Solutions

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Solutions

• Chapters 20 and 21

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Properties of Solutions, Suspensions, and Colloids
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Solutions

• Solutions a homogeneous mixture of two or more
  substances in a single phase of matter.
• In a simple solution where, for example, salt is
  dissolved in water, the particles of one
  substance are randomly mixed with the particles
  of another substance.

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• The dissolving medium (water) in a solution is
  called the solvent.
• The substance dissolved (salt) in a solution is
  called the solute.
• The solute is generally designated as that
  component of a solution that is of lesser
  quantity.

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• If we had a mixture of 25 mL of ethanol and 75 mL
  of water, the ethanol would be the solute and
  water would be the solvent.
• If we had a 50 to 50 ratio, it would be
  unnecessary to designate solvent or solution.

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• 4 simple solution situations can be considered.
• When deciding what type of solvent to use with a
  given solute it is important to identify what
  types of substances you have.
• 1. Ionic or Polar
• (partial or charges)
• 2. Nonpolar (equal sharing of e-)
• Remember the rule
• Like Dissolves Like

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Solvent-Solute Combinations
Solvent Type Solute Type Is solution likely?
Polar Polar Yes
Polar Nonpolar No
Nonpolar Polar No
Nonpolar Nonpolar Yes

• Remember
• LIKE DISSOLVES LIKE

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The Dissolving Process
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• Water is a polar solvent and is attracted to
  polar solutes.
• Salt is polar (ionic).
• Water molecules surround and isolate the surface
  ions. The ions become hydrated and move away from
  each other in a process called dissociation.

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Suspensions

• If the particles in a solvent are so large that
  they settle out unless the mixture is constantly
  stirred or agitated, the mixture is called a
  suspension.
• Think of a sedimentator (models a river bed).
• If left undisturbed, the larger, denser particles
  sink due to gravity. Particles over 1000nm in
  diameter form suspensions.

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• These particles can be filtered out of the
  heterogeneous mixtures.
• Remember our snow globe lab?

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Colloids

• Particles that are intermediate in size between
  those in solutions and suspensions form mixtures
  known as colloidal dispersions.
• Particles between 1nm and 1000 nm in diameter may
  form colloids. After the larger particles settle
  out (suspensions), the water may still be cloudy
  because colloidal particles remain dispersed in
  the water.
• Milk is an example of a colloid.

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Solutions
Colloids
Suspensions
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The Tyndall Effect

• Many colloids appear homogeneous because the
  individual particles cannot be seen. The
  particles are, however, large enough to scatter
  light.
• Tyndall effect is a property that can be used to
  distinguish between a solution and a colloid.

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When a laser is passed through a solution and a
colloid at the same time, it is evident which
glass contains the colloid.
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Solutions Colloids Suspensions
Homogeneous Heterogeneous Heterogeneous
Particle size 0.01-1 nm can be atoms, ions, molecules Particle size 1-1000 nm, dispersed can be large molecules Particle size Over 1000 nm, suspended can be large particles
Do not separate on standing Do not separate on standing Particles settle out
Cannot be separated by filtration Cannot be separated by filtration Can be separated by filtration
Do not scatter light Scatter light (Tyndall effect) Not transparent May scatter lite
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Solution Concentration

• Molarity is simply a measure of the "strength" of
  a solution.  A solution that we would call
  "strong" would have a higher molarity than one
  that we would call "weak".
• If you ever made or drank a liquid made from a
  powdered mix, such as Kool-Aid or hot cocoa, you
  probably are familiar with the difference between
  what is called a "weak" solution or a "strong"
  solution. 

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• To make Kool-Aid of "normal" strength
• 4 scoops of powder
• -----------------------                    
                            2 quarts of water
•                                                   
                       
• To make Kool-Aid twice the "normal" strength
• 8 scoops of powder  4 scoops of
  powder
• -------------------- or
  --------------------  
• 2 quarts of water   1 quart water
                                                   
               

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Solution Concentration

• Molarity
• One-molar (M) 1 mole solute
• 1Liter solvent
• One mole of NaCl (22.99 35.45 58.44 grams) is
  dissolved in enough water (1 Liter) to make a 1M
  solution.

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Molarity Calculations

• Calculate the molarity of
• 35.2 grams of CO2 in 500. mL.
• Step 1 convert 35.2 g of CO2 into moles
• 35.2g (1 mole ) 0.800 mol
• 1 (44.01 g)
• Step 2 divide moles by volume in liters
• 0.800 mol CO2 1.60 M CO2
• 0.500 L

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Molarity Calculations

• One of the most common molarity calculations is
  to find the grams of solute necessary to produce
  a known quantity of a desired molarity.
• Your teachers do this for most solutions used in
  lab because we purchase our solutes in salt form.
  It is more economical to do a little math and
  make the solutions ourselves.

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• Calculate grams of NaOH needed to make 0.500 L of
  a 1.50 M NaOH solution.
• Since M moles of solute
• liters of solution
• 1.50molNaOH( 0.500L)(40.00g)
• 1L ( 1 ) ( 1mol )
• 30.0 g
• NaOH
• needed

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• Not as common but testable is to calculate volume
  if grams or moles of solute is known for a
  desired molarity.
• Calculate volume of solution if 20.0 g of NaOH is
  used to produce a 0.750 M NaOH solution.
• 20.0 g NaOH ( 1mol ) ( 1L )
• 1 (40.00g) (0.750mol)
• 0.667 L
• solution

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Making Diluted Solutions

• Often in chemistry, you will have to make a
  dilute solution using a more concentrated
  solution.
• Use the following equation
• M1V1 M2V2
• M1 and M2 are the initial and final molar
  solutions.(M1highest molarity)
• V1 and V2 are the initial and final volumes of
  solutions.

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• Initial M and V are based on the most
  concentrated substance.
• Final M and V are based on the diluted substance
  desired.

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• When calculating sulfuric acid, H2SO4, you must
  adjust M1 and/or M2 because sulfuric acid
  dissociates into two moles of H ions. Multiply
  molarities by 2.
• EX (12M x 2) 24M

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If you needed 200. mL of a 0.10 M solution, how
many mL would you use of the following?

• 1.0 M HC2H3O2
• EX M1V1 M2V2
• M1 1.0 M HC2H3O2 (more concentrated)
• V1 ?
• M2 0.10 M HC2H3O2 (diluted)
• V2 200. mL

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• Isolate unknown (V1) from M1V1 M2V2
• because you are finding the volume of the higher
  molarity solution you will need to make the
  diluted solution.
• V1 M2V2
• M1
• 0.10M HC2H3O2 x 200. mL
• 1.0MHC2H3O2

20. mL 1.0 M HC2H3O2 180 mL of H2O
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• In order to obtain a 200. mL volume of the
  diluted solution you would
• Pipet 20 mL of 1.0 M HC2H3O2 into a 200 mL to
  250 mL flask that contains 180. mL of water.
  (final volume is 200. mL)
• Always add acid to water.

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Solubility

• Solubility is defined as the amount of a
  substance that can be dissolved in a given
  quantity of solvent.
• Any substance whose solubility is less than 0.01
  mol/L will be referred to as insoluble.
• We can predict whether a precipitate (insoluble
  substance) will form when solutions are mixed if
  we know the solubilities of different substances.

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• Experimental observations have led to the
  development of a set of empirical solubility
  rules for ionic compounds.
• EX Experiments demonstrate that all ionic
  compounds that contain the nitrate anion, NO3-,
  are soluble in water.

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Solubility Rules

• 1. All common compounds of Group I and ammonium
  ions are soluble.
• 2. All nitrates, acetates, and chlorates are
  soluble.
• . All binary compounds of the halogens (other
  than F) with metals are soluble, except those of
  Ag, Hg(I), and Pb. (Pb halides are soluble in hot
  water.)

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• 4. All sulfates are soluble, except those of
  barium, strontium, calcium, lead, silver, and
  mercury (I). The latter three are slightly
  soluble.
• 5. Except for rule 1, carbonates, hydroxides,
  oxides, silicates, and phosphates are insoluble.
• . Sulfides are insoluble except for calcium,
  barium, strontium, magnesium, sodium, potassium,
  and ammonium.

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Ionic Equations

• An ionic equation is a chemical equation in which
  electrolytes (soluble ions that conduct
  electricity) are written as dissociated ions.
  Ionic equations are used for single and double
  replacement reactions which occur in aqueous
  solutions.
• In an aqueous reaction ions that are found as
  both reactants and products are not part of a
  reaction. They are termed spectator ions and
  essentially cancel out of the ionic equation.

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• To write net ionic equations, follow these simple
  rules
• Write a balanced equation.
• 2. Repeat the equation with reactant(s) and
  product(s) dissociated where appropriate.
• Salts written in ionic form if soluble, and in
  undissociated form if insoluble. Know the
  solubility rules.
• EX KCl ? K Cl-

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• Strong Acids/Bases written in dissociated form.
  HCl, HBr, HI, HClO3, HClO4, H2SO4, HNO3
• Bases (OH-) -Follow solubility rules
• Weak Acids/Bases written in undissociated form.
• EX HF, H3PO4, Mg(OH)2(s)

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• Oxides Oxides are always written in molecular or
  undissociated form. EX MgO(s), H2O(l)
• Exception Group 1 metal oxides
• Gases Gases are always written in molecular
  form.
• EX SO2(g), NH3, H2, O2
• 3. Cancel all spectator ions and rewrite the
  remaining net ionic equation.

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• Step 1
• AgNO3 NaCl ? AgCl NaNO3
• Step 2
• Ag NO3- Na Cl- ? AgCl(s) Na NO3-
• Step 3
• Ag Cl- ? AgCl(s)

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What Does This Represent?

• NaCl(aq) NaCl(aq)
• CCCCCCC
• Saline
• Saline
• Over
• The 7 Seas

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Colligative Properties

• Colligative comes from the Greek word kolligativ
  meaning glue together.
• We use this term for the properties of substances
  (solutes and solvents) together.
• Colligative properties of solutions is used to
  describe the effects of antifreeze/summer
  coolant.

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Saturated Solutions

• A solution at equilibrium with undissolved solute
  is said to be saturated.
• Additional solute will not dissolve if added to
  this solution.
• It is possible to dissolve less solute than
  needed to form a saturated solution. These
  solutions are unsaturated.
• A supersaturated solution can be made by
  dissolving the solute under high temps and then
  carefully cooling them. These are unstable
  solutions.

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Saturated Solution ?

• Supersaturated Solution

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Factors Affecting Solubility

• Solubility depends on the nature of both the
  solvents and solutes, temperature, and for gases,
  on pressure.
• The solubility of most solid solutes in water
  increases as the temp of the solution increases.
• This means that more sucrose C12H22O11 can be
  dissolved in hot water than cold, the basis for
  making rock candy.

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• The graph
• represents the
• solubility of substances, including NaCl,
• NaNO3, and KNO3 at different temps.
• Notice that when
• temp increases,
• solubility increases for most substances.

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Solubility of Gases

• Solubility of gases increases if the pressure is
  increased and/or the temperature decreases.
• Think about the carbonation in a warm soda vs a
  cold soda. The cold soda stays carbonated longer.
• Based on the solubility curve, the solubility of
  NH3 and SO2 (both gases) decreases as temperature
  increases.

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Molality

• Recall the units for Molarity
• moles
• L
• Molality is the measure of the number of moles of
  a solute per 1000g of solvent.
• moles solute
• 1000 g ( 1kg) solvent
• Molality is best used to describe colligitive
  properties and is represented by m.

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Boiling Point and Freezing Point

• Review the phase diagram of a pure substance.
• How will the phase diagram of a solution
  (freezing and boiling points) differ from those
  of a pure solvent?

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• The addition of a nonvolatile solute will require
  a higher temperature in which to reach boiling
  point, thus
• Boiling point elevation
• The addition of a nonvolatile solute will require
  a lower temperature in which to reach freezing
  point, thus
• Freezing point depression

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• Pure water Water with NaCl
• The water with the solute of NaCl has fewer
  liquid molecules becoming gases.
• This will increase the temp needed to change the
  state from (l) ? (g)

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Calculating Freezing and Boiling Points

• The following table contains the molal (K)
  Boiling Point Elevations, Kb, and Freezing Point
  Depressions, Kf.

Solvent Normal boiling pt (C) Kb (C/m) Normal freezing pt (C) Kf (C/m)
Water 100.0 0.52 0.0 1.86
Benzene 80.1 2.53 5.5 5.12
Ethanol 78.4 1.22 -114.6 1.99
CCl4 76.8 5.02 -22.3 29.8
Chloroform 61.2 3.63 -63.5 4.68
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• The data for the table was found by doing
  experiments.
• It has been found that 1 mole of a nonvolatile
  solute particles will raise the boiling
  temperatures of 1 kg of water by 0.52 C.
• The same concentration of solute will lower the
  freezing point of 1 kg of water by 1.86 C.
• These two figures are the molal boiling point
  constant (Kb) and the molal freezing point
  constant (Kf).

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• A 1m solution of sugar in water contains 1 mol of
  solute particles per 1 kg of solvent.
• A 1m solution of NaCl in water contains 2 mol of
  solute (because NaCl is an ion, it will
  dissociate in water into Na and Cl- ions) per 1
  kg of solvent.
• How many mol of solute would a 1m calcium nitrate
  have per 1 kg of solvent?
• Thats right!
• 3 mol because of the Ca2 and the two NO3- ions.

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Calculating Changes in Kb and Kf

• Boiling point elevation is
• ?Tb Kbm
• 
  (molality)
• (change in boiling point) (boiling point
• 
  constant)
• Freezing point depression
• ?Tf Kfm
• 
  (molality)
• (change in freezing point) (freezing point
• 
  constant)

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• If 55.0 grams of glucose (C6H12O6) are dissolved
  in 525 g of water, what will be the change in
  boiling and freezing points of the resulting
  solution?
• Step 1 Convert g of glucose to moles
• 55.0 g (1 mol)
• 1 (180.18 g) 0.305 mol
• Step 2 Convert g of water to kg
• 525g ? 0.525kg
• Step 3 Calculate m
• 0.305mol/0.525kg 0.581 m

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• Step 4 Obtain molal Kb and Kf from table.
• Step 5 Place values into equation
• ?Tb Kbm
• ?Tb (0.52C/m)(0.581m) 0.302 C
• This means that the boiling point will be
  elevated by 0.302 C.
• (100 C 0.302 C)
• This solution will reach boiling point at 100.302
  C.
• Now calculate the change in freezing.

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• Calculate the change in freezing point of 24.5g
  nickel(II) bromide dissolved in 445 g of water.
  (assume 100 dissociation)
• Step 1 Convert g of NiBr2 into moles
• 24.5g ( 1 mol)
• 1 (218.49 g) 0.112mol
• Step 2 Convert solvent to kg
• 445 g ? 0.445kg
• Step 3 Calculate m
• 0.112 mol/0.445kg 0.252m

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• We now have to take 0.252 and multiply by 3
  because the dissociation of the ionic compound
  makes 3 moles of ions (solute) per kg of solvent
• NiBr2 ? Ni2 2Br
• 0.252 x 3 0.756m
• Step 4 Obtain molal Kf from table.
• Step 5 Place values into equation
• ?Tf (1.86C/m)(0.756m) 1.41 C
• (0 C - 1.41 C)
• Freezing point has been depressed to
• -1.41 C.

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• Coolant is used because it takes higher
  temperatures to reach boiling point.
• Antifreeze needs lower temperatures in order to
  freeze.
• This also why salt is used on frozen roads and
  walkways. The salt dissolves in the water and
  lowers the freezing point of water. It now takes
  colder temps to turn the water into ice.
• A 10-percent salt solution freezes at 20 F (-6
  C), and a 20-percent solution freezes at 2 F (-16
  C).

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• Practice problems Compute both boiling and
  freezing points of these solutions (assume 100
  dissociation of all ionic compounds)
• 27.6 g Mg(ClO4)2 in 100.g of water.
• 100.0 g of C10H8 (naphthalene) in 250. g of C6H6
  (benzene).
• 25.9 g of C7H14BrNO4 (3-bromo-2-nitrobenzoic
  acid) in 150. g of benzene.
• 55.6 g of Al2(SO4)3 in 500. g of water.
• 1500.g of NaCl in 4500. g of water.

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Brief Summary

• Heterogeneous liquid mixtures are classified as
  suspensions (large particles that settle out), or
  colloids (small particles that stay dispersed).
• Homogeneous mixtures are solutions made of a
  solute dissolved in a solvent.
• Solutes and solvents must be alike in polarity in
  order to produce a solution.
• The concentration of a solution is molarity
  (molar) and has the unit M, which includes moles
  of solute per unit volume of solvent.

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• When preparing a dilute solution from a
  concentrated solution, use the formula
• M1V1 M2V2
• Where initial volume and molarity of concentrated
  solution (EX 12M HCl) is compared to final
  volume and molarity of diluted solution (EX 6M
  HCl).
• Solubility of solutes can be reflected in a
  solubility graph.
• Solubility of solid substances generally
  increases as temperature increases.
• Solubility of gases decreases with increased
  temperature.

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• Ionic equations can be written to express the net
  reaction occurring in a system after the
  spectator ions have been removed.
• Solubility rules for substances have been
  experimentally determined. They indicate what
  substances are or are not water soluble.
• Colligative properties demonstrate the properties
  of the solution rather than solute and solvent
  independently.

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• A solution with undissolved solute is termed
  unsaturated.
• A solution with undissolved solute is termed
  saturated.
• A solution that has more dissolved solute at a
  particular temp due to being dissolved at a
  higher temp is termed supersaturated.
• Boiling points and Freezing points of solutions
  can be calculated using molality.

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• Molality (molal) is described by unit m and
  expresses moles of solute per kg of solvent.
• When calculating BP and FP differences use
  equation Kfp or bp Kb or f x m
• Kb or f is a standard and must be given
• m must be calculated and adjusted to express
  moles contributed.
• -molecules contribute only 1 mol.
• -ionic compounds contribute the number of moles
  they dissociate into.
• EX K2SO4 ? 2 mole K 1 mole SO4-2

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• After calculating difference, refer to normal BP
  and FP of solvents and
• Boiling Point Elevation ? add difference to
  normal BP.
• Freezing Point Depression ? subtract difference
  from normal FP.