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Solutions Ch 15

Solutions Ch 15. Mixtures (review). Mixture – has variable composition, different types of parts. For example: air, water from the tap, hot chocolate, paint, 14k gold, stainless steel (metal mixtures are called alloys ) Heterogeneous “hetero” means different

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Solutions Ch 15

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  1. SolutionsCh 15

  2. Mixtures (review) • Mixture – has variable composition, different types of parts. For example: • air, • water from the tap, • hot chocolate, • paint, • 14k gold, stainless steel (metal mixtures are called alloys) • Heterogeneous “hetero” means different • Homogenous “homo” means same • Solution = homogenous mixture Homogenized milk Non-homogenized milk

  3. Solutions • The solvent is the substance that the solute dissolves in • So, if I stir sugar into my coffee, which is which? • Once the solute is dissolved, as long as there’s none left at the bottom, what do you call the new coffee? • A homogenous mixture, or solution • What if there’s sugar left at the bottom? That would be called…? • Saturated – there’s more solute than the solvent can dissolve

  4. Solutions • Solution – mixture of two or more substances in a single physical state. • Characteristics: • Homogeneous • Does not settle out • Cannot be separated by filtering

  5. Solutions • Solutions properties are: • the same throughout the solution • different from the individual properties of their constituents (parts that make them up) • i.e., salt and water have different individual properties than salt water. • Solubility refers to a solute’s ability to be dissolved by the solvent. • A substance is said to be insoluble if it stays in its original state in the solvent

  6. Properties of Solutions Solutions can be made up of any combination of states of matter for their solutes and solvents Gas in gas: Oxygen in nitrogen (71 % of air) Solid in liquid: Salt in water (= sea water) Solid in solid: Copper in mercury to make dental amalgam All metal solutions are called ____ Alloys • Can you think of an example for each? • Liquid in liquid: • concentrated syrup in water to make soda mix • Gas in liquid: • CO2 in soda • (less gas can dissolve into a warm soda than a cold soda – that’s why warm sodas loose their fizz)

  7. Properties of Solutions • concentration = amt. of solute in a solvent. A concentrated solution has a relatively high amount of solute. • There is a limit to the amount of solute that can be dissolve into a solution. • Eventually, the solution will contain so much solute that no more solute can dissolve into it. • At this point, the solution is considered to be • Saturated - the solution is at its max. concentration for the given temperature and pressure. • You can tell a solution is saturated when the solute precipitates out.

  8. Properties of solutions • Unsaturated solutions still have “room” for more solute to dissolve into them. • Anyone who has ever tried to dissolve sugar into iced tea knows that it is much harder to do than if the tea were hot. • Iced tea simply has a lower saturation point than hot tea does. • So the saturation points can vary with temperature and pressure. • Usually, the hotter the solution, the higher the saturation point. This is not true for gases dissolved into liquids, though – why?

  9. Properties of solutions • A supersaturated solution actually holds more solute than its saturation point would dictate. • A solution can become supersaturated when it is brought to saturation, heated to raise the saturation point, and then allowed to cool carefully. • At the cooler temperature, any slight disturbance will trigger the precipitation of the solid back out of the solution.

  10. Solubility • Water is the universal solvent because of its polarity • Substances that are dissolved in water are said to be aqueous. • Most Ionic compounds are soluble in water • Why? • Ionic compounds are polar • The polarity of the water molecules pulls the ionic compounds apart.

  11. solubility • Some covalent compounds dissolve in water because of their polarity. • Sucrose (table sugar) is polar, and dissolves. • Covalent substances that are nonpolar are insoluble in water • Oil and other fats are non-polar and don’t dissolve in water. • Soap is non-polar, so dissolves grease. • Like dissolves like. • Polar dissolves polar. • Nonpolar dissolves nonpolar

  12. Rate of Dissolving • Affected by 3 factors: • Temperature • Agitation (mixing) • Surface area of solute

  13. molarity • Number of moles of solute dissolved in a liter of solution • Represented by M • i.e. To make a 1M solution of aqueous NaCl, measure a mole (58.44g) of NaCl and add water to 1 Liter mark. • Example problem: 5.7 g KNO3 in 233 mL solution. What is the solution molarity?

  14. molarity • Example problem: 5.7 g KNO3 in 233 mL solution. What is the solution molarity? • Convert grams to mole of solute • 5.7 g KNO3 mole KNO3 = 0.056 mol KNO3 • 101.11 g KNO3 • .056 mol KNO3 1000mL = 0.24 M KNO3 • 233mL 1 L

  15. molality • Number of moles of solute dissolved in 1 kg of solvent • m = mol/kg • Useful for studying colligative properties • 1 liter of water = 1 kg • All other solvents must be weighed, as they have a different specific gravity and molality is about the mass of solvent.

  16. molality • To make a 1m aqueous solution of NaCl, measure 1 kg (=1 L) of water (solvent) and add in 1 mole (58.44g) NaCl

  17. molality • What is the molality of a dentist’s amalgam? Amalgam is made up of 70% mercury and 30% copper by mass. • Determine which is solute and which is solvent. Mercury is solvent, since there is more mass of it. • So convert the copper to moles • 30 g Cu mol = .47 mol Cu • 63.55 g • Now convert to molality units of mol/kg • .47 mol Cu 1000g = 6.7 mol = 6.7 m • 70 g Hg kg kg

  18. Properties of Solutions • Some physical properties of solutions are dependent upon the solute to solvent ratio, or the molality of the solution. • Freezing point and melting point are examples of these colligative properties • Another factor affecting the freezing and boiling points is the dissociation factor of the solute • The dissociation factor is the number of particles a solute breaks into when it dissolves. • If the solute dissociates, it has a greater effect on the temperature change. • The freezing point is always lowered • And the boiling point is always raised

  19. Calculating changes in boiling and freezing points • The amount of change of the freezing point is Δ TF • The amount of change of the boiling point is Δ TB • The dissociation factor is d.f. • i.e. NaCl’sd.f. is 2 • Covalent compounds’ d.f.’s are 1, since they stay whole in solutions. • In addition, every substance has a certain constant which has been experimentally determined and is always given, kF and kB • The constant is related to the properties of the solvent

  20. Calculating colligative property • To calculate the change in boiling or freezing point based on the molality: • Δ TF = m d.f. kF • Δ TB = m d.f. kB • What is the new boiling point of water if 100. g of glucose (C6H12O6) is dissolved into 750. mL of H2O?

  21. Calculating Colligative Property • What is the new boiling point of water if 100. g of glucose (C6H12O6) are dissolved into 750. mL of H2O? • Δ TB = m d.f. kB , kBof H2O = 0.52 °C/m • First calculate the molality • m = moles solute/mass solvent • mol = 100 g glucose 1 mol = 0.554 mol glucose • 180.56 g • d.f. of sugar is 1, as it is a covalent molecule • Δ TB = 0.554 mol x 1 x 0.52 °C L = 0.384 °C • 0.750 L 1 mol

  22. Dilution • Usually acids come from chemical supply companies in highly concentrated forms. • And you have to dilute them to a lower molarity. • You NEVER add the water to acid • So you have to have a way to calculate how much water to pour the concentrated acid into to dilute it.

  23. Dilution • When we dilute a substance, it is always assumed that we are adding water, the universal solvent. • The actual number of moles of the solute never changes. • Only the amount of water changes to reduce the molarity of the solution. • Extra dilution calculations: • Text p. 556 #33 & 34

  24. Dilution Calculations • When you need to dilute a solution, you know 3 of the following 4 things: • the molarity of what you started with (M1), • the new molarity of the solution you want to make (M2) • The volume of the solution you start with (V1) • The volume of the new solution (V2) • Use the dilution equation M1V1= M2V2 to solve for the last variable.

  25. Dilution Calculations • Remember that V1 is the volume of original concentrated acid that must be added to water. • And that V2 is the amount of diluted solution. • So the amount of water to add is: • final volume (V2) – amt. of concentrated solution (V1)

  26. Dilution Calculations • #31a in text p. 556 • Calculate the molarity that results when 250 mL of water is added to 125 mL of .251 M HCl. • First assign variables: • V1 = 125 mL • V2 = 125 mL + 250 mL = 375 mL • M1 = .251 mol/liter • M2 = ? • Then plug into formula and solve for unknown • M2 = M1V1/V2 = (.251 M)(125 mL)/375 mL • M2 = .0837 M

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