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Unit 11- Solubility

Unit 11- Solubility. Water & Solutions. I. Water. A. The Molecule 1. O—H bond is highly polar 2. Bond angle 105 ° making it Bent shaped 3. Water Molecule as a whole is polar 4. Attracted to each other by intermolecular hydrogen bonds. I. Water (cont.). B. Important Properties

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Unit 11- Solubility

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  1. Unit 11- Solubility Water & Solutions

  2. I. Water A. The Molecule 1. O—H bond is highly polar 2. Bond angle 105° making it Bent shaped 3. Water Molecule as a whole is polar 4. Attracted to each other by intermolecular hydrogen bonds

  3. I. Water (cont.) B. Important Properties 1. High surface tension 2. low vapor pressure • hydrogen bonds hold molecules to one another, tendency to escape surface is low 3. high specific heat capacity • 4.18 J/g×°C 4. high melting and boiling points • 0°C and 100°C

  4. I. Water (cont.) C. Surface Tension – inward force, or pull, that tends to minimize the surface area of a liquid • Surfactant – wetting agent such as soap or detergent that decreases the surface tension

  5. I. Water (cont.) D. Atypical Ice 1. As a typical liquid cools, density increases b/c Volume decreases as the mass stays constant 2. As water cools it first behaves like a typical liquid until it reaches 4°C 3. Below 4°C the density of water starts to decrease **Ice is one of only a few solids that float in their own liquid.

  6. Atypical Ice Do not need to write. °

  7. Atypical Ice Why does ice behave differently? Open framework arranged like a honeycomb. Framework collapses, molecules packed closer together, making it more dense

  8. II. The Solution Process A.Solution - homogeneous mixture Solute- substance being dissolved Solvent- dissolving medium

  9. II. The Solution Process B. Solvation– the process of dissolving 1st solute particles are surrounded by solvent particles First... 2nd solute particles are separated and pulled into solution Then...

  10. II. The Solution Process C. “Like Dissolves Like” Polar solvents dissolve polar molecules and ionic compounds Nonpolar solvents dissolve nonpolar compounds

  11. A. Electrolytes – compounds that conduct an electric current in solutions III. Electrolytes • All ionic compounds are electrolytes • Compounds that don’t conduct an electric current are called nonelectrolytes – not composed of ions, includes many molecular compounds (covalent bonds)

  12. - + - - + + acetic acid salt sugar Non- Electrolyte Weak Electrolyte Strong Electrolyte solute exists as ions and molecules solute exists as ions only solute exists as molecules only solute exists as ions and molecules solute exists as ions only solute exists as molecules only

  13. IV. Heterogeneous Mixtures A. Suspensions – mixtures from which particles settle out upon standing and the average particle size is greater than 100 nm in diameter. • Clearly identified as two substances • Gravity or filtration will separate the particles B. Colloids – heterogeneous mixtures containing particles that are between 1 nm and 100 nm in diameter • Appear to be homogeneous but particles are dispersed through medium • Ex: paint, aerosol spray, smoke, marshmallow, whipped cream

  14. IV. Heterogeneous Mixtures C. Tyndall Effect - phenomenon observed when beam of light passes through a colloid or suspension • Colloids exhibit the Tyndall effect Colloid Solution

  15. V. Solubility • defined as the maximum grams of solute that will dissolve in 100 g of solvent at a given temperature • based on a saturated solution

  16. Solubility SATURATED SOLUTION max amount no more solute dissolves SUPERSATURATED SOLUTION over max amount becomes unstable, crystals form UNSATURATED SOLUTION capable of dissolving more solute • Supersaturated solutions are not in equilibrium with the solid substance and can quickly release the dissolved solids. • Saturated solution is one that is in equilibrium with respect to the dissolved substance. These conditions can quickly change with temperature. Concentration Increasing

  17. V. Solubility (cont.) A. Factors Affecting Solubility 1. Stirring (agitation) • Increases solubility b/c if fresh solvent is brought in contact with the surface of the solute

  18. V. Solubility (cont.) 2. Temperature • Increases solubility by increasing kinetic energy, which increases the collisions between molecules of solvent and the surface of the solute

  19. V. Solubility (cont.) 3. Surface Area • A smaller particle size dissolves more rapidly than larger particles size • Surface phenomenon • More surface area exposed, faster rate of dissolving

  20. V. Solubility (cont.) B. Solubility Curve • shows the dependence of solubility on temperature • Note: the solubility of the gases are greater in cold water than in hot water Out of the solids, which has the lowest solubility at 40°C? KClO3 Which has the highest? NaNO3

  21. Output side 1. Which of the following compounds dissolved the highest at 20°C? 2. The lowest at 20°C? 3. Overall which compound dissolved the fastest? 4. Name a compound in the graph that is a gas? How do you know it’s a gas?

  22. V. Solubility (cont.) • Solids are more soluble at... • high temperatures. • Gases are more soluble at... • low temperatures & • high pressures (Henry’s Law). • EX: the “bends” & soda When opened partial pressure of CO2 liquid decreases and the concentration of dissolved CO2 decreases.

  23. V. Solubility (cont.) C. Henry’s Law – states that at a given temperature the solubility (S) of a gas in a liquid is directly proportional to the pressure (P) S1 = S2 P1 P2 If the solubility of a gas in water is 0.77 g/L at 3.5 atm of pressure, what is its solubility at 1.0 atm of pressure? (The temperature is held at 25°C) S1 = 0.77 g/L = X P1 = 3.5 atm P2 = 1.0 atm 0.22 g/L

  24. VI. Concentrations of Solutions • Concentrationof a solution is a measure of the amount of solute that is dissolved in a given quantity of solution. • Dilute solution – contains a low concentration of solute. • Concentrated solution – contains a high concentration of solute. • Molarity (M)– number of moles of a solute dissolved per liter of solution • a.k.a. molar concentration

  25. A. Molarity • Calculate the number of moles in 1 L of the solution Molarity (M) = moles of solute liters of solution Example 1 Calculate the molarity when 2 mol of glucose is dissolved in 5 L of solution, divide the number of moles by the volume in liters. 2 mol glucose 5 L solution = 0.4 mol/L = 0.4 M

  26. Example 2 How many moles of solute are present in 1.5 L of 0.24M Na2SO4? • volume of solution = 1.5L • Solution concentration = 0.24 M Unknown: moles solute = ? mol M = n of solute L of solution “Triangle Trick” Divide 0.24M = x 1.5 L X 1.5L 0.24M X = 0.36 mol Multiply

  27. Example 3 A saline solution contains 0.90 g NaCl in exactly 100mL of solution. What is the molarity of the solution? 1. CONVERT GRAMS TO MOLES!! 2. CONVERT mL to L. UNKNOWN: Molarity 0.90g NaCl 1 mol = molar mass M = 0.016 mol 0.1 L M = n of solute L of solution 100 mL 1L = 0.1 L 1000mL 0.016 mol 58g Molarity = 0.16 M Na 23 + Cl 35

  28. 0.650M = n 2.40 L Question 4 How many grams of solute are in 2.40L of 0.650M HClO2? M = n L L = 2.40L M = 0.650M n = 1.56 mol 68g 1.56 mol HClO2 = 106g 1 mol

  29. Molarity ProblemsYou do not have to write the problem. You MUST show your work. 1. A solution has a volume of 2.0L and contains 36.0g of glucose. If the molar mass of glucose is 180 g, what is the molarity of the solution? 2. A solution has a volume of 250 mL and contains 0.70 mol NaCl. What is its molarity? 3. How many moles of ammonium nitrate are in 335 mL of 0.425M NH4NO3? 4. How many moles of solute are in 250 mL of 2.0M CaCl2? How many grams of CaCl2 is this?

  30. VI. Concentrations of Solutions (cont.) B. Making Dilutions • You can make a solution less concentrated by diluting it with solvent. • A dilution reduces the moles of solute per unit volume, however, the total moles of solute in solution does not change Moles of solute before dilution = moles of solute after dilution M1× V1 = M2 × V2 • M1 & V1 are initial and M2 & V2 are final of the SAME substance • Volumes can be in L or mL, as long as the same units are used for both V1 & V2

  31. Example 1 How many milliliters of a stock solution of 2.00M MgSO4 would you need to prepare 100.0 mL of 0.400M MgSO4? M1 = 2.00M MgSO4 M2 = 0.400M MgSO4 V2 = 100 mL of 0.400M MgSO4 Unknown = V1 M1× V1 = M2 × V2 V1 = 0.400M × 100 mL 2.00M × V1= 20 mL

  32. Example 2 You need 150 mL of 0.40M NaCl and you have a 1.0M of NaCl solution. Calculate the volume of the NaCl solution. V1 = 150 mL of 0.40M NaCl M1 = 0.400M NaCl M2 = 1.0M NaCl Unknown = V2 M1× V1 = M2 × V2 0.400M × 150 mL = 1.0M × V2 60. mL = V2

  33. Example 3 What volume of 5.00M sulfuric acid is required to prepare 25.00L of 0.400M sulfuric acid? M1 = 5.00M V2 = 25.00L M2 = 0.400M M1× V1 = M2 × V2 5.0M × V1 = 0.40M × 25 L V1 = 2.0 L

  34. Title: Dilution ProblemsYou do not have to write the problem. You MUST show your work. You have the following stock solutions available: 2.00M NaCl, 4.00M KNO3 and 0.50M MgSO4. Calculate the volumes you must dilute to make the following solution. 1. 250.0 mL of 0.300M NaCl 2. 75.0 mL of 0.200M KNO3 3. 5.0 L of 0.2M MgSO4

  35. VI. Concentrations of Solutions (cont.) C. Percent Solutions • If both solute & solvent are liquids Percent by volume (% v/v) = volume of solute × 100% solution volume • If a solid is dissolved in a liquid Percent (mass/volume) (%(m/v)) = mass of solute (g) × 100% solution volume (mL) Must be the same unit: mL or L Must be this unit

  36. % (v/v) = volume of solute × 100% volume of solution Example 1 What is the percent by volume of ethanol (C2H6O) or ethyl alcohol, in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water? Volume of solute = 85 mL Volume of solution = 250 mL % (v/v) = 85 mL ethanol × 100% 250 mL solution = 34% ethanol

  37. Example 2 How many grams of glucose (C6H12O6) would you need to prepare 2.0 L of 2.8% glucose (m/v) solution? Solution volume = 2.0 L → change to mL Percent by mass = 2.8% • Percent (mass/volume) (%(m/v) = mass of solute (g) × 100% • solution volume (mL) 2.0L 1000mL = 2,000 mL 1L 2.8% = mass of solute (g) × 100% 2,000 mL 0.028 = X 2,000 mL X = 56 g of solute 100% 100%

  38. Percent Solution ProblemsYou do not have to write the problem. You MUST show your work. 1. What is the concentration, in percent (m/v), of a solution with 75g K2SO4 in 1500mL of solution? 2. A bottle of hydrogen peroxide antiseptic is labeled 3.0% (v/v). How many mL H2O2 are in a 400.0 mL bottle of this solution? 3. Calculate the grams of solute required to make 250 mL of 0.10% MgSO4 (m/v).

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