Chapter 7: Solutions and Colloids

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Chapter 7Solutions and Colloids
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Solutions Solute and Solvent

• Solutions
• Are homogeneous mixtures of two or more
  substances.
• Consist of a solvent and one or more solutes.

3
Nature of Solutes in Solutions

• Solutes
• Spread evenly throughout the solution.
• Cannot be separated by filtration.
• Can be separated by evaporation.
• Are not visible, but can give a color to the
  solution.

4
Examples of Solutions

• The solute and solvent in a solution can be a
  solid, liquid, and/or a gas.

5
Water as a polar solvent

• Polar molecules have an asymmetrical shape
• Have an electron rich area (often containing lone
  pairs)
• Have an electron poor area
• Water is polar
• O is electron rich
• H is electron poor

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Water

• Water
• Is the most common solvent.
• Is a polar molecule.
• Forms hydrogen bonds between the hydrogen atom in
  one molecule and the oxygen atom in a different
  water molecule.

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Formation of a Solution

• Na and Cl- ions
• On the surface of a NaCl crystal are attracted to
  polar water molecules.
• Are hydrated in solution with many H2O molecules
  surrounding each ion.

8
Equations for Solution Formation

• When NaCl(s) dissolves in water, the reaction
  can be written as
• H2O
• NaCl(s) Na(aq) Cl- (aq)
• solid
  separation of ions

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Water and Potassium Cyanide
Potassium Cyanide is a powerful poison and ionic
compound
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Like Dissolves Like

• Two substances form a solution
• When there is an attraction between the particles
  of the solute and solvent.
• When a polar solvent, such as water, dissolves
  polar solutes, such as sugar, and ionic solutes,
  such as NaCl.
• When a nonpolar solvent, such as hexane (C6H14),
  dissolves nonpolar solutes, such as oil or
  grease.

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Like Dissolves Like

• Solvents Solutes
• Water (polar) Ni(NO3)2
• CH2Cl2 (nonpolar) (polar)
  
• I2
  (nonpolar)

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Solutes and Ionic Charge

• In water,
• Strong electrolytes produce ions and conduct an
  electric current.
• Weak electrolytes produce a few ions.
• Nonelectrolytes do not produce ions.

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Strong Electrolytes

• Strong electrolytes are ionic compounds as well
  as strongly acidic/basic molecules
• Dissociate in water producing positive and
  negative ions.
• Conduct an electric current in water.
• H2O 100 ions
• NaCl(s) Na(aq) Cl- (aq)
  
• H2O
• CaBr2(s) Ca2(aq) 2Br- (aq)

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Weak Electrolytes

• Weak electrolyte are weakly acidic/basic
  molecules
• Dissociates only slightly in water.
• In water forms a solution of a few ions and
  mostly undissociated molecules.
• HF(g) H2O(l) H3O(aq)
  F- (aq)
• NH3(g) H2O(l) NH4(aq) OH-
  (aq)

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Nonelectrolytes

• Nonelectrolytes are nonacidic/ nonbasic
  molecules.
• Dissolve as molecules in water.
• Do not produce ions in water.
• Do not conduct an electric current.

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Unsaturated Solutions

• Unsaturated solutions
• Contain less than the maximum amount of solute.
• Can dissolve more solute.

Dissolved solute
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Saturated Solutions

• Saturated solutions
• Contain the maximum amount of solute that can
  dissolve.
• Generally contain undissolved solute in the
  container.

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Effect of Temperature on Solubility

• Solubility
• Depends on temperature.
• Of most solids increases as temperature
  increases.
• Of gases decreases as temperature increases.

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Solubility and Pressure

• Henrys Law states
• The solubility of a gas in a liquid is directly
  related to the pressure of that gas above the
  liquid.
• At higher pressures, more gas molecules dissolve
  in the liquid.

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Soluble and Insoluble Salts

• Ionic compounds that
• Dissolve in water are soluble salts.
• Do not dissolve in water are insoluble salts.

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Solubility Rules

• Soluble salts
• Typically contain at least one ion from Groups
  1A(1) or
• NO3-, or C2H3O2- (acetate).

TABLE 8.7
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Formation of a Solid

• When solutions of salts are mixed,
• A solid forms if ions of an insoluble
  combinations are present.

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Equations for Forming Solids

• A molecular equation shows the formulas of the
• compounds.
• Pb(NO3)(aq) 2NaCl(aq) PbCl2(s)
  2NaNO3(aq)
• An ionic equation shows the ions of the
  compounds.
• Pb2(aq) 2NO3-(aq) 2Na(aq) 2Cl-(aq)
• PbCl2(s) 2Na(aq)
  2NO3-(aq)
• A net ionic equation shows only the ions that
  form a solid.
• Pb2(aq) 2Cl-(aq) PbCl2(s)

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Concentration

• The concentration of a solution
• Is the amount of solute dissolved in a specific
  amount of solution.
• amount of solute
• amount of solution

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Mass Percent Concentration

• Mass percent (m/m) concentration is the
• Percent by mass of solute in a solution.
• mass percent (m/m) g of solute
  x 100
• g of
  solute g of solvent
• Amount in g of solute in 100 g of solution.
• mass percent g of solute x 100
• 100 g of solution

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Volume Percent

• The volume percent (v/v) is
• Percent volume (mL) of solute (liquid) to volume
  (mL) of solution.
• Volume (v/v) mL of solute x 100
  mL of
  solution
• Solute (mL) in 100 mL of solution.
• volume (v/v) mL of solute x 100
  
  100 mL of solution

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Mass/Volume Percent

• The mass/volume percent (m/v) is
• Percent mass (g) of solute to volume (mL) of
  solution.
• mass/volume (m/v) g of solute x 100
  
• mL of
  solution
• Solute (g) in 100 mL of solution.
• mass/volume (m/v) g of solute
  x 100 100 mL of
  solution

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Using Percent Concentration (m/m) Factors

• How many grams of NaCl are needed to prepare
• 225 g of a 10.0 (m/m) NaCl solution?
• STEP 1 Given 225 g solution 10.0 (m/m)
  NaCl
• Need g of NaCl
• STEP 2 g solution g NaCl
• STEP 3 Write the 10.0 (m/m) as conversion
  factors.
• 10.0 g NaCl and 100 g solution
  
• 100 g solution 10.0 g NaCl
• STEP 4 Set up to cancel g solution.
• 225 g solution x 10.0 g NaCl 22.5 g
  NaCl 100 g solution

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Using Percent Concentration(m/v) Factors

• How many mL of a 4.20(m/v) will contain 3.15 g
  KCl?
• STEP 1 Given 3.15 g KCl(solute) 4.20 (m/v)
  KCl
• Need mL of KCl solution
• STEP 2 Plan g KCl mL KCl solution
• STEP 3 Write conversion factors.
• 4.20 g KCl and 100 mL
  solution
• 100 mL solution 4.20 g KCl
• STEP 4 Set up the problem
• 3.15 g KCl x 100 mL KCl solution
  75.0 mL KCl 4.20 g KCl

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Molarity (M)

• Molarity (M)
• Is a concentration term for solutions.
• Gives the moles of solute in 1 L solution.
• M moles of solute
• liter of solution

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Calculation of Molarity

• What is the molarity of 0.500 L NaOH solution if
  it
• contains 6.00 g NaOH?
• STEP 1 Given 6.00 g NaOH in 0.500 L solution
• Need molarity (mole/L)
• STEP 2 Plan g NaOH mole NaOH
  molarity

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Calculation of Molarity (cont.)

• STEP 3 Conversion factors 1 mole NaOH 40.0 g
• 1 mole NaOH and 40.0 g NaOH
• 40.0 g NaOH 1 mole NaOH
• STEP 4 Calculate molarity.
• 6.00 g NaOH x 1 mole NaOH 0.150 mole
• 40.0 g NaOH
• 0.150 mole 0.300 mole 0.300 M
  NaOH
• 0.500 L 1 L

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Molarity Conversion Factors

• The units of molarity are used to write
  conversion factors for calculations with
  solutions.

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Calculations Using Molarity

• How many grams of KCl are needed to prepare 125
  mL
• of a 0.720 M KCl solution?
• STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl
• Need Grams of KCl
• STEP 2 Plan L KCl moles KCl
  g KCl

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Calculations Using Molarity

• STEP 3 Conversion factors
• 1 mole KCl 74.6 g
• 1 mole KCl and 74.6 g KCl
• 74.6 g KCl 1 mole KCl
• 1 L KCl 0.720 mole KCl
• 1 L and 0.720 mole KCl
• 0.720 mole KCl 1 L
• STEP 4 Calculate g KCl
• 0.125 L x 0.720 mole KCl x 74.6 g KCl
  6.71 g KCl
• 1 L
  1 mole KCl

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Dilution

• In a dilution,
• Water is added.
• Volume increases.
• Concentration decreases.

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Initial and Diluted Solutions

• In the initial and diluted solution,
• The moles of solute are the same.
• The concentrations and volumes are related by the
  following equations
• C1V1 C2V2
• initial diluted

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Dilution Calculations

• What is the molarity (M) of a solution prepared
• by diluting 0.180L of 0.600 M HNO3 to 0.540 L?
• Prepare a table
• C1 0.600 M V1 0.180 L
• C2 ? V2 0.540 L
• Solve dilution equation for unknown and enter
  values
• C1V1 C2V2
• C2 C1V1 (0.600 M)(0.180
  L) 0.200 M
• V2 0.540 L

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Molarity in Chemical Reactions

• In a chemical reaction,
• The volume and molarity of a solution are used to
  determine the moles of a reactant or product.
• molarity ( mole ) x volume (L) moles
• 1 L
• If molarity (mole/L) and moles are given, the
  volume (L) can be determined
• moles x 1 L volume (L)
• moles

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Using Molarity of Reactants

• How many mL of 3.00 M HCl are needed to
  completely
• react with 4.85 g CaCO3?
• 2HCl(aq) CaCO3(s) CaCl2(aq) CO2(g)
  H2O(l)
• STEP 1 Given 3.00 M HCl 4.85 g CaCO3
• Need volume in mL
• STEP 2 Plan
• g CaCO3 mole CaCO3 mole HCl
  mL HCl

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Using Molarity of Reactants (cont.)

• 2HCl(aq) CaCO3(s) CaCl2(aq)
  CO2(g) H2O(l)
• STEP 3 Equalitites
• 1 mole CaCO3 100.1 g 1 mole CaCO3
  2 mole HCl
• 1000 mL HCl 3.00 mole HCl
• STEP 4 Set Up
• 4.85 g CaCO3 x 1 mole CaCO3 x 2 mole HCl x 1000
  mL HCl
• 100.1 g CaCO3 mole CaCO3 3.00 mole
  HCl
• 32.3 mL HCl required

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Solutions

• Solutions
• Contain small particles (ions or molecules).
• Are transparent.
• Do not separate.
• Cannot be filtered.
• Do not scatter light.

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Colloids

• Colloids
• Have medium size particles.
• Cannot be filtered.
• Can be separated by semipermeable membranes.
• Scatter light (Tyndall effect).

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Examples of Colloids
TABLE 8.11
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Suspensions

• Suspensions
• Have very large particles.
• Settle out.
• Can be filtered.
• Must be stirred to stay suspended.
• Examples include blood platelets, muddy water,
  and Calamine lotion.

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Osmosis

• In osmosis, water (solvent)
• flows from the lower solute
• concentration into the
• higher solute concentration.
• The level of the solution with
• the higher concentration
• rises.
• The concentrations of the two
• solutions become equal with
• time.

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Osmosis

• A semipermeable membrane separates a 4
• starch solution from a 10 starch solution.
  Starch is a
• colloid and cannot pass through the membrane, but
• water can. What happens?
• semipermeable membrane

10 starch
4 starch
H2O
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Water flow Equalizes

• The 10 starch solution is diluted by the flow of
  water out of the 4 and its volume increases.
• The 4 solution loses water and its volume
  decreases.
• Eventually, the water flow between the two
  becomes equal.

7 starch
7 starch
H2O
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Osmotic Pressure

• Osmotic pressure is
• Produced by the solute particles dissolved in a
  solution.
• Equal to the pressure that would prevent the flow
  of additional water into the more concentrated
  solution.
• Greater as the number of dissolved particles in
  the solution increases.

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Osmotic Pressure of the Blood

• Red blood cells
• Have cell membranes that are semipermeable.
• Maintain an osmotic pressure that cannot change
  or damage occurs.
• Must maintain an equal flow of water between the
  red blood cell and its surrounding environment.

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Isotonic Solutions

• An isotonic solution
• Exerts the same osmotic pressure as red blood
  cells.
• Is known as a physiological solution.
• Of 5.0 glucose or 0.90 NaCl is used medically
  because each has a solute concentration equal to
  the osmotic pressure equal to red blood cells.

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Hypotonic Solutions

• A hypotonic solution
• Has a lower osmotic pressure than red blood
  cells.
• Has a lower concentration than physiological
  solutions.
• Causes water to flow into red blood cells.
• Causes hemolysis RBCs swell and may burst.

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Hypertonic Solutions

• A hypertonic solution
• Has a higher osmotic pressure than RBCs.
• Has a higher concentration than physiological
  solutions.
• Causes water to flow out of RBCs.
• Causes crenation RBCs shrinks in size.

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Dialysis

• In dialysis,
• Solvent and small solute particles pass through
  an artificial membrane.
• Large particles are retained inside.
• Waste particles such as urea from blood are
  removed using hemodialysis (artificial kidney).

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Hemodialysis

• When kidneys fail, an artificial kidney uses
  hemodialysis to remove waste particles such as
  urea from blood.