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If you are not part of the SOLUTION, you’re part of the PRECIPITATE! SOLUTIONS Chapter 15
SOLUTION OBJECTIVES Differentiate between colloids, solutions, and suspensions Define solvent and solute and be able to identify them in a solution. Describe the process of solvation, dissociation, and dissolving. Be sure to study polarity of compounds to be able to predict solubility. Differentiate between saturated, unsaturated, and supersaturated. State and discuss the factors affecting the rate of solubility. State and discuss the factors that affect solubility. Be able to read a solubility curve graph. Relate the enthalpy of solution to endothermic and exothermic dissolving processes. Differentiate between molarity and molality. Solve problems involving molarity, molality, mole fraction mass percent, volume percent, and making solutions.
THREE TYPES OF MIXTURES SOLUTION COLLOID SUSPENSION
DEMO SOLUTION: Salt and water COLLOID: Shaving gel SUSPENSION: Cornstarch in water TYNDALL EFFECT: particles dispersed in mixture are big enough to scatter light.
SOLUTIONS A solution is a homogeneous mixture of two or more substances in a single physical state. SOLVENT: the substance doing the dissolving and in the greater amount. SOLUTE: the substance being dissolved and in less abundant.
DEMO If I combine 50 m. L beads and 50. 0 m. L sand, what will the final volume be? If I combine 50. 0 m. L water (green) and 50. 0 m. L alcohol, what will the final volume be?
NINE BASIC TYPES OF SOLUTIONS Liquid Solutions: solid in liquid (salt water) liquid in liquid (vinegar) gas in liquid (carbonated drink) Gaseous Solutions: All gaseous mixtures are solutions solid in gas (soot in air) liquid in gas (humid air) gas in gas (air)
NINE BASIC TYPES OF SOLUTIONS Solid Solutions Homogeneous mixtures of solids are usually made from liquid solutions that have been mixed and then solidified (frozen) solid in solid (alloys – brass, bronze) liquid in solid (dental fillings) gas in solid (charcoal gas mask)
SOLUTION EQUILIBRIUM - A solution is in dynamic equilibrium when the number of solute particles returning to the crystal surface is equal to the number of solute particles leaving the crystal surface.
SOLUTION EQUILIBRIUM A SATURATED solution is a solution with the maximum amount of solute dissolved at a given temperature (Any more added goes to the bottom). It has reached dynamic equilibrium. An UNSATURATED solution has the ability to dissolve more solute at a given temperature If a solution is SUPERSATURATED, then a hot solution is saturated and cooled. An unstable condition results because the solution holds more solute that it normally does at a given temperature.
LASERDISC SOLUTIONS – Chapter 20 1. 2. 3. What are ion-dipole attractions? During the process of solvation, how do ion-dipole attractions compare to the ionic bonds inside an ionic crystal? Describe the processes that are balanced when solubility equilibrium has been reached. If more solute is added to a solution in equilibrium, will this always result in more dissolved solute particles in the solution? Explain. How does a solution become supersaturated? After crystals precipitate out of a supersaturated solution, would you still call it supersaturated? Explain.
THE DISSOLVING PRACTICE Dissolving occurs when the solute is pulled apart by the solvent. This takes place at the surface of the solute. The solvent surrounds the solute. This process of surrounding the solute is called SOLVATION. When the surrounding is done by water, this is called HYDRATION, a particular type of solvation. When ionic compounds separate into their ions in a solvent, DISSOCIATION occurs.
DISSOCIATION OF Na. Cl
Solute-Solvent Combinations 1. Polar Solvent – Polar Solute: The polar solvent is attracted to the polar solute. The solvent gradually surrounds the solute. The particles attach themselves due to polar attraction. Solvation occurs. Like dissolves like. Ex. Salt and water
Solute-Solvent Combinations 2. Polar Solvent – Nonpolar Solute: Polar solvent particles are attracted to each other and not the solute. Solvation does not occur and a solution is unlikely. Ex. oil and water DEMO: Marbles
Solute-Solvent Combinations 3. Nonpolar Solvent – Polar Solute: Nonpolar solvent particles have little attraction to the polar solute. Solvation does not occur and a solution is unlikely. Ex. Salt and oil
Solute-Solvent Combinations 4. Nonpolar solvent – nonpolar solute: Random motion of solute particles causes them to leave the surface of the solute and become evenly dispersed in the nonpolar solvent. Solvation occurs. Like dissolves like.
LASERDISC Polar and Nonpolar Solvents – Chapter 30 1. 2. 3. 4. Why are some liquids immiscible? Explain in terms of intermolecular forces? How would you predict whether carbon tetrachloride is a polar or nonpolar solvent? What evidence have you observed that supports your prediction? Based on chemical formula alone, can you tell whether iodine, I 2, is a polar or nonpolar solute? What evidence have you observed that supports your prediction? Based on the chemical formula alone, can you tell whether copper (II) chloride, Cu. Cl 2, is a polar or nonpolar solute? What evidence have you observed that supports your prediction?
DO NOW Pick up handout – due Tuesday Turn in Superaturation lab Book projects due today: William B. , Marwa J. , Quinn R. , Devon L. , Andrew C. , Madison B. , Peter D. , Seth T. , Conner R. , Kat M. , Tatum N. , Sam W. , Dan N. , Claire W. , and Jamal R.
SOLUBILITY Solubility is how much solute can dissolve in a given amount of solvent. It is measured in g/L or mol/L. It is usually the grams of solute per 100 g of solvent. A CONCENTRATED solution is said to have a high ratio of solute to solvent. A DILUTE solution is the opposite of this. For solids: solubility usually increases with increased temperature For gases: solubility usually decreases with increased temperature
SOLUBILITY How do you determine saturated, unsaturated, and supersaturated for a given substance?
FACTORS AFFECTING SOLUBILITY 1. Nature of the solvent and solute - “like dissolves like”. This means that polar solvent dissolves a polar solute and nonpolar solvent dissolves nonpolar solute. But polar does not dissolve nonpolar. In other words – will it even dissolve?
FACTORS AFFECTING SOLUBILITY 2. Temperature - increase the temperature and solubility increases (except gases).
FACTORS AFFECTING SOLUBILITY 3. Pressure - increase the pressure and you increase solubility (only with gases). Gases ONLY 4. How much is already dissolved – saturated versus unsaturated versus supersaturated…
Factors Affecting RATE of Solubility (How fast something will dissolve) Agitation (shaking, stirring) Increased temperature (except gases) Smaller Particle size Each allows more solvent to come in contact with the solute faster.
HEAT OF SOLUTION Energy is released and absorbed as substances dissolve. ENDOTHERMIC: The solute particles must separate and the solvent particles must separate to make room for the solute. EXOTHERMIC: When the solute and the solvent mix, the particles are attracted to each other. The overall change in energy is the heat of solution. Most solutions are endothermic.
CONCENTRATION Concentration is NOT dependent upon the sample size and it can be measured. Grams/100. 0 grams measures solubility Parts per million (ppm) measures small concentrations Parts per billion (ppb) measures pollutants Molarity used in lab chemistry Molality used for special calculations We will learn to calculate molarity, molality, mass percent, volume percent, and mole fraction
MOLARITY, M This is the ratio between the moles of dissolved substance and the volume of the solution expressed in liters. Molarity, M = moles of solute volume of solution in liters A one-molar (1 M) solution of HCl contains one mole of HCl in one liter of water. (Which means it contains 36. 46 g of HCl in 1 liter of water. )
SAMPLE PROBLEM Sandy dissolves 45. 0 g of Na. Cl in 2. 5 liters of solution. What is the concentration in molarity of Na. Cl? Mass = 45. 0 g Molar Mass = 58. 44 g/mol V = 2. 5 L Molarity = 45. 0 g Na. Cl 1 mol Na. Cl 2. 5 L 58. 44 g Na. Cl = 0. 31 M
PRACTICE What is the molarity of 58. 5 g of Na. Cl dissolved in 2. 0 L of solution?
MOLALITY, m This is concentration expressed in terms of moles of solute per kilogram of solvent. Volume is not a factor. Molality, m = moles solute Kg solvent A 1. 0 molal aqueous sugar soln: 1 mole sugar 1 kg water
SAMPLE PROBLEM Calculate the molality of 98. 0 g Rb. Br in 0. 824 Kg water. m = 98. 0 g Rb. Br 1 mol Rb. Br 0. 824 Kg H 2 O 165. 38 g Rb. Br = 0. 719 m
PRACTICE Calculate the molality of 85. 2 g Sn. Br 2 in 140. 0 g water.
MASS PERCENT Scientists frequently express the concentration of solutions in mass percent. Mass % = g of solute x 100 g of solute + g of solvent A 5% solution of Na. OH contains 5 g Na. OH in each 100 g of solution (95 g solvent and 5 g of solute).
VOLUME PERCENT Consumer products frequently express their concentration of solutions in volume percent. Volume % = L of solute x 100 L of solute + L of solvent A 5% solution of sodium hydroxide contains . 05 L Na. OH in each 1. 0 L of solution (0. 95 L solvent and 0. 05 L of solute).
MOLE FRACTION The concentration of solution can also be expressed in mole fractions. Mole Fraction = mole of solute + mole of solvent
PREPARING SOLUTIONS A 3 M solution of HCl is not bought but made from 12 M stock solutions. In addition, a 1. 0 M solution of Na. OH is made from a calculated amount of solid Na. OH added to water. It is important to know how to make different concentration of solutions.
PREPARING SOLUTIONS Prepare 1. 0 liter of a 1. 0 M aqueous solution of Na. OH M = mol 1. 0 M = mol L 1. 0 L mol = (1. 0 M)(1. 0 L) = 1. 0 mol Na. OH 40. 00 g Na. OH = 40. 00 g 1 mol Na. OH So, you will put 40. 00 g Na. OH in a flask, add 1. 0 L water, and mix well.
Preparing Solutions Prepare 1. 0 L of a 6. 0 M aqueous solution of KCl. The molar mass of KCl is 74. 55 g/mol.
DILUTING SOLUTIONS To dilute a solution, you can form a ratio between molarity and volume. M 1 V 1 = M 2 V 2
DILUTING SOLUTIONS Diluting a 12. 0 M solution of HCl to 6. 0 M HCl What volume would you use to make 0. 500 L of 6. 0 M HCl solution? V 1 = M 2 V 2 = (6. 0 M HCl)(0. 500 L HCl) M 1 (12. 0 M) = 0. 250 L HCl Thus you would need 0. 250 L HCl and 0. 250 L water to make the solution.
PRACTICE Preparing a 0. 1 M solution of HCl What volume would you use of 12. 0 M HCl to make 1. 0 L of 0. 1 M HCl?
ANSWER Preparing a 0. 1 M solution of HCl M 1 = 12. 0 M HCl V 1 = ? M 2 = 0. 1 M HCl V 2 = 1. 0 L V 1 = M 2 V 2 M 1 V 1 = (0. 1 MHCl)(1. 0 L) (12. 0 M HCl) = 0. 0083 L M 1 V 1 = M 2 V 2 Add 8. 3 m. L 12. 0 M HCl to 991. 7 m. L water
COLLIGATIVE PROPERTIES “colligative” – depending upon the collection These are properties that depend on the concentration (the number of) of the solute particles. Three Factors: Vapor pressure reduction Boiling Point Elevation Freezing Point Depression
LASERDISC Colligative Properties (chapter 32) 1. Support one of the following hypotheses about why it helps to add salt to the water before cooking the pasta: a) The salt brings the water to a boil sooner; or b) The salt brings the water to a boil at a higher temperature. 2. Which antifreeze solution boils at the highest temperature? Which solution would you want in your car on a hot summer day? Explain your reasoning. 3. On cold winter days, unprotected radiator water can freeze and expand, which can ruin an engine by cracking the engine block. Which antifreeze mixture would you want in your car on a cold winter day?
Vapor Pressure: A Review
VAPOR PRESSURE REDUCTION
VAPOR PRESSURE REDUCTION
VAPOR PRESSURE REDUCTION The vapor pressure of a solvent containing a non-volatile solute is LOWER than the vapor pressure of the pure solvent. WHY: The solute takes up space at the surface of the liquid and prevents some solvent molecules from leaving the liquid.
RAOULT’S LAW The vapor pressure of a solution of a non-volatile solute is equal to the vapor pressure of the pure solvent at that temperature multiplied by its mole fraction.
RAOULT’S LAW Po is the vapor pressure of the pure solvent at a particular temperature. xsolv is the mole fraction of the solvent. So if the solution is 20% solute and 80% solvent, the mole fraction for the solvent is 0. 8. So the Vpsoln = 0. 8 vpsolvent
RAOULT’S LAW Because changes in state depend upon vapor pressure, the presence of a solute also affects the freezing point and the boiling point of a solvent. **It depends on the number and not the identity of the solute particles
BOILING POINT REVIEW The boiling point of a substance is the temperature at which the vapor pressure of a liquid is equal to the external pressure on its surface (usually atmospheric pressure).
BOILING POINT ELEVATION An addition of solute lowers that vapor pressure, therefore a higher temperature is necessary to get the vapor pressure up to the external (atmospheric) pressure so that the solution will boil. Tb is the difference between the boiling point of the solution and the boiling point of the pure solvent. It is directly proportional to the number of solute particles per mole of solvent particles.
BOILING POINT ELEVATION Dissociation factor molality Tb = (Kb)(df)(m) constant (molal BP elevation constant) The dissociation factor, df, is how many particles the solute breaks up into. For a nonelectrolyte, the df is 1 Water’s “Kb” value is 0. 515 C/m
FREEZING POINT: A REVIEW The freezing point of a substance is the temperature at which the vapor pressure of the solid and liquid phases are the same.
FREEZING POINT DEPRESSION Club Soda demo Animation for Freezing Point depression
FREEZING POINT DEPRESSION When a dissolved solute lowers the freezing point of its solution, you have freezing point depression. Tf is directly proportional to the molality of the solution. Tfp = (Kfp)(df) (m) constant (molal fp depression constant) Water’s “Kfp” value is 1. 853 C/m
CHEMICALS USED TO MELT ICE Formula Lowest Temp Pros Cons (NH 4)2 SO 4 -7. 0°C fertilizer Damages concrete Ca. Cl 2 -29. 0°C Melts ice faster than Na. Cl Attracts moisture; surfaces can be slippery Mg. Cl 2 -15. 0°C Melts ice faster than Na. Cl Attracts moisture CH 3 COOK -9. 0°C biodegradable corrosive KCl -7. 0°C fertilizer Damages concrete Na. Cl -9. 0°C Keeps sidewalks dry Corrosive; damages concrete and vegetation CMA -9. 0°C Safest for concrete and vegetation Works better at preventing re-icing
CALCULATIONS: NON-IONIC Determine the freezing and boiling point for 85. 0 g of sugar (C 12 H 22011) in 392 g water. Need these things: 1. Molality, m 2. number of particles (should be one) 3. Change in temperature,
CALCULATIONS: NON-IONIC Molality 392 g H 2 O 1 kg = 0. 392 Kg 1000 g m = 85. 0 g C 12 H 22011 1 mol C 12 H 22011 0. 392 Kg H 2 O 342. 34 g C 12 H 22011 = 0. 633 m 1. 2. df = 1 (non-ionic solute)
CALCULATIONS: NON-IONIC Change in Freezing point temperature: Tfp = (Kfp)(df) (m) = 1. 853 C 1 0. 633 m m = 1. 17 C New Freezing Point: 0. 00 C – 1. 17 C = -1. 17 C Freezing point of pure water
CALCULATIONS: NON-IONIC Change in boiling point temperature: Tbp = (Kbp)(df) (m) = 0. 515 C 1 0. 633 m m = 0. 326 C New Boiling Point: 100. 000 C + 0. 326 C = 100. 326 C Boiling point of pure water
PRACTICE What is the freezing point for a solution of 210. 0 g of glycerol dissolved in 350. 0 g of water? (The molecular mass of glycerol is 92. 11 g/mol) FP = ? Solute = 210. 0 g glycerol Solvent = 350. 0 g H 2 O
PRACTICE What is the freezing point for a solution of 210. 0 g of glycerol dissolved in 350. 0 g of water? (The molecular mass of glycerol is 92. 11 g/mol) 1. m = 210. 0 g gly 1 mole gly 1000 g 350. 0 g H 2 O 92. 11 g gly 1 kg = 6. 51395 = 6. 514 m 2. Df = 1 3. Tf = (Kf)(df)(m) = (1. 853ºC/m)(1)(6. 514 m) = 12. 070 = 12. 07ºC New FP = 0. 000ºC – 12. 07ºC = -12. 07ºC
CALCULATIONS: IONIC Determine the freezing and boiling point for 21. 6 g Ni. SO 4 in 100. 0 g water. Need these things: 1. molality 2. number of ions 3. change in temperature
CALCULATIONS: IONIC Calculate molality: 100. 0 g H 2 O 1 kg = 0. 1000 Kg 1000 g m = 21. 6 g Ni. SO 4 1 mol Ni. SO 4 0. 1000 Kg H 2 O 154. 76 g Ni. SO 4 1. = 1. 396 = 1. 40 m
CALCULATIONS: IONIC 2. Determine the number of ions: Substance is Ni. SO 4 There are two ions, Ni+2 and SO 4 -2, so the df is 2
CALCULATIONS: IONIC 3. Calculate the change in temperature and the new freezing point. Tfp = 1. 853 C 2 1. 40 m 0. 00 C – 5. 19 C = -5. 19 C = 5. 19 C
CALCULATIONS: IONIC Tbp = 0. 515 C 2 1. 40 m m = 1. 44 C 100. 00 C + 1. 44 C = 101. 44 C
CALCULATIONS: IONIC What is the boiling point of a solution containing 34. 3 g of Mg(NO 3)2 dissolved in 0. 107 kg of water? (The formula mass of magnesium nitrate is 148. 32 g/mol. ) Bp = ? Solute 34. 3 g Mg(NO 3)2 Solvent 0. 107 kg
CALCULATIONS: IONIC 1. m = 34. 3 g Mg(NO 3)2 1 mole Mg(NO 3)2 0. 107 kg 148. 32 g Mg(NO 3)2 = 2. 16127796 = 2. 16 m 2. Df = 3 Mg+2 NO 3 -1 3. Tb = (Kb)(df)(m) = (0. 515ºC/m)(3)(2. 16 m) = 3. 3372 = 3. 34ºC New BP = 100. 0000ºC + 3. 34ºC = 103. 34ºC
Do Now Pick up handout – due Monday as a homework check Ice Cream Supplies due NEXT Tuesday Book Projects due Thursday: Hayden R. , Matt H. , Bob Wiley, Sean R. , Sam K. , Cameron H. , Joe F. , Marshall C. , Josh M. , Kainath M. , Grace M. , Sarah. , Adiba K. , Alexander G. , and Stilian I.
RECAP A solution reduces the vapor pressure of the pure solvent. Vapor pressure reduction results in freezing point depression (lowering) and boiling point elevation. To solve these problems, you need: Molality Dissociation factor (Number of ions) Molal BP/FP constant (given to you)
RECAP Calculating freezing point change: Tfp = (Kfp)(df)(m) Calculating boiling point change: Tb = (Kb)(df)(m) Final step is to subtract from true freezing point or add to true boiling point
DETERMINING MOLAR MASS Colligative properties provide a useful means to experimentally determine the molar mass (molecular mass in one mole) of an unknown substance. Steps: 1. Solve for molality. 2. Solve for moles of solute. 3. Solve for molar mass This is basically going backwards
DETERMINING MOLAR MASS A 10. 0 g sample of an unknown organic compound is dissolved in 0. 100 kg water. The boiling point of the solution is elevated to 0. 433 o. C above the normal boiling point of water. What is the molar mass? solute = 10. 0 g Solvent = 0. 100 kg water Tbp = 0. 433 o. C df = 1
DETERMINING MOLAR MASS Step One: determine molality Tbp = Kbp df m m = Tbp = 0. 433 o. C = 0. 841 m (mol/kg) Kbp 0. 515 o. C/m Assume df = 1 because it is an unknown ORGANIC compound (not ionic)
DETERMINING MOLAR MASS Step Two: Detemine moles of solute m = mol solute Kg solvent mol solute = m x kg solvent = 0. 841 mol/kg x 0. 100 kg = 0. 0841 mol
DETERMINING MOLAR MASS Step Three: Determine Molar Mass mol solute = mass solute molar mass solute molar mass = mass solute = 10. 0 g mol solute 0. 0841 mol = 118. 90606 = 119 g/mol
DETERMINING MOLAR MASS PRACTICE: A solution of 3. 39 g of an unknown in 10. 00 g of water has a freezing point of – 7. 31 o. C. The solution is a nonelectrolyte. What is the molar mass of the compound? (The molal freezing point constant of water is 1. 853 C/m. ) Mass solute = 3. 39 g Mass solvent = 10. 00 g FP = – 7. 31 o. C ΔTfp = 7. 31 o. C df = 1
DETERMINING MOLAR MASS PRACTICE: A solution of 3. 39 g of an unknown in 10. 00 g of water has a freezing point of – 7. 31 o. C. The solution is a nonelectrolyte. What is the molar mass of the compound? (The molal freezing point constant of water is 1. 853 C/m. ) Step One: determine molality Tfp = Kfp df m m = Tfp = 7. 31 o. C = 3. 94 m Kbp df (1. 853 o. C/m)(1)
DETERMINING MOLAR MASS PRACTICE: A solution of 3. 39 g of an unknown in 10. 00 g of water has a freezing point of – 7. 31 o. C. The solution is a nonelectrolyte. What is the molar mass of the compound? (The molal freezing point constant of water is 1. 853 C/m. ) Step Two: Determine moles of solute m = mol solute Kg solvent mol solute = m x kg solvent = 3. 44 m x 0. 01000 kg = 0. 0394 mol
DETERMINING MOLAR MASS PRACTICE: A solution of 3. 39 g of an unknown in 10. 00 g of water has a freezing point of – 7. 31 o. C. The solution is a nonelectrolyte. What is the molar mass of the compound? (The molal freezing point constant of water is 1. 853 C/m. ) Step Three: Determine Molar Mass mol solute = mass solute molar mass solute molar mass = mass solute = 3. 39 g mol solute 0. 0394 mol = 86. 040609 = 86. 0 g/mol
SEIZURE - - Hiorns began by reinforcing the walls and ceiling, and tanking the flat (apartment)with plastic sheeting. Then 70 -80, 000 liters of hot copper sulfate solution was pumped in through a hole in the ceiling from the flat above. Weeks went by, until the temperature of the solution dropped, and the crystals began to precipitate. Finally, any remaining liquid was pumped back out, to be recycled by the chemical industry.