Solution homogeneous mixture Solute substance being dissolved Solvent present in greater amount
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1 Pre-AP Chemistry 1
2 Solution homogeneous mixture Solute substance being dissolved Solvent present in greater amount 2
3 Make-up Laundry detergent Motor oil Gasoline Food preservatives Deodorant Lawn fertilizers & weed killers Shampoo Air fresheners Furniture polish Toothpaste and mouthwash Oven cleaner Glass cleaner etc 3
4 Alloy a solid solution of metals Bronze = Cu + Sn Brass = Cu + Zn Soluble will dissolve in Insoluble will not dissolve in Miscible refers to two gases or two liquids that form a solution More specific than soluble e.g. food coloring and water Immiscible refers to two gases or liquids that will not form a solution Suspension appears uniform while being stirred, but settles over time 4
5 Solute Solvent Solution Gaseous Solutions gas Liquid gas gas air (nitrogen, oxygen, argon gases) humid air (water vapor in air) Liquid Solutions gas liquid solid liquid liquid liquid carbonated drinks (CO 2 in water) vinegar (CH 3 COOH in water) salt water (NaCl in water) Solid Solutions liquid solid solid solid dental amalgam (Hg in Ag) sterling silver (Cu in Ag) 5
6 Solute a substance in a smaller amount dissolved in a larger amount of another substance (the solvent) Concentration the number of moles present in a certain volume of solution Expressed as the amount of solute dissolved in a given amount of solution. An intensive quantity Molarity (M) expresses the concentration in units of moles of solute per liter of solution Can be used as a conversion factor between volume of solution and amount (mol) of solute M mol solute L solution 6
7 A. mass % = mass of solute mass of solution B. parts per million (ppm) (also, ppb and ppt) commonly used for minerals or contaminants in water supplies C. molarity (M) = moles of solute L of solution used most often in this class D. molality (m) = moles of solute kg of solvent 7
8 Concentration = # of moles volume (L) V = 250 ml n = 8 moles [ ] = 32 molar V = 1000 ml n = 2 moles [ ] = 2 molar V = 1000 ml n = 4 moles [ ] = 4 molar V = 5000 ml n = 20 moles [ ] = 4 molar 8
9 LITERS OF GAS AT STP Molar Volume (22.4 L/mol) MASS IN GRAMS Molar Mass (g/mol) MOLES particles/mol NUMBER OF PARTICLES Molarity (mol/l) LITERS OF SOLUTION 9
10 1. Hydrobromic acid (HBr) is a solution of hydrogen bromide gas in water. Calculate the molarity of hydrobromic acid solution if L contains 1.80 mol of hydrogen bromide. 2. How many moles of KI are in 84 ml of 0.50 M KI? 10
11 3. How many grams of solute are in 1.75 L of M sodium hydrogen phosphate (Na 2 HPO 4 )? 4. How many liters of 3.30 M sucrose (C 12 H 22 O 11 ) contain 135 g of solute? 11
12 5. You have 10.8 g potassium nitrate. How many ml of solution will make this a 0.14 M solution? 12
13 1. Given the reaction Pb(NO 3 ) 2 (aq) + KI (aq) PbI 2 (s) + KNO 3 (aq), what volume of 4.0 M KI solution is required to yield 89 g PbI 2? 2. How many ml of a M CuSO 4 solution will react w/excess Al to produce 11.0 g Cu? Aluminum sulfate is also produced. 13
14 3. Given the unbalanced reaction Cu + AgNO 3 Ag + Cu(NO 3 ) 2, how many grams of Cu are required to react with 1.5 L of 0.10M AgNO 3? g of zinc react with 0.90 L of 2.5M HCl to form zinc chloride and hydrogen gas. Identify the limiting and excess reactants. How many liters of hydrogen are formed at STP? 14
15 5. A common antacid contains magnesium hydroxide, which reacts with HCl to form water and magnesium chloride solution. As a government chemist testing commercial antacids, you use 0.10 M HCl to simulate the acid concentration in the human stomach. How many liters of stomach acid react with a tablet containing 0.10 g of magnesium hydroxide? 6. Mercury (II) nitrate reacts with sodium sulfide solution to produce solid mercury (II) sulfide and sodium nitrate solution. In a laboratory simulation, L of M mercury (II) nitrate reacts with L of 0.10 M sodium sulfide. How many grams of mercury (II) sulfide form? 15
16 The volume of a solution is made up of solute and solvent. To prepare a solution of a specific molarity: Weigh the solid (solute) needed. Transfer the solid to a volumetric flask that contains about half the final volume of solvent. Dissolve the solid thoroughly by swirling. Add solvent until the solution reaches its final volume. To dilute the solution to a lesser molarity, add only solvent to increase the volume. 18
17 19
18 1. Assume that 1 L of 3 M NaOH solution is needed for a class lab. Determine the mass of sodium hydroxide required to create the solution. 2. What mass of salt is required to prepare 500 ml of 1.54 M NaCl solution? 20
19 Concentration a measure of solute to solvent ratio Concentrated lots of solute Dilute not much solute ( watery ) To dilute solutions, add more solvent (often water) to solution. Moles of solute remain the same. 21
20 22
21 Dilution of solutions Acids (and sometimes bases) are purchased in concentrated form (concentrate) and are easily diluted to any desired concentration. Safety Tip: When diluting, add acid or base to water (A&W!) Dilution Equation: M C V C =M D V D C concentrate D dilute 23
22 1. Concentrated H 3 PO 4 is 14.8 M. What volume of concentrate is required to make L of M H 3 PO 4? 2. What volume of 15.8M HNO 3 is required to make 250 ml of a 6.0M solution? 24
23 3. Isotonic saline is a 0.15 M aqueous solution of NaCl that simulates the total concentration of ions found in many cellular fluids. Its uses range from a cleansing rinse for contact lenses to a washing medium for red blood cells. How would you prepare 0.80 L of isotonic saline from a 6.0 M stock solution? 25
24 4. If 25.0 ml of 7.50 M sulfuric acid are diluted to exactly 500. ml, what is the mass of sulfuric acid per milliliter? 26
25 Beer s Law relates the absorption of light to the properties of the material through which the light is traveling. There is a dependence between the absorbance of light by a substance and the product of the molar absorptivity coefficient, a, the distance the light travels through the material, b, and the molar concentration, c. A abc Also called the Beer-Lambert Law, the more concentrated a solution is, the more light it will absorb and the darker it will appear. 27
26 Beer s law is typically used in a process called spectroscopy. In this process, a beam of light is sent through a small sample of a solution. A detector on the other side of the solution records the amount of light that passed through the solution. 28
27 1. A red light is passed through a blue solution as shown. The molar absorptivity constant of the solution is 1.3x10 6 M -1 cm -1. The absorbance reading is Determine the concentration of the solution. 1 cm If water is added to the solution in question #1, what do you think will happen to the concentration and the absorbance reading? 29
28 3. A green light is passed through a purple solution as shown. The molar absorptivity constant of the solution is 1.8x10 4 M -1 cm -1. The absorbance reading is Determine the concentration of the solution. 1 cm List at least two laboratory procedures you could do to increase the absorbance reading for the purple solution in #3. 30
29 Absorbance Experiments with Beer s Law typically create a calibration curve by recording absorbance values for solutions with known concentrations. Then, the absorbance for a solution of unknown concentration can be measured and the concentration may be calculated using the equation from the calibration curve. An example of a calibration curve is shown below. 1.2 Calibration Curve for Crystal Violet y = x R² = Concentration (M) 31
30 Covalent bonds between H and O have unequal sharing. Electrons spend more time closer to O, creating a slightly negative pole. Water molecule has bent shape; atoms form angle. The distribution of its bonding electrons and its overall shape makes water an ionizing solvent. 32
31 To be soluble, the attraction between each ion and the water molecules must outweigh the attraction of the oppositely charged ions. If the electrostatic attraction among ions in the compound is greater than the attraction between ions and water molecules, the substance is insoluble. 33
32 Water interacts strongly with dissolved reactants and can affect their bonds. Electrolyte a substance that conducts a current when dissolved in water Soluble ionic compounds dissociate completely into ions and create a large current; called strong electrolytes. Solvated ions are surrounded by solvent molecules. Oppositely charged ions separate when dissolved in water, become surrounded by water molecules, and spread randomly throughout the solution. H O KBr ( s) 2 K ( aq) Br ( aq) The H 2 O above the arrow indicates that water is required but is not a reactant in the usual sense. 34
33 Na + ions Water molecules Cl - ions NaCl(s) + H 2 O Na + (aq) + Cl - (aq) 35
34 Animation 36
35 Many covalent compounds dissolve in water. Do not dissociate into ions Remain intact molecules Do not conduct an electric current; called nonelectrolytes. Acids are H-containing covalent compounds that do dissociate into ions in water. 37
36 Water interacts most strongly with the hydrogen cation (H + ) H + is just a proton H + attracts the negative pole of surrounding water molecules so strongly that it forms a covalent bond to one of them H 3 O + (hydronium ion) 38
37 H + hydrogen ion (a proton) H 2 O water H 3 O + hydronium ion 39
38 1. Nitric acid is a major chemical in the fertilizer and explosives industries. In aqueous solution, each molecule dissociates and the H becomes a solvated H + ion. What is the molarity of H + (aq) in 1.4 M nitric acid? 2. How many moles of H + (aq) are present in 451 ml of 3.20 M hydrobromic acid? 40
39 3. How many moles of H + (aq) are present in 585 ml of 3.50 M H 3 PO 4? 4. What is the molarity of the H + ion in 1.6 M sulfuric acid? 41
40 Dissociation: separation of an ionic solid into aqueous ions e.g. NaCl(s) Na + (aq) + Cl - (aq) Ionization: breaking apart of polar molecules into aqueous ions e.g. HNO 3 (aq) + H 2 O(l) H 3 O + (aq) + NO 3 (aq) Molecular Solvation: molecules (covalent) dissolve, but remain intact e.g. C 6 H 12 O 6 (s) C 6 H 12 O 6 (aq) 42
41 Predict the ionization or dissociation for the following chemical species: a. Sodium hydroxide b. Hydrochloric acid c. Sulfuric Acid d. Acetic Acid e. Potassium fluoride f. Calcium nitrate 43
42 Strong Electrolytes exhibit nearly 100% dissociation e.g. Weak electrolytes exhibit little dissociation e.g. NaCl Na + + Cl NOT in water: In aq. solution: CH 3 COOH CH 3 COO + H + NOT in water: In aq. solution: Strong or Weak is a property of the substance. One cannot be changed into the other. When written in a chemical equation, a strong electrolyte is shown as individual ions while a weak electrolyte is shown in molecular form. 44
43 Electrolytes - solutions that carry an electric current strong electrolyte weak electrolyte nonelectrolyte NaCl(aq) Na + + Cl - HF(aq) H + + F - 45
44 Temperature: as T increases, rate increases Particle Size: as particle size decreases, rate increases Mixing: more mixing/stirring, rate increases Nature of solvent or solute: identity determines whether a substance will dissolve and to what extent 46
45 UNSATURATED SOLUTION more solute dissolves SATURATED SOLUTION no more solute dissolves SUPERSATURATED SOLUTION becomes unstable, crystals form increasing concentration 47
46 Solids dissolved in liquids Gases dissolved in liquids Sol. Sol. T o As T o, solubility T o As T o, solubility 48
47 Solubility: how much solute dissolves in a given amount of solvent at a given temperature Unsaturated: Solution can hold more solute; below line Saturated: Solution has just the right amount of solute; on line Supersaturated: Solution has too much solute dissolved in it; above the line KNO 3 (s) Solubility (g/100 g H 2 O) KCl (s) HCl (g) Temp. ( o C) 49
48 Solubility (grams of solute/100 g H 2 O) KI Solubility vs. Temperature NaNO 3 gases solids HCl NH 3 KNO 3 NH 4 Cl KCl NaCl KClO 3 SO Temperature C 50
49 Solubility (grams of solute/100 g H 2 O) KI NaNO 3 gases Describe each situation below according to saturation and appearance using the given solubility table: A. Per 100 g H 2 O,100 g NaNO 60 C KNO 3 solids B. Per 100 g H 2 O, 20 g KClO 80 C HCl NH 3 NH 4 Cl KCl C. Cool solution (B) very slowly to 30 C NaCl KClO 3 D. Quench solution (B) in an ice bath to 30 C E. Per 100 g H 2 O, 70 g KNO 50 C SO Temperature C 51
50 In a gas, the energy of attraction is small relative to the energy of motion. On average, the particles are far apart. Large interparticle distance has several macroscopic consequences: A gas moves randomly throughout its container and fills it. Gases are highly compressible (can be squeezed and shrunk). Gases flow and diffuse through one another easily. 52
51 In a liquid, the attractions are stronger because the particles are in contact. But their kinetic energy allows them to tumble randomly over and around each other. Therefore, a liquid conforms to the shape of its container but has a surface. With very little free space between the particles, liquids compress only very slightly. Liquids flow and diffuse but much more slowly than gases. 53
52 In a solid, the attractions dominate the motion to such an extent that the particles remain in position relative to one another. With the particles very close together and positions fixed, a solid has a specific shape. Solids compress even less than liquids. Solids do not flow significantly. 54
53 Phase changes are determined by the interplay between kinetic energy and intermolecular forces. As the temperature increases, the average kinetic energy increases as well so the faster moving particles can overcome the attractions more easily. Lower temperatures allow the forces to draw the slower moving particles together. The process by which a gas changes into a liquid is called condensation. A liquid changing to a gas is called vaporization. The process by which a liquid changes into a solid is called freezing. A solid changing to a liquid is called melting or fusion. The process by which a solid becomes a gas (without becoming a liquid first) is called sublimation. A gas changing to a solid is called deposition. 55
54 Phase diagrams are used to depict the phase changes of a substance at various conditions of temperature and pressure. A phase diagram has these four features: Regions of the diagram: Each region corresponds to one phase of the substance. A particular phase is stable for any combination of pressure and temperature within its region. If any other phase is placed under those conditions, it will change to the stable phase. Lines between regions: The lines separating the regions represent the phase transition curves. Any point along a line shows the pressure and temperature at which the two phases exist in equilibrium. The critical point: The liquid-gas line ends at the critical point. Beyond the critical temperature, a supercritical fluid exists rather than separate liquid and gaseous phases. The triple point: The three phase transition curves meet at the triple point: the pressure and temperature at which three phases are in equilibrium. 56
55 57
56 Most substances have a solid phase that is more dense than the liquid phase, giving a positive slope to the solid-liquid line. (Because the liquid occupies slightly more space than the solid, an increase in pressure favors the solid phase, in most cases.) Unlike almost every other substance, solid water is less dense than liquid water (due to hydrogen bonds causing the solid phase to have a greater volume because of large spaces between molecules). An increase in pressure favors the phase that occupies less space, which for water is the liquid phase. Therefore, the solid-liquid line for water has a negative slope. 58
57 H 2 O CO 2 59
58 Using the phase diagram for water, determine what phase water is in at the following conditions: a. 2 atm, 55 C b. 1 atm, 75 C c. 1 atm, 200 C d. 5 atm, -100 C 60
59 Colligative properties refer to the physical properties that are changed by the presence of any solute particles. The chemical identity of the solute is not important. Only the concentration of the solute particles is significant in altering physical properties. The physical properties that change when a solute is present include vapor pressure, boiling point, melting point, and osmotic pressure. 61
60 The lowering of vapor pressure caused by a nonvolatile nonelectrolyte solute leads to an increase in the boiling point and a decrease in the freezing point of the solvent. Higher temperatures are needed to make the vapor pressure of the system equal atmospheric pressure (boiling point). Lower temperatures are needed to overcome the interference with the crystallization process caused by the attractive forces between the solute and solvent. The changes in the boiling and melting points of a solution depend on the solvent and on the molal concentration of the solute. 62
61 Boiling Point Elevation / Freezing Point Depression The increase in the boiling point is given by T ik b m m is the molality of the solution k b is the boiling point elevation constant for the solvent i represents the number of particles produced by each part of the solute The decrease in the melting (freezing) point is given by T ik f m The negative sign indicates a decrease in temperature k f is the freezing point depression constant 63
62 Chemists use the freezing point depression to make cooling baths below 0 C by dissolving large quantities of salt in water and then adding ice. Antifreeze in a car s engine lowers the freezing point of water by adding a high concentration of ethylene glycol. Antifreeze also increases the boiling point of water to allow engines to run above 100 C without boiling over. A common cooking use is adding salt to water to increase the boiling point, allowing the food to cook at a higher temperature than otherwise possible. The most important use is the determination of molar masses of solutes. 64
63 Example #1: Which of the following, when added to 1.00 kg H 2 O, is expected to give the greatest increase in the boiling point of water? (k b = C/m) a mol sucrose (C 12 H 22 O 11 ) b mol iron (III) nitrate c mol ammonium chloride d mol calcium sulfate e mol acetic acid 65
64 Three types of equations: Molecular Total Ionic Net Ionic Atoms and charges must balance in ionic equations. Molecular equations show all reactants and products as if they were intact, undissociated compounds Total ionic equations show all the soluble ionic substances dissociated into ions. Spectator ions appear in the same form on both sides of the equation and are not involved in the actual chemical change. Net ionic equations eliminate the spectator ions and show the actual chemical change taking place 66
65 Molecular Equation 2NaCl(aq) + Pb(NO 3 ) 2 (aq) PbCl 2 (s)+ 2NaNO 3 (aq) Total Ionic Equation 2Na + (aq)+2cl - (aq) + Pb 2+ (aq) + 2NO 3- (aq) PbCl 2 (s) + 2Na + (aq)+2no 3- (aq) Net Ionic Equation 2Cl - (aq) + Pb 2+ (aq) PbCl 2 (s) Spectator Ions 67
66 Molecular Equation 2 3 AgNO 3( aq) Na2CrO4 ( aq) Ag 2CrO4 ( s) 2NaNO ( aq) Total Ionic Equation 2 2Ag 3 ( aq) 2NO3 ( aq) 2Na ( aq) CrO4 ( aq) Ag 2CrO4 ( s) 2Na ( aq) 2NO ( aq) Net Ionic Equation 2 2Ag 4 ( aq) CrO4 ( aq) Ag 2CrO ( s) 68
67 Write the net ionic equations for the following molecular equations and identify the spectator ions: a. Li 2 SO 4 (aq)+ca(oh) 2 (aq) CaSO 4 (s)+2lioh(aq) b. 2NaI(aq) + Pb(NO 3 ) 2 (aq) 2NaNO 3 (aq) + PbI 2 (s) 69
68 Write the net ionic equations for the following molecular equations and identify the spectator ions: c. AgNO 3 (aq) + KBr(aq) AgBr(s) + KNO 3 (aq) d. H 2 SO 4 (aq) + NaOH(aq) Na 2 SO 4 (aq) + H 2 O (l) 70
69 Write the net ionic equations for the following molecular equations and identify the spectator ions: e. Al 2 (SO 4 ) 3 (aq)+6nh 4 OH(aq) 3(NH 4 ) 2 SO 4 (aq) +2Al(OH) 3 (s) 71
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