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Solutions Chapter 13

Solutions Chapter 13. Tro, 2 nd ed. SOLUTIONS & PHYSICAL PROPERTIES. A solution is a system in which one or more substances are homogeneously mixed or dissolved in another substance. The solute is the component that is dissolved or is the least abundant component of the solution.

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Solutions Chapter 13

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  1. SolutionsChapter 13 Tro, 2nd ed.

  2. SOLUTIONS & PHYSICAL PROPERTIES A solution is a system in which one or more substances are homogeneously mixed or dissolved in another substance. The solute is the component that is dissolved or is the least abundant component of the solution. The solvent is the dissolving agent or the most abundant component in the solution. If the solution has only one phase, it is homogeneous. (A mixture which has two phases is heterogeneous.)

  3. Similar to table 13.1 Metals in metals = alloys

  4. Properties of True Solutions The solute remains uniformly distributed throughout the solution and will not settle out with time. The solute can generally be separated from the solvent by purely physical means such as evaporation, usually of the solvent. The solute particles of a true solution are molecular or ionic in size. The solution can be either colored or colorless but is always clear or transparent.

  5. Formation of a Solution solute solvent solution

  6. Solvated Ions When materials dissolve, the solvent molecules surround the solvent particles due to the solvent’s attractions for the solute. The process is called solvation. Solvated ions are effectively isolated from each other. (Also called hydrated ions if the solvent is water.)

  7. TERMS THAT DESCRIBE THE SOLUBILITY OF LIQUIDS: Solubility (limit) is the maximum amount of solute that can be dissolved in a given amount of solvent at a given fixed T, at equilibrium. Saturatedsolutions contain the maximum amount of solute, so any additional solute appears as a precipitate or a gas, or a separate liquid phase. Unsaturated means that more solute can be added to the solution. Supersaturated is a temporary condition where more solute has dissolved, but add just 1 crystal to this, many crystals will precipitate from solution.

  8. Supersaturated Solution A supersaturated solution has more dissolved solute than the solvent can hold. When disturbed, all the solute above the saturation level comes out of solution.

  9. TERMS THAT DESCRIBE THE SOLUBILITY OF LIQUIDS Miscible: liquids that are capable of mixing and forming solutions. methanol and water Immiscible: liquids that are insoluble in each other. oil and water

  10. Solubilities of substances vary widely.

  11. Solubility of Various Common Ions in Cold Water See back of blue Periodic Table for rules and table. Similar to Table 7.2 14.2

  12. Electrolytes: substances whose aqueous solution is a conductor of electricity Strong electrolytes:all the electrolyte units are dissociated into ions (salts, strong bases, strong acids) Nonelectrolytes:none of the units are dissociated into ions (molecular solutes) Weak electrolytes: a small percentage of the units are dissociated into ions (weak bases, weak acids) Electrolytes

  13. Factors affecting solubility: • Natural inclination of the universe towards disorder so substances do mix • Strength of Force of Attraction between solute particles, between solvent particles and between solute & solvent (nature of solute and solvent) • Temperature • Pressure NOTE: Gases are always completely miscible!

  14. Factors affecting solubility:Nature/Intermolecular Forces Liquids: Similar liquid molecules will dissolve in each other - heptane in octane - both have only London forces involved at about the same strength, both nonpolar. Different IP forces - octane and water. H-bonding very strong in water, very different from London forces in octane; water would have to give up H-bonding for weaker force. THIS WON'T HAPPEN. The less dense liquid will rise and stay in separate phase on top of water.

  15. SolventPolarity Factors affecting solubility:Nature/Intermolecular Forces Polar compounds tend to be more soluble in polar solvents than nonpolar solvents. NaCl (sodium chloride) is an ionic compound which is: • soluble in water • slightly soluble in ethanol • insoluble in ether and benzene • General rule is “like dissolves like”

  16. The hydrated ions slowly diffuse away from the crystal to become dissolved in solution. As the attraction between the ions weakens, the ions move apart and become surrounded by water dipoles. Polar water molecules are attracted to Na+ and Cl- ions in the salt or crystal, weakening the attraction between the ions. Dissolution of sodium chloride in water. 14.3

  17. SolventPolarity Factors affecting solubility:Nature/Intermolecular Forces Nonpolar compounds tend to be more soluble in nonpolar solvents than in polar solvents. Benzene is a nonpolar organic compound which is: • insoluble in water • soluble in ether

  18. Classifying Solvents Similar to Table 13.2

  19. Practice with solutes and solvents • Which solvent will ethanol dissolve in more readily, water or octane? Why? • What about liquid C6H13OH? Water or octane? • What about solid glucose? (Draw it.) Water or octane? • What about solid I2? Will it dissolve more readily in water or octane?

  20. Practice with solutes and solvents • Water, because of the polarity of both with their –OH groups. • Octane, because most of the molecule is nonpolar like octane is. • Solids must also be "like" the solvent. Glucose dissolves in water because of extensive H-bonding w/ its many -OH groups. (See Lewis structure) 4. Solid I2 - held by nonpolar London forces; octane also London forces. I2 dissolves in octane better than water.

  21. Effect of Temperature on Solubility For most solids dissolved in a liquid, an increase in temperature results in increased solubility. The solubility of a gas in water usually decreases with temperature.

  22. large increase in solubility with temperature slight increase in solubility with temperature decrease in solubility with increasing temperature FIGURE 13.4

  23. Effect of Pressure on Solubility Small pressure changes have little effect on the solubility of solids in liquids or liquids in liquids Small pressure changes have a great effect on the solubility of gases in liquids - The solubility of a gas in a liquid is directly proportional to the pressure of that gas above the liquid

  24. Rate of Dissolving Solids: 4 factors • Particle Size A solid can dissolve only at the surface that is in contact with the solvent. Smaller crystals have a larger surface to volume ratio than large crystals. Smaller crystals dissolve faster than larger crystals.

  25. Rate of Dissolving Solids: 4 factors 2.Temperature In most cases, the rate of dissolving of a solid increases with temperature. This occurs because solvent molecules strike the surface of the solid more frequently and harder, causing the solid to dissolve more rapidly.

  26. Δt Δc Δt Rate of Dissolving Solids: 4 factors 3. Concentration of the Solution As solution concentration increases, the rate of dissolving decreases. Δc The rate of dissolving is at a maximum when solute and solvent are first mixed.

  27. Rate of Dissolving Solids: 4 factors 4. Agitation or stirring. When a solid is first put into water, it comes in contact only with water. The rate of dissolving is a maximum. As the solid dissolves, the amount of dissolved solute around the solid increases and the rate of dissolving decreases. Stirring distributes the dissolved solute throughout the water; more water is in contact with the solid causing it to dissolve more rapidly.

  28. Concentration of Solutions The concentration of a solution expresses the amount of solute dissolved in a given quantity of solvent or solution. The terms dilute and concentrated are qualitative expressions of the amount of solute present in a solute. You will need to learn the mathematical representations of concentration: mass-percent, mass/volume percent, Molarity, etc.

  29. Mass Percent Solution Mass percent expresses the concentration of solution as the percent of solute in a given mass of solution. %-mass = (g solute/(g solute + g solvent)) * 100 = (g solute/g solution) * 100

  30. Example of Mass Percent How do you prepare 425 g of a 2.40%-wt solution of sodium acetate? Solve for grams of solute, then grams of water: (X g/425 g) x 100 = 2.40% x = 10.2 g NaC2H3O2 425 – 10.2 = 414.8 g H2O You try: What is the mass-percent of a solution that has 13.6 g of NaCl in 250.0 g of solution? What is the mass-percent of a solution that has 15.0 g of ethanol in 35.0 g of water?

  31. Mass/Volume Percent (m/v) Mass /volume percent expresses the concentration as grams of solute of solute per 100 mlsolution. %-mass/vol = g solute/mL solution * 100

  32. A 3.00 %-mass/vol H2O2 solution is commonly used as a topical antiseptic to prevent infection. What volume of this solution will contain 10.0 g of H2O2? Solve the mass/volume equation for grams of solute. %-mass/vol = (g solute/mL solution) * 100 mL solution = (g solute/%-mass/vol) * 100 = (10.0 g/3.00%) * 100 = 333 mL

  33. Volume Percent Solutions that are formulated from liquids are oftenexpressed as volume percent with respect to the solute. The volume percent is the volume of a liquid in100 ml of solution. %-vol = (solute vol/solution vol) * 100

  34. Volumes are not necessarily additive. A bottle of rubbing alcohol reads 70% by volume. The alcohol solution could be prepared by mixing 70 mL of alcohol with water to make a total volume of 100 mL of solution. 30 m L of water could not be added to 70 mL of alcohol because the volumes are not necessarily additive.

  35. Practice Find both the mass-percent and volume percent of a solution that has 10.0 g of ethanol (D = 0.7893 g/mL) and 90.0 g of water (D = 0.9987 g/mL). Assume volumes are additive.

  36. Molarity Molarity of a solution is the number of moles of solute per liter of solution. Molarity = M = moles solute/liter of solution = mol/L Sometimes n represents moles, then M = n/V.

  37. Molarity Calculations How many moles of NaOH are present in 25.0mL of 0.555 M NaOH? M = mol/vol, therefore mol = M * V (must be in Liters) 0.555 mol/L * (25.0mL * 1L/103mL) = 0.0138 mol NaOH

  38. Molarity Calculations What is the Molarity of a solution that has 13.6 g of NaCl in 250.0 mL of solution? What is the volume of 1.25 M HCl solution that will provide 0.4414 moles of HCl?

  39. Solubility g/100g grams of solute/100 grams of solvent Normality N molarity * (# of H+ or OH-) Yes, you have to know all of these!

  40. NORMALITY: EXAMPLES 1.0 M HCl * 1 H+ = 1.0 N 1.0 M H2SO4 * 2 H+ = 2.0 N 1.0 M NaOH * 1 OH- = 1.0 N 1.0 M Ba(OH)2 * 2 OH- = 2.0 N

  41. Dilution Problems If a solution is diluted by adding pure solvent: - the volume of the solution increases. - the number of moles of solute remain the same.

  42. Dilution Problems When the moles of a solute in a solution before and after dilution are the same, the moles before and after dilution may be set equal to each other. Remember M = n/V, or n = M * V. If V changes, M will change to keep moles, n, constant. Dilution Equation: M1V1 = M2V2 MEMORIZE THIS!!!

  43. Practice How would you make 500.0 mL of a 0.500 M solution of NaOH from 6.00 M NaOH? 0.500 M * 500.0 mL = 6.00 M * V2 V2 = 41.7 mL

  44. Practice Making Solutions How would you make 250.0ml of a 0.555 M standard NaOH solution? Use two methods: (1) Weigh out dry chemical NaOH solid and dissolve it. (2) Dilute a more concentrated solution of 6.00 M NaOH. A. Weighing a solid and using M = n/V. Find n first, then grams of solid. 250.0mL(1L/103mL)(0.555 mol/L) = 0.13875 mol NaOH 0.13875 mol * 39.997 g/mol = 5.55 g NaOH Get a vol. flask and add water about halfway up. Add solid. Mix til dissolved, then dilute to mark on flask.

  45. Practice Making Solutions B. Dilution: Given 6.00 M NaOH. Use the Dilution Equation, M1V1=M2V2 6.00 mol/L * V1 = 0.555 mol/L * 0.2500 L V1 = O.O231 L or 23.1 mL Obtain 23.1 mL in a graduated cylinder. Get a vol. flask and add water about halfway up. Add the correct amount of 6.00 M NaOH. Mix until dissolved, then dilute to mark on flask.

  46. Preparation of a 1 molar solution Start with some water in the flask! 14.7

  47. Molarity, Dissociation andSolution Inventories When strong electrolytes dissolve, all the solute particles dissociate into ions. By knowing the formula of the compound and the molarity of the solution, it is easy to determine the molarity of the dissociated ions simply multiply the salt concentration by the number of ions. It’s like doing stoichiometry!

  48. Find the solution inventory for each ofthese strong electrolytes • 0.25 M MgBr2 • 0.33 M Na2CO3 • 0.0750 M Fe2(SO4)3 1. MgBr2(aq) → Mg2+(aq) + 2 Br-(aq) I: 0.25 M 0 0 R: -0.25 M +0.25 +0.50 F: 0 0.25 M 0.50 M So we say: [Mg2+] = 0.25 M, [Br-] = 0.50 M 2. Na2CO3(aq) → 2 Na+(aq) + CO32-(aq) I: 0.33 M 0 0 R: -0.33 +0.66 +0.33 F: 0 0.66 M 0.33 M [Na+] = 0.66 M, [CO32-] = 0.33 M 3. [Fe3+] = 0.150 M, [SO42-] = 0.225 M

  49. STOICHIOMETRY WITH AQUEOUS SOLUTIONS: A. GAS-FORMING: use regular stoichiometry and concentration information B. PRECIPITATION: use solubility rules (see examples on following pages) C. NEUTRALIZATION: use concentration information and stoichiometry steps. (Do NOT use dilution equation! Use stoichiometry.) acid + base  salt + water TAKE GOOD NOTES – THE TEXT ASSUMES YOU KNOW HOW TO DO THIS!

  50. STOICHIOMETRY OF SOLUTION PRECIPITATION REACTIONS: How many grams of silver will precipitate if excess copper is added to 500.0 ml OF 0.100 M AgNO3? 1: Balanced chemical equation Cu(s) + 2 AgNO3(aq) 2 Ag(s) + Cu(NO3)2(aq) What you know & what you want to know 500.0 ml OF 0.100 M AgNO3. Want grams Ag. 2: 500.0 ml * 1 L * 0.100 moles = 0.0500 mol AgNO3 103 ml 1.00 L 3: 0.0500 mol AgNO3 * 2 mol Ag/2 mol AgNO3 = 0.0500 mol Ag 4: 0.0500 mol Ag * 107.9 g/mol = 5.40 g Ag

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