![%100
solutionsgrams
solutegrams
W
W
% ×=
6.2ConcentrationBasedon
Mass
Weight/Weight Percent
• Weight/weight percent is most useful for
solutions of 2 solids whose masses are
easily obtained
• Calculate % (W/W) of platinum in gold
ring with 14.00 g Au and 4.500 g Pt
[4.500 g Pt / (4.500 g Pt + 14.00 g Au)] x 100%
= 4.500 g / 18.50 g x 100% = 24.32% Pt](https://image.slidesharecdn.com/mecchapter6-120814200155-phpapp01-180120104443/85/Mecchapter6-120814200155-phpapp01-14-320.jpg)
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- 1. Chapter 6 Solutions Denniston Topping Caret 7th Edition Copyright© The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
- 2. 6.1 Properties of Solutions • Solution - homogeneous mixture • Solute - the substance in the mixture present in lesser quantity • Solvent - the substance present in the largest quantity • Aqueous solution - solution where the solvent is water • Solutions can be liquids as well as solids and gases
- 3. Examples of Solutions • Air - oxygen and several trace gases are dissolved in the gaseous solvent, nitrogen • Alloys - brass and other homogeneous metal mixtures in the solid state • Focus on liquid solutions as many important chemical reactions take place in liquid solutions 6.1PropertiesofSolutions
- 4. • Clear, transparent, no visible particles • May have color • Electrolytes are formed from solutes that are soluble ionic compounds • Nonelectrolytes do not dissociate • Volumes of solute and solvent are not additive – 1 L ethanol + 1 L water does not give 2 L of solution 6.1PropertiesofSolutions )(-Cl)(Na)NaCl( OH2 aqaqs +→ + General Properties of Liquid Solutions
- 5. 6.1PropertiesofSolutions Solutions and Colloids • Colloidal suspension - contains solute particles which are not uniformly distributed – Due to larger size of particles (1nm - 200 nm) – Appears identical to solution from the naked eye – Smaller than 1 nm, have solution – Larger than 1 nm, have a precipitate
- 6. 6.1PropertiesofSolutions Degree of Solubility • Solubility - how much of a particular solute can dissolve in a certain solvent at a specified temperature • Factors which affect solubility: 1 Polarity of solute and solvent • The more different they are, the lower the solubility 2 Temperature • Increase in temperature usually increases solubility 3 Pressure • Usually has no effect • If solubility is of gas in liquid, directly proportional to applied pressure
- 7. 6.1PropertiesofSolutions Saturation • Saturated solution - a solution that contains all the solute that can be dissolved at a particular temperature • Supersaturated solution - contains more solute than can be dissolved at the current temperature • How is this done? • Heat solvent, saturate it with solute then cool slowly • Sometimes the excess will precipitate out • If it doesn’t precipitate, the solution will be supersaturated
- 8. 6.1PropertiesofSolutions Solubility and Equilibrium • If excess solute is added to a solvent, some dissolves • At first, rate of dissolution is large • Later, reverse reaction – precipitation – occurs more quickly • When equilibrium is reached the rates of dissolution and precipitation are equal, there is some dissolved and some undissolved solute • A saturated solution is an example of a dynamic equilibrium
- 9. 6.1PropertiesofSolutions Solubility of Gases: Henry’s Law • Henry’s law – the number of moles of a gas dissolved in a liquid at a given temperature is proportional to the partial pressure of the gas above the liquid • Gas solubility in a liquid is directly proportional to the pressure of the gas in the atmosphere in contact with the liquid • Gases are most soluble at low temperatures • Solubility decreases significantly at higher temperatures – Carbonated beverages – CO2 solubility less when warm – Respiration – facilitates O2 and CO2 exchange in lungs
- 10. 6.2 Concentration Based on Mass • Concentration - amount of solute dissolved in a given amount of solution • Concentration of a solution has an effect on – Physical properties • Melting and boiling points – Chemical properties • Solution reactivity 6
- 11. • Amount of solute = mass of solute in grams • Amount of solution = volume in milliliters • Express concentration as a percentage by multiplying ratio by 100% = weight/volume percent or % (W/V) %100 solutionofsmilliliter soluteofgrams V W % ×= solutionofamount soluteofamount ionconcentrat = Weight/Volume Percent 6.2ConcentrationBasedon Mass
- 12. 6.2ConcentrationBasedon Mass Calculating Weight/Volume Percent Calculate the percent composition or % (W/V) of 2.00 x 102 mL containing 20.0 g sodium chloride 20.0 g NaCl, mass of solute 2.00 x 102 mL, total volume of solution % (W/V) = 20.0g NaCl / 2.00 x 102 mL x 100% = 10.0% (W/V) sodium chloride
- 13. 6.2ConcentrationBasedon Mass Calculate Weight of Solute from Weight/Volume Percent Calculate the number of grams of glucose in 7.50 x 102 mL of a 15.0% solution 15.0% (W/V) = Xg glucose/7.50 x 102 mL x 100% Xg glucose x 100% = (15.0% W/V)(7.50 x 102 mL) Xg glucose = 113 g glucose %100 solutionofsmilliliter soluteofgrams V W % ×=
- 14. %100 solutionsgrams solutegrams W W % ×= 6.2ConcentrationBasedon Mass Weight/Weight Percent • Weight/weight percent is most useful for solutions of 2 solids whose masses are easily obtained • Calculate % (W/W) of platinum in gold ring with 14.00 g Au and 4.500 g Pt [4.500 g Pt / (4.500 g Pt + 14.00 g Au)] x 100% = 4.500 g / 18.50 g x 100% = 24.32% Pt
- 15. 6.3 Concentration of Solutions: Moles and Equivalents • Chemical equations represent the relative number of moles of reactants producing products • Many chemical reactions occur in solution where it is most useful to represent concentrations on a molar basis
- 16. • The most common mole-based concentration unit is molarity • Molarity – Symbolized M – Defined as the number of moles of solute per liter of solution Molarity6.3MolesandEquivalents solutionL solutemoles =M
- 17. • Calculate the molarity of 2.0 L of solution containing 5.0 mol NaOH • Use the equation • Substitute into the equation: MNaOH = 5.0 mol solute 2.0 L solution = 2.5 M Calculating Molarity from Moles6.3MolesandEquivalents solutionL solutemoles =M
- 18. • If 5.00 g glucose are dissolved in 1.00 x 102 mL of solution, calculate molarity, M, of the glucose solution • Convert from g glucose to moles glucose – Molar mass of glucose = 1.80 x 102 g/mol 5.00 g x 1 mol / 1.80 x 102 g = 2.78 x 10-2 mol glucose – Convert volume from mL to L 1.00 x 102 mL x 1 L / 103 mL = 1.00 x 10-1 L • Substitute into the equation: Mglucose = 2.78 x 10-2 mol glucose 1.00 x 10-1 L solution = 2.78 x 10-1 M Calculating Molarity From Mass6.3MolesandEquivalents solutionL solutemoles =M
- 19. solutionL solutemoles =M 6.3MolesandEquivalents Dilution Dilution is required to prepare a less concentrated solution from a more concentrated one – M1 = molarity of solution before dilution – M2 = molarity of solution after dilution – V1 = volume of solution before dilution – V2 = volume of solution after dilution moles solute = (M)(L solution)
- 20. • In a dilution will the number of moles of solute change? – No, only fewer per unit volume • So, • Knowing any three terms permits calculation of the fourth M1V1 = M2V2 6.3MolesandEquivalents Dilution
- 21. • Calculate the molarity of a solution made by diluting 0.050 L of 0.10 M HCl solution to a volume of 1.0 L – M1 = 0.10 M molarity of solution before dilution – M2 = X M molarity of solution after dilution – V1 = 0.050 L volume of solution before dilution – V2 = 1.0 L volume of solution after dilution • Use dilution expression • X M = (0.10 M) (0.050 L) / (1.0 L) 0.0050 M HCl OR 5.0 x 10-3 M HCl M1V1 = M2V2 6.3MolesandEquivalents Calculating Molarity After Dilution
- 22. 6.3MolesandEquivalents Representation of Concentration of Ions in Solution Two common ways of expressing concentration of ions in solution: 1. Moles per liter (molarity) • Molarity emphasizes the number of individual ions 2. Equivalents per liter (eq/L) • Emphasis on charge
- 23. iononchargesofnumber (g)ionofmassmolar ionanofequivalentOne = 6.3MolesandEquivalents Comparison of Molarity and Equivalents 1 M Na3PO4 • What would the concentration of PO4 3- ions be? • 1 M • Equivalent is defined by the charge • One Equivalent of an ion is the number of grams of the ion corresponding to Avogadro’s number of electrical charges
- 24. 6.3MolesandEquivalents Molarity vs. Equivalents – 1 M Na3PO4 • 1 mol Na+ = 1 equivalent Na+ • 1 mol PO4 3- = 3 equivalents PO4 3- • Equivalents of Na+ ? – 3 mol Na+ = 3 equivalents of Na+ • Equivalents of PO4 3- ? – 1 mol PO4 3- = 3 equivalents of PO4 3-
- 25. 6.3MolesandEquivalents Calculating Ion Concentration • Calculate eq/L of phosphate ion, PO4 3- in a solution with 5.0 x 10-3 M phosphate • Need to use two conversion factors: – mol PO4 3- mol charge – mol charge eq PO4 3 5.0 x 10-3 mol PO4 3- x 3 mol charge x 1 eq 1 L 1 mol PO4 3- 1mol charge • 1.5 x 10-2 eq PO4 3- /L
- 26. 6.4 Concentration-Dependent Solution Properties • Colligative properties - properties of solutions that depend on the concentration of the solute particles, rather than the identity of the solute • Four colligative properties of solutions 1. vapor pressure lowering 2. boiling point elevation 3. freezing point depression 4. osmotic pressure
- 27. Vapor Pressure of a Liquid Consider Raoult’s law in molecular terms • Vapor pressure of a solution results from escape of solvent molecules from liquid to gas phase • Partial pressure of gas phase solvent molecules increases until equilibrium vapor pressure is reached • Presence of solute molecules hinders escape of solvent molecules, lowering equilibrium vapor pressure 6.4Concentration-Dependent SolutionProperties
- 28. 6.4Concentration-Dependent SolutionProperties Vapor Pressure Lowering • Raoult’s law - when a nonvolatile solute is added to a solvent, vapor pressure of the solvent decreases in proportion to the concentration of the solute • Solute molecules (red below) serve as a barrier to the escape of solvent molecules resulting in a decrease in the vapor pressure
- 29. 6.4Concentration-Dependent SolutionProperties Freezing Point Depression and Boiling Point Elevation • Freezing point depression may be explained considering the equilibrium between solid and liquid states – Solute molecules interfere with the rate at which liquid water molecules associate to form the solid state • Boiling point elevation can be explained considering the definition as the temperature at which vapor pressure of the liquid equals the atmospheric pressure – If a solute is present, then the increase in boiling temperature is necessary to raise the vapor pressure to atmospheric temperature
- 30. • Freezing point depression (∆Tf) - is proportional to the number of solute particles – Solute particles, not just solute • How does an electrolyte behave? – Dissociate into ions • An equal concentration of NaCl will affect the freezing point twice as much as glucose (a nonelectrolyte) • Each solvent has a unique freezing point depression constant or proportionality factor ∆Tf=kf m 6.4Concentration-Dependent SolutionProperties Freezing Point Depression
- 31. • Boiling point elevation (∆Tb) - is proportional to the number of solute particles • An electrolyte will affect boiling point to a greater degree than a nonelectrolyte of the same concentration • Each solvent has a unique boiling point elevation constant ∆Tb=kb m 6.4Concentration-Dependent SolutionProperties Boiling point elevation
- 32. Osmotic Pressure • Some types of membranes appear impervious to matter, but actually have a network of small holes called pores • These pores may be large enough to permit small solvent molecules to move from one side of the membrane to the other • Solute molecules cannot cross the membrane as they are too large • Semipermeable membrane - allows solvent but not solute to diffuse from one side to another 6.4Concentration-Dependent SolutionProperties
- 33. 6.4Concentration-Dependent SolutionProperties Osmotic Pressure • Osmosis - the movement of solvent from a dilute solution to a more concentrated solution through a semipermeable membrane • Requires pressure to stop this flow
- 34. • Osmotic pressure (π) - the amount of pressure required to stop the flow across a semipermeable membrane • Osmolarity - the molarity of particles in solution – Osmol, used for osmotic pressure calculation π=MRT 6.4Concentration-Dependent SolutionProperties Osmotic Pressure
- 35. 6.4Concentration-Dependent SolutionProperties Tonicity and the Cell • Living cells contain aqueous solution and these cells are also surrounded by aqueous solution • Cell function requires maintenance of the same osmotic pressure inside and outside the cell • Solute concentration of fluid surrounding cells higher than inside results in a hypertonic solution causing water to flow into the surroundings, causing collapse = crenation • Solute concentration of fluid surrounding cells too low, results in a hypotonic solution causing water to flow into the cell, causing rupture = hemolysis • Isotonic solutions have identical osmotic pressures and no osmotic pressure difference across the cell membrane
- 37. Pickling Cucumber in Hypertonic Brine Due to Osmosis 6.4Concentration-Dependent SolutionProperties
- 38. 6.5 Water as a Solvent • Water is often referred to as the “universal solvent” • Excellent solvent for polar molecules • Most abundant liquid on earth • 60% of the human body is water – transports ions, nutrients, and waste into and out of cells – solvent for biochemical reactions in cells and digestive tract – reactant or product in some biochemical processes