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Solutions and Equilibrium - Horton High School

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Solutions and Equilibrium - Horton High School Powered By Docstoc
					 Solutions (homogeneous mixtures) are a part
  of our everyday life. Do you use mouthwash,
  toothpaste, cough syrup?
 My swimming pool is one very large solution.
  The presence of algae makes my pool green
  and slimy. I then need to add an algicide
  which is another solution. Bromine is in
  solution as well as other chemicals (total
  alkalinity).
 Solubility is of fundamental importance in a
  large number of practical applications
  ranging from ore processing to the use of
  medicines and the transport of pollutants.
 Solubility is often said to be one of the
  “characteristic properties of a substance”.
 used to describe a substance
 indicates its polarity
 helps distinguish it from another substance
 becomes a guide for the applications of that
  substance
 You will often see in a compounds
  description: “is insoluble in water or alcohol
  but soluble in concentrated sulfuric acid”
 Solubilityof a substance is useful when
 separating mixtures. The synthesis of
 chemical compounds, by the milligram in a
 lab or by the ton in industry makes use of the
 relative solubility of the end product.

 Cooks,chemists, farmers, pharmacists and
 gardeners need to know which compounds
 are soluble and which are insoluble.

 Youwill separate a mixture or five
 substances for first lab using the property of
 solubility as an important property.
 What properties do solutions have?
 How do we describe the concentrations?
 Why do some substances dissolve while
  others do not?
 Solution – a homogeneous mixture that is
 uniform throughout. The same substances
 will be in the same relative amounts.
 Solutions are not pure substances because
 they can have variable composition (example
 salt water).

 Solvent – any substance that has other
 substances dissolved in it. The substance that
 is present in the largest amount (by volume,
 mass or number of moles) is usually the
 solvent.
 Solute  – a substance dissolved into the
  solvent.
 Concentrated – a solution with a high level of
  solute ( > 1.0 M)
 Dilute – a solution with a low level of solute
  (< 1.0 M)
 A solution can be a gas, solid (alloy) or a
  liquid.
 Aqueous – a liquid solution in which water is
  the liquid (aq)
 Miscible – liquids which will dissolve in each
  other
 Immiscible – liquids which do not dissolve in
  each other.
 The  ability of a solvent to dissolve a solute
  depends on the forces of attraction between
  the particles. There is always some
  attraction between the solvent and the
  solute.
 Solubility – the mass of the solute that
  dissolves in a given quantity of solvent, at a
  certain temperature. May be g/mL or molar
  solubility (moles of solute in 1 L of solution).
 Saturated – no more solute will dissolve in a
  solution. Excess solute can be seen.
 Unsaturated    – a solution which is not yet
  saturated. More solute can dissolve.
 Sparingly soluble or slightly soluble – the
  solubility is between insoluble (less than 0.1g
  per 100 mL) and soluable (1g per 100 mL).
 The solubility of gasses do not use the same
  terms as that for solids and liquids. An
  example is the amount of dissolved oxygen in
  water. 0.0009 g/100 mL is the solubility of
  oxygen at 20oC in fresh water. This is very
  soluble.
   Substances of “like” polarity tend to be more soluble in each
    other than substances that differ in polarity.
   Remember – polarity is a measure of how electrons are shared
    between bonding elements. It also is dependent on shapes of
    molecules determined by VESPR.
 “Like dissolves like.”
Polar solvents dissolve other polar molecules
  and ionic compounds.
Nonpolar molecules dissolve other nonpolar
  molecules.
Alcohols which have characteristics of both
  tend to dissolve both polar and nonpolar but
  will not dissolve ionic solids.
 Step  1 – the forces between the particles in
  the solid must be broken. Either between the
  ions or between the molecules. This requires
  energy.
 Step 2 – Some of the intermolecular forces
  between the particles in the liquid must be
  broken. This also requires energy.
 Step 3 – There is an attraction between the
  particles of the solid and the particles of the
  liquid. This step gives off energy.
 Sugar (polar) in water (polar).
 Iodine (nonpolar) in benzene (nonpolar)
 Electronegativity:

 Electronegativity   Type of
 Difference          Bond



 0.0 – 0.4           nonpolar   H-H    0.0
 0.4 – 2.0           polar      H-Cl   0.9
 > 2.0               ionic      NaCl   2.1
 Polar,   nonpolar or ionic

a)   N and H       3.0 – 2.1   0.9   polar
b)   F and F       4.0 - 4.0   0     nonpolar
c)   Ca and O      3.5 -1.0    2.5   ionic
d)   Al and Cl     3.0 – 1.5   1.5   polar
    all substances dissolve in the same
 Not
 manner or amount.
 1. The solubility of most substances increases
  as the temperature of the solvent increases.
  For some substances, the opposite is true.
At 25oC NaCl 36.2 g/100g H2O
At 100oC NaCl 39.2 g/100g H2O

Sodium sulfate
At 40oC 50g/100g H2O
At 100oC 41g/100g H2O
 Gaseshave a higher solubility in cold water
 than in hot water.




 …….Sotemperature affects the rate at which
 something will dissolve.
 2.   Pressure

The solubility of a gas increases as the
 pressure of the gas above the solution
 increases. Soft drinks are bottled under high
 pressure.
Henry’s Law          At a given temperature the solubility of a
                         gas in a liquid (S) is directly proportional
                         to the pressure of the gas above the
                         liquid (P). S1 = S2

                                      P1 = P 2



                                  P then         S
   If the solubility of a gas in water is 0.77 g/L at
    3.5 atm of pressure, what is the solubility in g/L
    at 1.0 atm of pressure? The temperature is
    constant at 25oC.
 If    S1 = S2 then     S2 = S1 x P2
       ___ ____               _______
       P1    P2                  P1

                             = 0.77 g/l x 1.0 atm
                               _________________
                                         3.5 atm
                             = 0.22 g/L
 3.   Surface Area

 In order to dissolve, the solvent must come
 in contact with the solute. Smaller pieces
 increase the amount of surface of the solute
 interacting with the solvent.

  Term Used for Solubility   Solubility (g/L)
  Very Soluble               Greater than 100 g
  Soluble                    Between 10 g and 100 g
  Slightly Soluble           Between 1 g and 10 g
  Insoluble                  Less than 1 g (.1g /100 ml)
 When   a solution of NaOH is added to a
  solution of CuSO4 a blue solid precipitate
  forms. The mixture of solutions contains Na+,
  OH-, Cu 2+, and SO42- ions.
 When we consult a solubility table we find
  that the Cu(OH)2 is insoluble, the precipitate
  formed must be Cu(OH)2. An equation for this
  reaction can be written as:
  2NaOH (aq) + CuSO4 (aq)   Na2SO4 (aq) + Cu(OH)2 (s)
 In   solution these compounds exist as
    dissociated ions as shown below:
2Na+ (aq) + 2OH- (aq) + Cu2+ (aq) + SO42- (aq)
                 Na+ (aq) + SO42- (aq) + Cu(OH)2 (s)

   Notice that the Na+ (aq) and SO42- (aq) remain unchanged
    by the reaction. These ions are called spectator ions
    because they do not react. So if we show only the ions
    changed by the reaction, the equation becomes:
      Cu2+ (aq) + 2OH- (aq)             Cu(OH)2 (s)
      This is called the net ionic equation.
   Precipitation Reactions and Solubility Rules
    A solid dissolves in water because the water
    molecules around the solid’s surface collide with
    the particles of the solid. The attraction
    between polar water and a partially charged
    particle or ion enables the water molecules to
    pull these particles away from the crystal, a
    process called dissociation. This is followed by
    solvation.
    Predicting Precipitates
    We can predict the formation of a precipitate if
    we know the solubility of the compounds.
    Patterns in solubility have lead to the creation of
    solubility rules. Solubility rules are based on
    what happens when salts are placed into water,
    they either dissolve or they do not dissolve.
 Precipitation occurs because two SOLUBLE
  substances containing solvated ions are mixed
  and a pair of ions form a substance that is
  INSOLUBLE. The insoluble substance is the
  precipitate (solid) that is formed.
 When you mix two solutions containing ionic
  compounds you can predict the products based
  on DOUBLE REPLACEMENT reaction.
 If one of the predicted products is not soluble, it
  will precipitate out and a reaction occurs.
 If both predicted products are soluble, no
  reaction occurs (all ions remain solvated).
 Simple Rules for the Solubility of Ionic
  Compounds in Water
1. Most nitrate (NO3-) salts are soluble.
2. Most salts containing the alkali metals cations
  (Li+, Na+, K+, Rb+, Cs+) and the ammonium (NH4+)
  cation are soluble.
3. Most chloride, bromide, and iodide salts are
  soluble.
Except those containing Ag+, Pb2+, and Hg+
4. Most sulfate salts are soluble.
Except those containing Ba2+, Pb2+, Hg2+, Sr2+ and
  Ca2+.
5. Most hydroxide salts are insoluble except those
  of the alkali metal cations and a few alkaline
  earth metals: Ba2+, Sr2+, and Ca2+.
6. Most sulfides (S2-), carbonate (CO32-), chromate
  (CrO42-), and phosphate (PO43-) salts are
  insoluble.
Example:
Will a precipitation reaction occur when Ba(NO3)2 (aq) and Na2SO4
  (aq) are mixed?
Possible products from a double displacement reaction:
         BaSO4 (s) and NaNO3 (s)
   From the solubility table BaSO4 is not soluble so it would form a
    precipitate.
    Ba(NO3)2 (aq) + Na2SO4 (aq)         BaSO4 (s) + 2 NaNO3 (aq)
    We can also write a complete (total) ionic equation:
    Ba2+ (aq) + 2NO3- (aq) + 2Na+ (aq) + SO42- (aq)
                                 2Na+ (aq) + 2NO3- (aq) + BaSO4 (s)
 or finally we can write a net ionic equation which shows only the
  ions which precipitate from solution.
        Ba2+ (aq) + SO42- (aq)            BaSO4 (s)
   Example:
    Will a precipitation reaction occur if 150 mL of
    1.0M NaI (aq) is added to 200 mL of 0.5M AgNO3
    (aq)?




    The solubility table shows that AgI is insoluble.
    AgNO3 (aq) + NaI (aq)       AgI (s) + NaNO3 (aq)

Net ionic equation
     Ag+ (aq) + I- (aq)           AgI (s)
 Molarity  (M) is the most common method for
  describing the concentration of a solution,
  and is defined as the number of moles of
  solute present per litre of solution.
M = n/V (V in litres)
Example: Calculate the molarity of 75.0 g of
  KBr (s) dissolved in 500 mL of water.

M= mol solute/ V (in litres)
 Mol   KBr = 75.0 g / 119 g/mol
           = 0.630 mol

V solution = 500 mL / 1000 mL
         = 0.5 L
                    M = 0.630mol/0.5 L
                      = 1.26 M KBr (aq)

 Example:  What are the concentrations of
 Ca2+ (aq) and NO3- (aq) in 1.5 M Ca(NO3)2
 (aq)?
  Ca(NO3)2              Ca2+ (aq) +      2NO31-
 (aq)
   1.5 M              1.5 M     +      3.0 M
 Solutions prepared by dissolving a precise mass
  of solute in a precise volume of solvent are
  known as standard solutions. Volumetric flasks
  are used to prepare standard solutions.
 Dilutions
  It is often necessary to dilute a certain solution.
  Dilution is the process by which a solution is
  made less concentrated by the addition of more
  solvent. When a solution has been diluted its
  solute particles are spread out more. When
  solutions are diluted, the amount of solute
  particles remains the same. Since n=MV
  (concentration x volume) we can use the
  following equation:
                     M1V1 = M2V2
 Example:     200.0 mL of 3.0 M NaOH (aq) is
  diluted to 400 mL. What is the concentration
  of the diluted solution?
       Initail # moles = final # moles
              M1V1     =    M2V2
     (3.0 M x 0.2 L) = ( M2 x 0.4L)
               M2   = 1.5 M NaOH (aq)
 Example:       What volume of a stock solution
  of 6.0 M HCl (aq) is needed to make 900 mL
  of a 1.3 M solution?
               M1V1 = M2V2
  (6.0 M x V1) = (1.3 M x 0.9 L)
                  V1 = 0.2 L or 200 mL
 Molality  (m) is the number of moles of solute
  per kilogram of solvent.
 M= n/kg
 Example: A solution contains 15.5 g of
  NH2CONH2 in 74.3 g H2O. Calculate the
  molality.
 Molality = mol solute/kg solvent
   Mol NH2CONH2 = 15.5 g / 60.0 g/mol (molar mass)
                  = 0.258 mol
   Kg H2O = 74.3 g x 1 kg
                        1000 g
          = 0.0743 Kg


   m = mol solute
        Kg solvent
      = 0.258 mol
        0.0743 Kg
       = 3.48 m
 Mole fraction (no units) is the number of moles
  of one compound in a mixture divided by the
  total number of moles of all components in the
  mixture.
      Xa = na/ (na + nb)
 Example: What are the mole fractions of an
  ethanol and water solution prepared by
  dissolving 3.45 g C2H5OH in 21.1 g H2O?

    Mol ethanol = 3.45 g / 46.0 g/mol = 0.075 mol
    Mol of water = 21.0 g / 18.0 g/mol = 1.18 mol
    Total # moles = mol of ethanol + mol of water
                  = 0.075 mol + 1.18 mol
                  = 1.255 mol
 Molfraction of ethanol = 0.075 / 1.255
                         = 0.05976
 Mol fraction of water = 1.18 / 1.255
                        = 0.9402

 Mol   fraction ethanol + water = 1
   % by mass          mass of solute                  x 100
                  (mass of solute + mass of solvent)


   Example: A 5 % NaCl solution would consist of
    5.0 g of NaCl (s) and 95 g of water.
    ( I chose 100 g because it was easier to take 5%).
    1 mL of H2O = 1 g H2O.
     So 5.0 g NaCl (s) in 95 mL of H2O (l).

   Example: 24 g of NaCl (s) is dissolved in 152 g
    H2O (l).
       Calculate the % by mass of this solution.
    Mass % NaCl =       24 g             x 100
                      (24 g + 152g)
                  = 24/176 x 100 = 14 %
 Properties of solutions which depend on the
  number of solute particles but not on their
  nature.
 They are dependent on the number of solute
  particles dissolved in a given volume of
  solvent. They include: vapour pressure
  lowering, boiling point elevation, freezing
  point depression and osmotic pressure.
 Thepresence of a solute lowers the freezing
 point of a solution relative to that of the
 pure solvent. For example, pure water
 freezes at 0°C (32°F); if one dissolves 10
 grams of sodium chloride (table salt) in 100
 grams of water, the freezing point goes
 down to −5.9°C (21.4°F). If one uses sucrose
 (table sugar) instead of sodium chloride, 10
 grams in 100 grams of water gives a solution
 with a freezing point of −0.56°C (31°F).
 The reason that the salt solution has a lower
 freezing point than the sugar solution is that
 there are more particles in 10 grams of
 sodium chloride than in 10 grams of sucrose.
 Since sucrose, C 12 H 22 O 11 has a molecular
 weight of 342.3 grams per mole and sodium
 chloride has a molecular weight of 58.44
 grams per mole, 1 gram of sodium chloride
 has almost six times as many sodium chloride
 units as there are sucrose units in a gram of
 sucrose. In addition, each sodium chloride
 unit comes apart into two ions (a sodium
 cation and a chloride anion ). The freezing
 point depression of a solution containing a
 dissolved substance, such as salt dissolved in
 water, is a colligative property.
One can calculate the change in freezing point
(Δ T f ) relative to the pure solvent using the
 equation:
                       Δ T f= K fm
  where K f is the freezing point depression constant
 for the solvent (1.86°C·kg/mol for water) (can be
 looked up on a table) ,m is molality (the number of
 moles of solute in solution per kilogram) of solvent
 for ALL particles.

     Because the presence of a solute lowers the freezing
    point, the Department of Highways puts salt on our
    roads after a snowfall, to keep the melted snow
    from refreezing.
    Also, the antifreeze used in automobile heating and
    cooling systems is a solution of water and ethylene
    glycol (or propylene glycol); this solution has a lower
    freezing point than either pure water or pure
    ethylene glycol.
 ΔTf   = Kf x m

What is the freezing point of these solutions?

 1.4 mol of Na2SO4 in 1750 of H2O

       m = 1.4 mol
              1.75 kg
           = 0.8 m
3 particles are formed when Na2SO4 ionizes
                     3 x 0.8m = 2.4 m
  Δ Tf = Kf x m
       = 1.86 oC/m x 2.4 m
       = 4.46 oC
  0 oC – 4.5 oC = - 4.46 oC
 Whatis the freezing point of 0.60 mol of
 MgSO4 in 1300 g of H2O?


  m = 0.60 mol
        1.3 kg
       = 0.46 m
     2 particles x 0.46 m = 0.92 m

    ΔTf = 1.86 oC x 0.92 m
        = 1.71 oC

    0oC – 1.71 oC = - 1.71 oC
   The boiling point of a solution is higher than that
    of the pure solvent. The actual boiling point
    elevation is the difference in temperature
    between the boiling points of a solution and the
    pure substance. Accordingly, the use of a
    solution, rather than a pure liquid, in antifreeze
    serves to keep the mixture from boiling in a hot
    automobile engine. As with freezing point
    depression, the effect depends on the number of
    solute particles present in a given amount of
    solvent, but not the identity of those particles. If
    10 grams of sodium chloride are dissolved in 100
    grams of water, the boiling point of the solution
    is 101.7°C (215.1°F; which is 1.7°C (3.1°F)
    higher than the boiling point of pure water).
    The formula used to calculate the change in
    boiling point (Δ T b ) relative to the pure
    solvent is similar to that used for freezing
    point depression:
           Δ T b= K bm ,
      where K b is the boiling point elevation
    constant for the solvent (0.52°C·kg/mol for
    water), and m has the same meaning as in
    the freezing point depression formula.
      Note that Δ T b represents an increase in
    the boiling point, whereas Δ T f represents a
    decrease in the freezing point.
 Boiling   Point
  Δ Tb = Kb x m
  What is the boiling point of a solution that contains
    1.2 mol of sodium chloride in 800 g of water?

  m= 1.2 mol NaCl
      0.8 kg
    = 1.5 m
  Total molality = 1.5 m x 2 = 3m

  Δ Tb = Kb x m
       = 0.512 oC/m x 3 m
       = 1.54 oC
   100 oC + 1.54 oC = 101.54 oC
 What is the boiling point of a solution that
  contains 1.25 mol of CaCl2 in 1400 g of
  water?
m= 1.25 mol
   1.4 kg
 = 0.89 m

CaCl2 = 3 particles when it ionizes
       3 x 0.89 m = 2.67 m


Δ Tb = Kb x m
      = 0.512 oC/m x 2.67 m
      = 1.37 oC

100 oC + 1.37 oC = 101.37 oC
 A solution of 7.50 g of a nonvolatile compound in
  22.60 g of water boils at 100.78 oC . What is the
  molecular mass of the solute?
 Molality of the solution:
 100.78 oC – 100 oC = 0.78 oC ( ΔTb)
   If ΔTb = Kb x m then m = Δ Tb
                               Kb
    m=    0.78 oC
        0.512 oC/m
    m = 1.5 m
   1.5 m x 0.0226 kg = 0.034 mol

    Molecular mass of solute = mass of solute
                               moles of solute

                                 = 7.5 g
                             0.0344 mol
                            = 218 g/mol
 The vapor pressure of a liquid is the
 equilibrium pressure of gas molecules from
 the liquid (evaporation) above the liquid
 itself in a closed system. A glass of water
 placed in an open room will evaporate
 completely (and thus never reach
 equilibrium); however, if a cover is placed on
 the glass, the space above the liquid will
 eventually contain a constant amount of
 water vapor. How much water vapor is
 present depends on the temperature, but not
 on the amount of liquid that is present at
 equilibrium.
If, instead of pure water, an aqueous solution is
   placed in the glass, the equilibrium pressure will
   be lower than it would be for pure water.

A solution that contains a nonvolatile solute (it
  does not vaporize like sugar or salt) always has a
  lower vapour pressure than the pure solvent. In
  an aqueous solution of sodium chloride, there
  are sodium ions and chloride ions throughout the
  solution. The ions are surrounded by shells of
  water of solvation. This reduces the number of
  solvent molecules that have enough kinetic
  energy to escape as vapour. The result is a
  solution with a lower vapour pressure than the
  pure solvent.
 The decrease in the vapour pressure is
  proportional to the number of particles the
  solute makes in solution. A solute like sodium
  chloride which dissociates into several
  particles has a greater effect on the vapour
  pressure than the same concentration of a
  solute like glucose which does not dissociate.
 NaCl – 2 particles when dissociated
 Glucose – 1 molecule, no disassociation
 CaCl2 – 3 particles when dissociated
 Raoult’slaw states that the vapour pressure
 of the solvent over the solution is
 proportional to the fraction of solvent
 molecules in the solution, if two-thirds of the
 molecules are solvent molecules, the vapour
 pressure due to the solvent is approximately
 two-thirds of what it would be for pure
 solvent at that temperature.
If the solute has a vapour pressure of its own,
   then the total vapour pressure over the
   solution would be:

vapour pressure =   vapour pressure x mole fraction
  (solution)         (solvent)
   Osmosis is the process whereby a solvent passes
    through a semipermeable membrane from one
    solution to another (or from a pure solvent into a
    solution). A semipermeable membrane is a barrier
    through which some substances may pass (e.g., the
    solvent particles), and other species may not (e.g.,
    the solute particles). Important examples of
    semipermeable membranes are the cell walls in cells
    of living things (plants and animals). Osmosis tends to
    drive solvent molecules through the semipermeable
    membrane from the low solute concentrations to the
    high solute concentrations; thus, a "complete”
    osmosis process would be one that ends with the
    solute concentrations being equal on both sides of
    the membrane. Osmotic pressure is the pressure that
    must be applied on the high concentration side to
    stop osmosis.
 Osmosis is a very useful process. For
 example, meats can be preserved by turning
 them into jerky: The meat is soaked in a very
 concentrated salt solution, resulting in
 dehydration of the meat cells. Jerky does not
 spoil as quickly as fresh meat, since bacteria
 on the surface of the salty meat will fall
 victim to osmosis, and shrivel up and die.
 This process thus extends the life of the
 meat without the use of refrigeration. There
 are times when one wishes to prevent
 osmosis when two solutions (or a pure
 solvent and a solution) are on opposite sides
 of a semipermeable membrane.
. Osmosis can be prevented by applying
 pressure to the more concentrated solution
 equal to the osmotic pressure on the less
 concentrated side. This can be accomplished
 either physically, by applying force to one
 side of the system, or chemically, by
 modifying a solute concentration so that the
 two solute concentrations are equal. (If one
 applies a pressure greater than the osmotic
 pressure to the higher concentration
 solution, one can force solvent molecules
 from the concentrated solution to the dilute
 solution, or pure solvent. This process,
 known as reverse osmosis, is often used to
 purify water.)
A hospital patient receiving fluids
 intravenously receives an intravenous (IV)
 solution that is isotonic with (i.e., at the
 same solute concentration as) his or her
 cells. If the IV solution is too concentrated,
 osmosis will cause the cells to shrivel; too
 dilute a solution can cause the cells to burst.
 Similar problems would be experienced by
 freshwater fish swimming in salt water, or
 saltwater fish swimming in freshwater. The
 osmotic pressure, like other colligative
 properties, does not depend on the identity
 of the solute, but an electrolyte solute will
 contribute more particles per formula unit
 than a nonelectrolyte solute.

				
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