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Ch Types of Mixtures
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Heterogeneous vs. Homogeneous Mixtures
Heterogeneous Mixture: mixture does not have a uniform composition. Ex: Milk and soil Homogeneous Mixture: entire mixture has the same or uniform composition. Ex: Salt water
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Solutions Soluble: capable of being dissolved.
Ex. Sugar is soluble in water. Sugar and water create a solution, or a homogeneous mixture of two or more substances in a single phase. Solvent: the thing that does the dissolving. Solute: the thing that is being dissolved.
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Copper in Nickel (alloy)
Solutions may exist as gases, liquids, or solids, and may also be combinations. Solute State Solvent State Example Gas Oxygen in Nitrogen Liquid CO2 in Water Alcohol in Water Solid Mercury in Silver & Tin Sugar in Water Copper in Nickel (alloy)
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Suspensions Suspension: When the particles in a solvent are so large that they settle out unless the mixture is constantly agitated. Ex: Muddy water The particles in a suspension can be separated by passing the mixture through a filter.
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Colloids Particles that are intermediate in size between those in solutions and suspensions form mixtures called colloids. These are also known as emulsions and foams and cannot be separated using a filter. Ex. Mayonnaise and Milk Tyndall Effect: when light is scattered by the particles in a colloid.
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Solutes: Electrolytes vs. Nonelectrolytes
Electrolyte: a substance that dissolves in water to give a solution that conducts an electric current. Nonelectrolyte: a substance that dissolves in water to give a solution that doesn’t conduct an electric current.
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Ch 12.2 The Solution Process
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Factors Affecting Dissolution Rate
The compositions of the solvent and the solute determine whether a substance will dissolve. Three factors that affect dissolving rate: Stirring (agitation) Temperature Surface area of the dissolving particles.
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Solubility Solution Equilibrium: the physical state in which the opposing processes of dissolution and crystallization of a solute occur at equal rates. Solubility tells us how much solute can dissolve in a certain amount of solvent at a particular temperature and pressure to make a saturated solution. Expressed in grams of solute per 100 grams of solvent
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Saturated Solution: the solution cannot hold any more solute.
Unsaturated Solution: the solution could still dissolve more solute. Supersaturated Solution: the solution is holding more than it should at the given temperature, and if you messed with the solution by shaking it or adding even one more crystal of solute, the whole thing would crystallize rapidly.
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Solubility Values: amount of substance required to form a saturated solution with a specific amount of solvent at a specified temperature. Solubility of sugar is 204 grams per 100 grams of water at 20°C.
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Solute-Solvent Interactions
“Like dissolves Like” Polar will dissolve other polar molecules and Nonpolar dissolves other nonpolar. Hydration: when water is used to dissolve an ionic solution.
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Liquid Solutes and Solvents
Miscible: two liquids that can dissolve in each other. Immiscible: the liquids don’t mix. Ex. Oil and vinegar
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Factors Affecting Solubility
Temperature affects the solubility of: Solid Solutes Liquid Solutes Gaseous Solutes
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Temperature Gas dissolved in a Liquid: as the temperature increases, the solubility decreases. Example: Warm soda loses its carbonation. Solid dissolved in a Liquid: as the temperature increases, the solubility increases. Example: Sugar in hot tea versus iced tea.
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Factors Affecting Solubility
Pressure affects the solubility of: Gaseous Solutes
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Pressure Gas dissolved in Liquid: As pressure increases, solubility increases. Example: Soda is carbonated under high pressure. Solid dissolved in Liquid: As pressure increases, solubility does not change! Since you cannot compress solids and liquids, pressure has no effect on solubility.
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Henry’s Law 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) of the gas above the liquid. So, as the pressure of the gas above the liquid increases, the solubility of the gas increases. S S2 P P2
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Calculating Solubility of a Gas
If the solubility of a gas in water is 0.77 g/L at 3.5 atm of pressure, what is its solubility (g/L) at 1.0 atm of pressure and a constant temperature? P1 = 3.5 atm S1 = 0.77 g/L P2 = 1.0 atm S2 = ? g/L 0.77 g/L = S atm atm S2 = 0.22 g/L
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Enthalpies of Solution
Solvated: when a solute particle is surrounded by solvent molecules. The formation of a solution is accompanied by an energy change, it can be released or absorbed. Enthalpy of solution: the net amount of energy absorbed as heat by the solution when a specific amount of solute dissolves in a solvent.
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Ch 12.3 Concentrations of Solutions
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Concentrations of Solutions
Concentration of a solution: a measure of the amount of solute that is dissolved in a given quantity of solvent. Solutions can be referred to as dilute or concentrated, but these are not very definite terms.
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Molarity Molarity (M): the number of moles of solute dissolved in one liter of solution. Note: it is the total volume in liters of solution, not the liters of solvent.
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Calculating Molarity of a Solution
IV Saline Solutions are 0.90 g NaCl in exactly 100 mL of solution. What is the molarity of the solution? Step 1: convert mL to L (divide by 1000) Step 2: convert the grams of NaCl to moles of NaCl using molar mass. Step 3: put moles of NaCl and L of solution into the molarity equation and divide.
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Finding Moles of Solute
Household bleach is a solution of sodium hypochlorite (NaClO). How many moles of solute are present in 1.5L of 0.70M NaClO? Moles Solute = M x L = mol/L x L Multiply the given volume in L by the molarity expressed in mol/L.
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Molality Another way to express solution concentration is Molality (m)
NOT THE SAME AS MOLARITY! Molality (m) is the concentration of a solution expressed in moles of solute per kilogram solvent.
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m = mol of solute = 0.171 mol of NaCl = 0.285 m NaCl
Calculating Molality of a Solution Calculate the molality of a solution prepared by dissolving 10.0g of NaCl in 600.g of water. 10.0g NaCl mol NaCl 600.0 g kg m = mol of solute kg of solvent = mol of NaCl 0.600 kg of water = m NaCl
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Finding Moles of Solute using molality.
How many moles of sodium fluoride are needed to prepare a 0.40m NaF solution that contains 750.0g of water? mol solute = m x kg of solvent m = mol of solute kg of solvent mol NaF= 0.40 mol x 0.75 kg = 0.30 mol kg
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Making Dilutions Diluting a solution reduces the number of moles of solute per unit volume, but the total number of moles of solute in solution does not change. M1 x V1 = M2 x V2
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Preparing a Dilute Solution
How many mL of 2.00M MgSO4 solution must be diluted with water to prepare 100.0mL of 0.400M MgSO4? Use the dilution formula and plug in the known values and then solve for the unknown. Volume can be in any unit, as long as they are both the same. (Just like gas laws). 0.400 M MgSO4 x mL = 2.00 M MgSO4 x V2 V2 = 20.0 mL
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Chapter 13: Ions in Aqueous Solutions and Colligative Properties
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Section 1: Compounds in Aqueous Solutions
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1 mol of Sodium Ion and 1 mol of Chloride Ion
Dissociation When an ionic compound dissolves in water, the ions separate. To find how many moles of ions are produced, we write a balanced dissociation equation and look at the coefficients in front of the ions. NaCl Na+ + Cl- 1 mol of Sodium Ion and 1 mol of Chloride Ion These are like decomposition reactions.
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Example 1 Write the equation for the dissolution of aluminum sulfate, Al2(SO4)3, in water. How many moles of Al ions and SO4 ions are produced by dissolving 1 mol of Al2(SO4)3? What is the total number of moles of ions produced? Al2(SO4)3 2Al3+ + 3SO42- 2 mol Al3+ and 3 mol SO42- Total moles = = 5 moles
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Example 2 Do the same thing as the last example, except now you are dissolving 2 mols of Al2(SO4)3. 2Al2(SO4)3 4Al3+ + 6SO42- 4 mol Al3+ and 6 mol SO42- Total moles = = 10 moles
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Precipitation Reactions
GENERAL SOLUBILITY GUIDELINES Sodium, potassium, and ammonium compounds are soluble in water. Nitrates, acetates, and chlorates are soluble. Most chlorides are soluble, except those of silver, mercury (I) and lead. Lead (II) chloride is soluble in hot water.
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Precipitation Reactions
GENERAL SOLUBILITY GUIDELINES CONT… Most sulfates are soluble, except those of barium, strontium, lead, calcium, and mercury. Most carbonates, phosphates, and silicates are insoluble, except those of sodium, potassium, and ammonium. Most sulfides are insoluble, except those of calcium, strontium, sodium, potassium, and ammonium.
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Example 3 Sodium Carbonate Calcium Phosphate Cadmium Nitrate
Look at the solubility chart to determine if the following are Soluble or Insoluble? Sodium Carbonate Calcium Phosphate Cadmium Nitrate Ammonium Sulfide
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Example 3 Sodium Carbonate Soluble Calcium Phosphate Cadmium Nitrate
Look at the solubility chart to determine if the following are Soluble or Insoluble? Sodium Carbonate Soluble Calcium Phosphate Cadmium Nitrate Ammonium Sulfide
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Example 3 Sodium Carbonate Soluble Calcium Phosphate Insoluble
Look at the solubility chart to determine if the following are Soluble or Insoluble? Sodium Carbonate Soluble Calcium Phosphate Insoluble Cadmium Nitrate Ammonium Sulfide
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Example 3 Sodium Carbonate Soluble Calcium Phosphate Insoluble
Look at the solubility chart to determine if the following are Soluble or Insoluble? Sodium Carbonate Soluble Calcium Phosphate Insoluble Cadmium Nitrate Soluble Ammonium Sulfide
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Example 3 Sodium Carbonate Soluble Calcium Phosphate Insoluble
Look at the solubility chart to determine if the following are Soluble or Insoluble? Sodium Carbonate Soluble Calcium Phosphate Insoluble Cadmium Nitrate Soluble Ammonium Sulfide Soluble
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Compounds are both soluble…so we continue to step 2.
Example 4 Will a precipitate form when solutions of cadmium nitrate and ammonium sulfide are combined? Step 1: Determine if the compounds are soluble, if soluble, continue to step 2. Compounds are both soluble…so we continue to step 2.
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(NH4)2S + Cd(NO3)2 CdS + 2NH4NO3
Example 4 Will a precipitate form when solutions of cadmium nitrate and ammonium sulfide are combined? Step 2: Write double-displacement reaction between the two compounds. (NH4)2S + Cd(NO3)2 CdS + 2NH4NO3
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CdS or Cadmium Sulfide is insoluble, so it is the precipitate.
Example 4 Will a precipitate form when solutions of cadmium nitrate and ammonium sulfide are combined? (NH4)2S + Cd(NO3)2 CdS + 2NH4NO3 Step 3: Determine if the newly formed compounds are soluble. If one is insoluble, then it is a precipitate. CdS or Cadmium Sulfide is insoluble, so it is the precipitate.
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Net Ionic Equations Includes only those compounds and ions that undergo a chemical change in a reaction in an aqueous solution. Basically, if the ions are part of a soluble product, they don’t end up in the final equation, only the ions for the precipitate that is formed, remain in the equation. The ions that do not take part in the chemical reaction are called spectator ions.
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Example 5 Write the net ionic equation for the production of ammonium nitrate and cadmium sulfide. 2NH4+ + 2NO3- + Cd2+ + S2- CdS + 2NO3- + 2NH4+ If ions show up on both sides of the equation, cross them out and rewrite the equation without them. Cd2+ + S2- CdS
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Ionization Ions are formed from solute molecules by the action of the solvent. Different from dissociation because it involves molecular compounds rather than ionic compounds. In order for ions to form, the strength of the bond within the solute molecule must be weaker than the attractive forces of the solvent molecules.
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The Hydronium Ion H3O+ When a compound ionizes in a solution and releases a H+ ion, it binds to the H2O and forms H3O+. H2O + HCl H3O+ + Cl-
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Electrolytes and Nonelectrolytes
Electrolyte: a compound that conducts an electric current when it is in an aqueous solution or in the molten (liquid) state. All ionic compounds are electrolytes because they dissociate into ions. Nonelectrolyte: a compound that does not conduct an electric current in either aqueous solution or the molten state.
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Strong Electrolyte: nearly all the ionic compound exists as separate ions.
Weak Electrolyte: only a fraction of the ionic compound exists as separate ions.
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Section 2: Colligative Properties of Solutions
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Vapor-Pressure Lowering
The addition of a nonvolatile substance will raise the boiling point and lower the freezing point. This has to do with vapor pressure of the solvent. As the number of solute particles increase, the proportion of solvent molecules decreases.
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Freezing-Point Depression.
When 1 mol of a nonelectrolyte solution is dissolved in 1 kg of water, the freezing point is °C instead of 0.0°C. If 2 mols are dissolved, it is 2 x -1.86°C. This is called the molal freezing-point constant (Kf) and changes for different solvents. Freezing-point depression Δtf is the difference between the two freezing points. Changes according to concentration.
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Example 6 What is the freezing-point depression of water in a solution of g of sucrose, C12H22O11, in 200 g of water? What is the actual freezing point of the solution? Find molality of the sugar solution. Multiply the Kf (of water) by the molality. Take normal freezing point 0°C + Δtf.
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Boiling-Point Elevation
When 1 mol of a nonelectrolyte solution is dissolved in 1 kg of water, the boiling point is °C instead of 100.0°C. An increase of 0.51°C This is called the molal boiling-point constant (Kb) and changes for different solvents. Boiling-point elevation Δtb is the difference between the two boiling points. Changes according to concentration.
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Example 7 What is the boiling-point elevation of a solution made from 20.1 g of a nonelectrolyte solute and grams of water? The molar mass of the solute is 62.0 g/mol.
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Osmotic Pressure A semipermeable membrane allows only water molecules to pass through during osmosis. This can cause an increase in volume on one side of the membrane. Osmotic pressure is the external pressure that must be applied to stop osmosis. The higher the concentration of a solution, the greater the osmotic pressure.
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Electrolytes & Colligative Properties
When electrolytes are dissolved in a solvent, the effects are greater than that of nonelectrolytes. This is because more moles of solute particles are formed when the compounds dissolve.
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For #2, look at the chart on page 448 once you solve for the Kf value.
Homework Ch 13.2 pg 456 #1-4 and 458 #14, 19, 25 For #2, look at the chart on page 448 once you solve for the Kf value.
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