1 / 44

Classification of Matter and Solutions Notes Part 1

Classification of Matter and Solutions Notes Part 1. I. Classification of Matter. Matter Can it be physically separated? Yes No Mixtures Pure Substances Is the composition uniform? Can it be decomposed by an ordinary chemical reaction?

axel
Download Presentation

Classification of Matter and Solutions Notes Part 1

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Classification of Matter and Solutions Notes Part1

  2. I. Classification of Matter Matter Can it be physically separated? Yes No MixturesPure Substances Is the composition uniform? Can it be decomposed by an ordinary chemical reaction? Yes NoYes No HomogeneousHeterogeneousCompoundsElements MixturesMixtures(water, sodium(gold, oxygen, (Solutions)(Suspensionschloride, sucrose) carbon) (air, sugar water,or Colliods) salt water)(granite, wood, muddy water)

  3. Mixtures: matter that can be physically separated into component parts. • a. homogeneous mixture –has uniform composition; also called a solution • b. heterogeneous mixture – does not have a uniform composition; suspensions or colloids

  4. Pure Substances: when component parts of a mixture can no longer be physically separated into simpler substances. Pure substances are either compounds or elements. • a. Compounds – can be decomposed by a chemical change. • b. Elements – cannot be decomposed by a chemical change.

  5. II. Types of Mixtures (Solutions, Suspensions, and Colloids) Table 13-3 page 398 • 1. Solution (homogeneous mixture)- any substance (solid, liquid, gas) that is evenly dispersed throughout another substance. page 398 (Not the same as a chemical reaction!!) • Ex: sugar water, salt water (do not scatter light) Components of a Solution • 1. Solute – substance dissolved • 2. Solvent – substance that does the dissolving (water is the universal solvent)

  6. 2. Suspensions (heterogeneous mixtures) – particles in a solvent are so large that they settle out unless the mixture is constantly stirred Ex: muddy water, vegetable soup, page 398 (may scatter light, but are transparent)

  7. 3. Colloids (heterogeneous mixtures) – particles are intermediate in size between those in solutions and suspensions. Example: After large soil particles settle out of muddy water the water is often still cloudy because colloidal particles remain dispersed in the water. Ex: milk, mayonnaise , page 398 (do scatter light – Tyndall Effect)

  8. III. The Solution Process (Solvation) Solvation is the process by which a solute dissolves in a solvent. Miscible: when solutes and solvents are soluble in each other (solvation occurs) Immiscible: when solutes and solvents are not soluble in each other (solvation does not occur) Aqueous solutions – solvent is water.

  9. What happens when: Ionic compound as the solute and water is the solvent? Dissociation of ionic compound occurs (ions separate). Water is then attracted to the positive and negative ions. When all molecules have been “surrounded” the molecule is called hydrated. Miscible

  10. What happens when: Polar covalent molecules are the solute and water is the solvent? Dissociation does NOT occur. Water is polar and its “oppositely charged poles” will be attracted to other polar molecules' “oppositely charged poles.” When a solution is made between two polar molecules it is called molecular solvation. Miscible

  11. What happens when: Nonpolar covalent molecules are the soluteand water is the solvent? A solution will NOT occur. Water and any nonpolar molecule will not mix! Think of putting water and oil together. Water is polar and oil is nonpolar. The polar water is not attracted to the oil, because the nonpolar oil does not have any oppositely charged poles! Immiscible

  12. IV. Like Dissolves Like We don’t always use water as the solvent! Solutions can be made from various substances – a rule of thumb to follow when trying to determine if two substances will form a solution is “like dissolves like.”

  13. Water and Alcohol : Miscible Polar Molecules + Polar Molecules Nonpolar Molec. + Nonpolar Molec. Ionic and Ionic Ionic + Polar Molecules Polar Molecules + Nonpolar Molecules Ionic + Nonpolar Molecules Oil and Hexane : Miscible Chemical Reaction Salt and Water : Miscible Water and Oil : Immiscible Salt and Oil: Immiscible

  14. V. Solubility - Now that we can figure out what we can mix to make a solution, how do we know how much solvent and how much solute to use?

  15. Solubility: the maximum amount of a substance that will dissolve in a solvent (at a specific temperature) According to solubility, solutions can be either: unsaturated – a solution that is able to dissolve more solute (not enough) saturated – a solution that cannot dissolve any more solute (just enough)

  16. supersaturated – a solution that contains more solute than can be dissolved (too much!!) The solubility of substances varies widely. For example 0.189 grams of Ca(OH)2 dissolves in 100 grams of water at 0C. 122 grams of AgNO3 dissolves in 100 grams of water at 0C. (page 404 in your book)

  17. VI. Factors Effecting Rate and Solubility • Factors Effecting Rate: 1. Agitation – stirring or mixing the solution will increase the rate or how fast the solute dissolves, but it will not change how much solute can be dissolved. If you add 35.9 grams of salt to water (at 20C) it will all eventually dissolve, but if you stir the solution it will dissolve much quicker. (As you stir the particles are constantly being moved, allowing for interactions between solute and solvent to occur more quickly.)

  18. Surface Area – increasing the surface area of the solute will increase the rate or how fast the solute dissolves, but it will not change how much solute can be dissolved. 3. Temperature – increasing temperature will increase the rate or how fast the solute dissolves in the solvent. (As temperature increases the particles begin to move faster and faster and collide with more particles quicker, which means the solute and solvent particles have an increased chance of coming into contact with each other.)

  19. B. Factors Effecting Solubility: 1. Increasing Temperature - solubility of a solid solute in a liquid solvent generally increases with an increase in temperature. At 20C 35.9 grams of salt will dissolve in 100 grams of water, but at 100 C 39.2 grams of salt will dissolve in 100 grams of water!

  20. 2. Decreasing Temperature-increases the solubility of a gaseous solute in a liquid solvent. What would you rather drink, a hot coke or a cold coke?

  21. 3. Pressure – The solubility of a gas increases as the pressure of the gas above the liquid increases. Carbonated drinks have CO2 dissolved in them. They are also bottled under a high pressure of CO2, which forces the CO2 into solution. When the bottle is opened, the pressure above the solution decreases, and bubbles of CO2 form in the liquid, then escape. Eventually, most of the CO2 escapes and the drink becomes “flat.”

  22. Henry’s Law- “At a given temperature, the solubility, S, of a gas in a liquid is directly proportional to the pressure, P, of the gas above the liquid.” S1 = S2 P1 P2

  23. VII. Electrolyte VS Nonelectrolyte 1. Electrolyte –compounds that conduct an electric current in an aqueous solution OR in the molten state. An electrolyte solution contains charged particles (ions), which can move. Any salt dissolved in water is an electrolyte: NaCl, KI, etc. Some polar molecules also conduct electricity (most acids are electrolytes because H is the only nonmetal that has a + charge).

  24. Types of Electrolytes 1. Strong electrolytes – a large portion of the solute exists as ions: a. aqueous solutions of all ionic compounds b. strong acids: have at least 2 oxygens per hydrogen (H2SO4, HNO3) c. strong bases – these are hydroxides from Group I and II, except Be. (NaOH, CsOH, etc)

  25. 2. Weak electrolytes – these are solutions in which only a small portion of the solute exists as ions a. weak acids: -all binary acids – HF, H2S, etc -weak acids that have less than 2 oxygen's per hydrogen b. weak bases – hydroxides of everything else not in Group I or II, including Be(OH)2

  26. 2. Non-Electrolytes- compounds that do NOT conduct electricity in either aqueous solution or when melted: • distilled water • gases • molecular compounds (2 nonmetals) • organic compounds – alcohols, sugars, etc. anything containing a Carbon

  27. Practice Problems: Tell whether each of the following aqueous solutions would be a STRONG, WEAK, or NON electrolyte. • NaCl 2. CH3Br (l) 3. HMnO4 • HC2H3O2 5. LiOH 6. HC6H7O6 7. CO2 (l) 8. HF

  28. VIII. Concentration: the concentration of a solution is a measure of the amount of solute in a given amount of solvent or solution.

  29. 1. Molarity: the number of moles of solute in one liter of solution. M = amount of solute (moles) volume of solution (liters) OR M = grams of solute/molar mass of solute L of solution

  30. A bottle labeled 6M HCl is pronounced 6 molar HCl and it was prepared by mixing 6 moles of HCl with enough water to make 1 liter of solution. Molarity is always moles/liter. Ex: 6M is 6 moles/ liter 6M HCl

  31. Molarity is probably the MOST IMPORTANT unit of concentration that we work with in chemistry. Knowing the correct technique for preparing molar solutions is extremely important. A volumetric flask MUST be used. Here is the technique: (Know it) Determine the correct mass of solute needed, add to a volumetric flask and fill flask with distilled water until it reaches the line.

  32. Example Molarity Problems: 1. Calculate the molarity, M, of a solution prepared by dissolving 11.85 g of potassium permanganate in enough water to make 750. mL of solution. 2. Calculate the mass of NaCl needed to prepare 175 ml of 0.500 M saline solution. 3. Calculate the volume (in mL) needed to prepare a 2.48 M sodium hydroxide solution containing 31.52 g of the dissolved solid. 4. How many grams of calcium chloride must be dissolved in water to make 350. mL of a l.75 M solution? 5. What is the molar mass of 55.0 grams of a solute that has been dissolved in enough solvent to make 500. mL of a l.5 M solution?

  33. 2. Molarity of Ions in solution – Most ionic solids when dissolved in water, ionize Ex: CaCl2 (aq)  Ca2+ + 2 Cl- Na3PO4 3 Na+ + PO43- Mg(OH)2 Mg2+ + 2 OH-

  34. Example Molarity of Ions Problems: Calculate the molarity of the ions in the following solutions: 1. 0.25 M calcium phosphide 2. 2.0 M Chromium (III) chloride 3. 0.25 M barium hydroxide 4. 0.55 M aluminum nitride

  35. 3. Dilutions – you need to make 800.0 mL of a 0.25 M solution of HCl. The only available HCl is concentrated (12 M). How would you do this? Being able to prepare dilutions is a very common application of chemistry. Our department buys concentrated acids, but normally uses more dilute solutions of these acids in our labs. Therefore, it is important to know how to correctly dilute. M1V1 = M2V2 where M1 = concentrated solution V1 = volume of concentrated solution M2 = dilute solution V2 = volume of dilute solution

  36. In the above problem, M1 = 12.0 M M2 = 0.25 M Vl = ? V2 = 800.0 ml therefore (12.0 M) (V1) = (0.25 M)(800 mL) and V1 = 16.67 ml. This 16.67 mL is the amount of concentrated acid we will take out, but we still have to make 800 mL of 0.25 M. The other 783.3 mL of solution must be water (800 – 16.67). Therefore we would place 16.67 mL of HCl into 783.3 mL of water and we would have our diluted solution

  37. Practice Dilutions Problems: l. How would you prepare 485 mL of 0.39 M solution of NaCl when a l.0 M solution of NaCl is all you can find? 2. If 300.0 mL of a 2.5 M solution of nitric acid is added to 500.0 mL of water, what is the molarity of the dilute solution? 3. Prepare 500. mL of a dilute solution (0.50 M ) of nitric acid from the l5.0 M stock solution. You will need 600. mL of the dilute solution. 4. How would you prepare 500. mL of a 0.250 M solution of NaCl from a 3.00 M stock solution? 5. Tell how you would prepare enough 0.75 M NaCl solution so that 78 students working in groups of 2 will have 12 mL of solution for a lab. The stock solution is 3.5 M NaCl.

  38. 4. Percent Solutions (2 types): 1. Percent mass or % (m/m)– used when a solid solute is dissolved in liquid, usually water. % (m/m) = grams of solute grams of solution

  39. Example 1: Prepare a 10.00 % NaCl solution using 50.0 g water and solid salt: 0.l0 = x / 50 + x and x = 5.67 g NaCl in 50 grams of water Example 2: How many grams of water must be added to 25.0 g salt in order to have a 2.00 % (by mass) salt solution? Example 3. Prepare 600.0 g of a 3.00 % saline solution (NaCl solution).

  40. 2. Percent volume or %(v/v) – used when a liquid solute is mixed with a liquid solvent. The units are mL or L, but are worked the same. % (v/v) = mL solute mL solution

  41. Example 1: Prepare a 20.00 % alcohol solution using 400.0 mL of water: 0.20 = x / 400 + x and x = 100 mL alcohol + 400 mL water Example 2: What is the percent (v/v) of ethanol in the final solution when 90.0 mL of it are diluted to a volume of 300. mL with water?

  42. 5. Molality: is the concentration of a solution expressed in moles of solute per kilogram of solvent. Molality is represented by a lower case m. molality = amount of solute (moles) mass of solvent (kg) OR molality = grams of solute/molar mass kg of solvent 5m NaOH is pronounced 5 molal NaOH solution. 5m = 5 moles of NaOH 1 kg of water 5m or 5moles/kg

  43. Example Molality Problems: 1. Calculate the molality of a solution prepared by dissolving 5.0l g sodium sulfate in 700.0 g water. 2. Prepare a solution that is 2.50 molal barium nitrate in 1500. grams of water. 3. A solution is prepared by dissolving 3.00 g of potassium chromate in 58.5 g of water. Calculate the molality of the solution. 4. How many kg of water must be added to 8.3 g of oxalic acid, H2C204, to prepare a 0.050 m solution?

  44. IX. Problems involving percent solution and density • density – ratio of mass to volume D = m / v Example: Calculate the mass of sodium hydroxide in 300.0 ml of solution that is 8.00 % NaOH. The density of the solution is 1.09 g/ml. (1) find the number of grams of solution by using the density of the solution 300 mL 1.09 g = 327 g solution mL (2) find the grams of NaOH by using the percent solution: 0.08 = x / 327 g x = 26.2 g

More Related