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Solutions

Solutions. Soluble : solute dissolves in solvent Insoluble: solute will not dissolve in solvent Solubility : amount of solid solute that dissolves in given quantity of solvent at specific temperature/ pressure Miscible : two liquids dissolve in one another/no separation of layers

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Solutions

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  1. Solutions

  2. Soluble: solute dissolves in solvent Insoluble: solute will not dissolve in solvent Solubility: amount of solid solute that dissolves in given quantity of solvent at specific temperature/ pressure Miscible: two liquids dissolve in one another/no separation of layers Immiscible: two liquids do not dissolve in one another/separation of layers

  3. http://college.hmco.com/chemistry/shared/media/animations/supersaturatedsodiumacet.htmlhttp://college.hmco.com/chemistry/shared/media/animations/supersaturatedsodiumacet.html

  4. http://college.hmco.com/chemistry/shared/media/animations/dissolutionofsolidinliquid.htmlhttp://college.hmco.com/chemistry/shared/media/animations/dissolutionofsolidinliquid.html

  5. “Like” dissolves “like” • Polar solvents dissolve ionic/polar compounds • Most ionic compounds dissolve in water • Exception: ions w/greater attraction for each other than charged water molecules don’t dissolve • Nonpolar solvents dissolve nonpolar compounds • Won’t dissolve in water • Not attracted to water molecules (sucrose polar, so dissolve in water)

  6. Collisions of solvent/solute’s surface particles causes solvation of solute particles Polar water/ions attract each other Solvated solute particles collide increasingly with remaining crystal and rejoin crystal Solvation rate > crystallization rate = continuing solvation When solvation rate = crystallization rate = dynamicequilibrium http://www.chem.iastate.edu/group/Greenbowe/sections/projectfolder/flashfiles/thermochem/solutionSalt.html Solvation http://college.hmco.com/chemistry/shared/media/animations/solutionequilibrium.html

  7. Factors that affect solubility • Nature of solvent/solute (like dissolves like) • Temperature (generally, ↑T  ↑ solubility) • Pressure (only gases in liquids) • Rate of solution: measure of how fast substance dissolves

  8. Particles • Size • Larger solute molecules = lowered solubility • More difficult to surround w/solvent molecules • Increasing total surface area of solute particle increases dissolving (only takes place at surface of each particle) • Contact of solute w/solvent increases rate • Shaking/stirring (agitation) brings solvent in contact w/surface of solute • Saturation • Less solute in solution allows faster dissolving to take place • As solution approaches saturation, dissolving slows

  9. Liquids/solid solutes ↑T ↑rate solute dissolves Most solid solutes more soluble in warmer solutions Increases KE of water molecules Increases frequency/ force of collisions with crystal surfaces Gases: reverse is true ↑T ↓solubility/rate of solution (cold soda-more carbonation than warm) Temperature

  10. Pressure • Solids/liquid solutes • ∆P has practically no effect on solubility • Gaseous solutes • Increasing P increases solubility • Decreasing P decreases solubility • Gas has more space to escape into

  11. Henry’s Law • At given T, solubility of gas (solute) in solution directly proportional to partial pressure of that gas above solution • As P of gas above liquid increases, solubility of gas increases • As P of gas decreases, solubility of gas decreases • S1/P1= S2/P2 • S: gas solubility in liquid • P: Pgas above liquid

  12. Example: The concentration of dissolved oxygen is 0.44g / 100 mL solution. The partial pressure of oxygen is 150 mm Hg. What is the predicted concentration if the partial pressure for oxygen is 56 mm Hg? Solution: P1 = 150 mm Hg C1 = 0.44 g O2 /100 mL solution (C is being used as concentration instead of solubility) P2 = 56 mm Hg C2 = ? C2 = 0.15 g O2 /100 mL solution

  13. Solubility = grams of solute/100 grams of solvent If 0.85 g of a gas at 4.0 atm of pressure dissolves in 1.0 L of water at 23oC, how much will dissolve in 1.0 L of water at 1.0 atm of pressure and the same temperature? 0.85/4.0 = S2/1.0 S2 = 0.21 g/L If 0.55 g of a gas dissolves in 1.0 L of water at 20.0 kPa of pressure, how much will dissolve at 110.0 kPa of pressure?

  14. Homework: Read 15.1, pp. 452-461 Q pg. 484, #52, 53, 64, 66, 67

  15. Concentration of a solution-qualitative description Measure of how much solute is dissolved in a specific amount of solvent or solution

  16. Percent Composition(by mass) • To calculate percent by mass of solute in solution, we need • Mass of solute in solution • Mass of solution • Calculate percent by mass • Concentration expressed as percent is ratio of measured amount of solute to measured amount of solution

  17. Example • In order to maintain a sodium chloride concentration similar to ocean water, an aquarium must contain 3.6 g NaCl per 100.0 g of water. What is the percent by mass of NaCl in the solution? 3.6 g NaCl 103.6 g NaCl + H2O x 100 = 3.5% • What is the percent by mass of NaHCO3 in a solution containing 20.0 g NaHCO3 dissolved in 600 mL H2O?

  18. Using percent by volume-both solute and solvent are liquids • Solute volume : solution volume as percent • Volumes of liquids not always additive • Sometimes volumes change when two liquids are mixed together • For example, mixing 70 ml of isopropyl alcohol and 30 ml of water will not give you exactly 100 ml of solution • Ethanol/water molecules interact differently with each other than they do with themselves

  19. Examples • Rubbing alcohol is 70% isopropyl alcohol. 70 volumes of isopropyl alcohol are dissolved in every 100 volumes of solution. So 30 volumes of water are in every 100 volumes of the rubbing alcohol. • What is % by volume of ethanol in a solution that contains 35 mL of ethanol dissolved in 115 mL of water? (23%) • If you have 100.0 mL of a 30.0% aqueous solution of ethanol, what volumes of ethanol and water are in the solution? (30 mL ethanol, 70 mL water) • What is % by volume of isopropyl alcohol in a solution that contains 24 mL of isopropyl alcohol in 1.1 L of water? (2%)

  20. More Examples • What is % concentration of a solution that you made by taking 5.85 g of NaCl and diluting to 100 mL with H20? • 5.85 g/100 mL x 100 = 5.85% W/V solution of NaCl • What is % concentration of a solution that you made by taking 40 g of CaCl2 and diluting to 500 mL with H20? • 40 g/500 mL x 100 = 8% • How would you make 250 mL of a 8.5% NaCl solution? • x/250 mL = 8.5% x = 21.3 g • weigh out 21.3 g NaCl and dilute to 250 mL with H20. • How much 0.85% NaCl may be made from 2.55 g NaCl? • 2.55 g/x = .85% x = 300 mL

  21. A solution is made up of two partssolute and solvent • Powdered ice tea mix-4 scoops powder for every 2 quarts H2O • Normal recipe for ice tea mix looks like: 4 scoops of powderTo make ice tea of "normal" strength = -----------------------                                                         2 quarts of water • Make ice tea 2x strong as normal-use 8 scoops of powder w/2 quarts of water, or 4 scoops of powder w/1 quart water: • To make ice tea 2x "normal" strength = 8 scoops of powder 4 scoops of powder ---------------------------- or ----------------------------- 2 quarts of water          1 quart of water

  22. Molarity • Need more exact strength of solution when doing quantitative analysis in lab  • Unit for molarity is M and read as "molar" (3 M = three molar) • "Moles" measures quantity of material • "molarity" measures concentration of that material • Concentration of substance: measure of how much solute is dissolved in specific amount of solvent or solution • Concentration of substance in solution: number of moles of that substance moles per unit volume of solution

  23. Example of molarity • Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into some water. It would dissolve and make sugar water. We keep adding water, dissolving and stirring until all the solid was gone. We then made sure that when everything was well-mixed, there was exactly 1.00 liter of solution. • What would be the molarity of this solution? • The answer is 1.00 mol/L. Notice that both the units of mol and L remain. Neither cancels. • A replacement for mol/L is often used. It is a capital M. So if you write 1.00 M for the answer, then that is correct.

  24. Example #2 • Suppose you had 2.00 moles of solute dissolved into 1.00 L of solution. What's the molarity? • The answer is 2.00 M. • Notice that no mention of a specific substance is mentioned at all. The molarity would be the same. It doesn't matter if it is sucrose, sodium chloride or any other substance. One mole of anything contains 6.022 x 1023 units.

  25. Example #3 • What is the molarity when 0.75 mol is dissolved in 2.50 L of solution? • The answer is 0.300 M.

  26. Example #4 • Now, let's change from using moles to grams. This is much more common. After all, chemists use balances to weigh things and balances give grams, NOT moles. • Example #4 - Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 L of solution. What would be the molarity of the solution? • The solution to this problem involves two steps which will eventually be merged into one equation. • Step One: convert grams to moles. • Step Two: divide moles by liters to get molarity. • 58.44 grams/mol is the molecular weight of NaCl. • Dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol. • Then, dividing 1.00 mol by 2.00 L gives 0.500 mol/L (or 0.500 M). Sometimes, a book will write out the word "molar," as in 0.500-molar.

  27. Example #5 • Calculate the molarity of 25.0 grams of KBr dissolved in 750.0 mL. • Note change from mL to L in the answer.

  28. Example #6 • 80.0 grams of glucose (C6H12O6, mol. wt = 180. g/mol) is dissolved in enough water to make 1.00 L of solution. What is its molarity?

  29. Example #7 • How many moles of CaCl2 would be used in the making of 5.00 x 102 cm3 of a 5.0M solution? Notice volume is given in cm3.  Since there are 1000 cm3 in 1 liter, 500 cm3 must be equal to 0.500 liters. • 5.00 x 102cm3 x 1 L/1000 cm3 = 0.500 L • 5 M = x mol/0.500 L = 2.5 mol CaCl2 • How many grams of CaCl2 in above question? • Molar mass = 111 g/mol 2.5 mol x 111 g/mol = 280 g CaCl2

  30. Example #8 • Calculate the molarity of 20 g of NaCl added to water to make 1 L of solution • Molar mass of NaCl • 22.99 + 35.45 = 58.44 g/mol • 20 g/1mol/58.44 g = 0.342 mol • M = 0.342 mol/1L = 0.342 M

  31. Example #9 • Calculate the molarity when 0.25 g of Pb(NO3)2 in enough water to make 500 mL of solution • Molar mass of Pb(NO3)2 =207.2+28.01+96.00 = 331.2 g/mol • 0.25 g x 1mol/331.2 g = 7.5 x 10-4 mol • 500 ml x 1 L/1000 ml = 0.500 L • M = 7.5 x 10-4 mol/0.500 L=1.5 x 10-3 M

  32. 1) Calculate the molarity when 75.0 grams of MgCl2 is dissolved in 500.0 mL of solution. • 2) 100.0 grams of sucrose (C12H22O11, mol. wt. = 342.3 g/mol) is dissolved in 1.50 L of solution. What is the molarity? • 3) 49.8 grams of KI is dissolved in enough water to make 1.00 L of solution. What is the molarity?

  33. When 2.00 grams of KMnO4 (molec. wt = 158.0 g/mol) is dissolved into 100.0 mL of solution, what molarity results? • How many grams of KMnO4 are needed to make 500.0 mL of a 0.200 M solution?

  34. Try Some More! • 1) 10.0 g of acetic acid (CH3COOH) is dissolved in 500.0 mL of solution. What molarity results? • 2) How many mL of solution will result when 15.0 g of H2SO4 is dissolved to make a 0.200 M solution?

  35. Now try this: • If you put a teaspoon of sugar, C12H22O11, or about 5.00 g, into a cup of water (about 250 ml) what would be the molar concentration? • Sugar has a molecular weight of 342 g/mole so 5.00 g is 1.46 x 10-2 moles.

  36. And this one… • If I dissolve 2.38 g of vanillin, C8H8O3, which is the essential ingredient in the vanilla you put into cookies and cakes, into 930 ml of ethyl alcohol, what would be the molar concentration? • Notice that we can use any liquid for the solution. Water is just the most common solvent you're likely to encounter. • Vanillin has a molecular weight of 152 g/mole, so 2.38 grams is 1.57 x 10-2 moles. The resulting solution would be:

  37. Making Dilutions • Concentrated solution: large amount of solute • Dilute solution: small amount of solute • To make dilutions • Increase solvent/solute remains constant • M1 V1 = M2V2 • 53.4 mL of a 1.50 M solution of NaCl is on hand, but you need some 0.800 M solution. How many mL of 0.800 M can you make? • (1.50 mol/L) (53.4 mL) = (0.800 mol/L) (x) • 100. mL-don’t have to convert to L because given in mL

  38. More Examples of Dilutions • 100.0 mL of 2.550 M KBr solution is on hand. You need 0.5500 M. What is the final volume of solution which results? • (2.500 mol/L) (100.0 mL) = (0.5500 mol/L) (x) • x = 4545. mL • How much more water is added? 4545 - 100.0 = 4445 mL • How do you prepare 100ml of 0.40M MgSO4 from a stock solution of 2.0M MgSO4? How much more water needs to be added? • M1 = 2.0M MgSO4         V1 = unknown M2 = 0.40M MgSO4       V2 = 100mL • 2.0 M x = (0.40 M) (100 L) x = 20 mL • Add 80 mL distilled water to 20 mL of 0.40M MgSO4solution

  39. And still more… • How do you prepare 100cm3 of 0.1M CoCl2 from a 0.52M solution of CoCl2? • Add 19.2cm3 of 0.52M CoCl2 solution to a graduate. Add distilled water to make the total volume 100cm3. • How many cubic centimeters of 0.69M Ba(NO3)2 are needed to prepare 200ml of 0.25M solution? • Add 72.5cm3 of 0.69M Ba(NO3)2 solution to a graduate. Add distilled water to make the total volume 200cm3.

  40. Given a standard solution of 2M NH4Br, describe the preparation of 500ml of 0.15M ammonium bromide solution. • Add 37.5ml of 2M NH4Br solution to a graduate. Add distilled water to make the total volume 500ml. • Many solutions are prepared in the laboratory from purchased concentrated solutions.  What volume of concentrated 17.8 M stock sulfuric acid solution would a laboratory technician need to make 2.00 L of 0.200 M solution by dilution of the original, concentrated stock solution? (22 mL)

  41. Molality, m • # moles of solute dissolved in exactly one kilogram of solvent • Need • Moles of solute present in solution • Mass of solvent (in kilograms) in solution

  42. Example • Suppose we had 1.00 mole of sucrose (it's about 342.3 grams) and proceeded to mix it into exactly 1.00 liter water. It would dissolve and make sugar water. We keep adding water, dissolving and stirring until all the solid was gone. We then made sure everything was well-mixed. • What would be the molality of this solution? Notice that my one liter of water weighs 1000 grams (density of water = 1.00 g / mL and 1000 mL of water in a liter). 1000 g is 1.00 kg, so: • The answer is 1.00 mol/kg. Notice that both the units of mol and kg remain. Neither cancels. • A replacement for mol/kg is often used. It is a lower-case m and is often in italics, m.

  43. Example • What is the molality when 0.75 mol is dissolved in 2.50 L of solvent? • The answer is 0.300 m.

  44. Example • Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 kg of pure water (the solvent). What would be the molality of the solution? • The solution to this problem involves two steps. • Step One: convert grams to moles. • Step Two: divide moles by kg of solvent to get molarity. • In the above problem, 58.44 grams/mol is the molecular weight of NaCl. • Dividing 58.44 grams by 58.44 grams/mol gives 1.00 mol. • Then, dividing 1.00 mol by 2.00 kg gives 0.500 mol/kg (or 0.500 m).

  45. Calculate: • Calculate the molality of 25.0 grams of KBr dissolved in 750.0 mL pure water. • 80.0 grams of glucose (C6H12O6, mol. wt = 180. g/mol) is dissolved in 1.00 kg of water. Calculate the molality.

  46. Calcuate the molality when 75.0 grams of MgCl2 is dissolved in 500.0 g of solvent. • 100.0 grams of sucrose (C12H22O11, mol. wt. = 342.3 g/mol) is dissolved in 1.50 L of water. What is the molality? • 49.8 grams of KI is dissolved in 1.00 kg of solvent. What is the molality?

  47. Phrase "of solution" shows up in some of these problems Molarity definition based on volume of solution Temperature-dependent Molality definition does not have volume in it Temperature independent Very useful concentration unit in colligative properties

  48. Mole Fraction, X, of component in solution • Ratio of # moles of that component to total # moles of all components in solution

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