrated solution V1 = volume required of more concentrated solution C2 = concentration of more dilute solution V2 = volume of more dilute solution We can use any units in this equation but they must be the same on both sides. 13.7 SOLUTION CONCENTRATION How would one prepare 50.0 mL of a 3.00 M solution of NaOH using a 7.10 M stock solution? C1V1 = C2V2 (7.10 M)V1 = (3.00 M) (50.0 mL) (7.10 M)V1 = (3.00 M) (50.0 mL) (7.10 M) (7.10 M) V1 = 21.1 mL This means that you add 21.1 mL of the concentrated stock solution to a 50.0 mL volumetric flask and add water until the bottom of the meniscus touches the line on the volumetric flask. 13.8 SOLUTION STOICHIOMETRY Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) 1. Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 13.8 SOLUTION STOICHIOMETRY Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) 1. Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution 13.8 SOLUTION STOICHIOMETRY Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) 1. Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution L solution 103 mL 13.8 SOLUTION STOICHIOMETRY Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) 1. Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution moles AgCl 215.35 g AgCl 33.2 mL moles AgNO3 moles AgCl L solution 103 mL 13.8 SOLUTION STOICHIOMETRY Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) 1. Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution moles AgCl 215.35 g AgCl 33.2 mL moles AgNO3 moles AgCl L solution 103 mL 13.8 SOLUTION STOICHIOMETRY Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) 1. Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution moles AgCl 215.35 g AgCl 33.2 mL moles AgNO3 moles AgCl L solution 103 mL 13.8 SOLUTION STOICHIOMETRY Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) 1. Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 0.100 moles AgNO3 L solution moles AgCl 215.35 g AgCl 33.2 mL moles AgNO3 moles AgCl L solution 103 mL = 0.715 g AgCl 13.8 SOLUTION STOICHIOMETRY Consider the following balanced equation: CaCl2 (aq) + AgNO3 (aq) → AgCl (s) + Ca(NO3)2 (aq) 1. Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and an excess of calcium chloride. 2. Find the mass of silver chloride formed from 33.2 mL of a 0.100 M solution of silver nitrate and 200.0 mL of a 0.200 M solution of calcium chloride solution. 3. Find the volume of the excess reactant. 13.9 COLLIGATIVE PROPERTIES Colligative properties are physical properties that depend on the number of particles, but not the nature of the particles. We will discuss three colligative properties: Boiling point elevation Freezing point depression Osmotic pressure 13.9 COLLIGATIVE PROPERTIES The vapor pressure above a solution is lower than that above the pure solvent. This has some interesting effects: • The boiling point of a solution is higher than the pure solvent (boiling point elevation). • The freezing point of a solution is lower than the pure solvent (freezing point depression). Molality Molality is another frequently used solution concentration, specifically for colligative properties. moles solute Molality (m) = Kg solvent 13.9 COLLIGATIVE PROPERTIES The change in boiling point and freezing point can be calculated using the following constants: Boiling Point Elevation = (i) 0.512 °C m °C Freezing Point Depression = - (i) 1.86° m (i) = moles of ions found in an ionic solid M = molality = moles of solute kg solvent 13.9 COLLIGATIVE PROPERTIES Lets do an example: What is the boiling point and freezing of a 0.1 m solution of NaCl? NaCl (s) → Na+ (aq) + Cl- (aq) n = 2 ions Freezing point: -1.86 °C m 2 0.1 m 13.9 COLLIGATIVE PROPERTIES Lets do an example: What is the boiling point and freezing of a 0.10 m solution of NaCl? NaCl (s) → Na+ (aq) + Cl- (aq) n = 2 ions Freezing point: -1.86 °C m 2 0.10 m = - 0.37 °C ( Amount Lowered) Boiling point: 0.512 °C m 2 0.10 m Lets do an example: What is the boiling point and freezing of a 0.10 M solution of NaCl? NaCl (s) → Na+ (aq) + Cl- (aq) n = 2 ions Freezing point: -1.86 °C m 2 0.10 m = - 0.37 °C ( Amount Lowered) F.P. = O °C + ( - 0.37 °C) = - 0.37 °C Boiling point:: 0.512 °C m 2 0.10 m = 0.102 °C B.P. = 10O °C + 0.102 °C) = 100.102 °C Lets do an example: What is the boiling point and freezing of a 0.10 M solution of NaCl? NaCl (s) → Na+ (aq) + Cl- (aq) i = 2 ions Freezing point: -1.86 °C m 2 0.10 m = - 0.37 °C ( Amount Lowered) F.P. = O °C + ( - 0.37 °C) = - 0.37 °C Boiling point:: 0.512 °C m 2 0.10 m = 0.102 °C B.P. = 10O °C + 0.102 °C) = 100.102 °C Osmotic Pressure Sailors adrift at sea, know that drinking sea water will cause dehydration, do you? Maybe a better question is why? The answer involves another colligative property called osmosis. Osmotic Pressure Osmosis being a colligative property is dependent upon particles passing through a semipermeable membrane. A semipermeable membrane will let only smaller particles pass through. A chain link fence is and example of a semipermeable membrane. The fence will block baseballs from passing through, but not air. The roots of a tree contain a semipermeable membrane, they allow water to flow into a tree, but not salt or sugar. On the next diagram two different solutions are separated by a semipermeable membrane. Osmotic Pressure mm The compartment on the right has more water molecules compared to the one on the left, thus water flows from the right to the left in increasing the height of the column on the left and decreasing the height on the right. Measuring the difference in heights gives the osmotic pressure of the solution. Calculation of osmotic pressure in atm. can be done using the ideal gas constant: 0.08206 L-atm/mole-K, notice L/mole in the constant. Osmotic Pressure Osmotic pressure coupled with capillary action is responsible for moving water from the ground up to the top of a tree. Osmotic pressure also affects our blood cells. Isotonic solution: When the solution concentration outside the cell is the same as inside. Here the migration of water into and out of the cell are equal. Thus the cell remains in its normal state and works normally. Hypotonic solution: When the solution concentration outside the cell is lower than inside the cell. In this state more water flows into the cell than out of the cell. Overall result is the cell expands and could rupture. Called hemolysis. Hypertonic solution: When the solution concentration outside the cell is greater than inside the cell. In this state more water flows out of the cell than into the cell, causing the cell to shrink. This is called crenation. Osmotic Pressure Crenation Hemolysis Now we can explain about drinking sea water. Since the solute concentration is greater in the stomach (less water molecules), then water flows from the cells to the stomach, thus causing dehydration. The End Ch#13 Solutions