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					Solution Chemistry
   (Chp. 7)

    Chemistry 2202
    Topics
 Molar Concentration (mol/L)
 Dilutions

 % Concentration (pp. 255 – 263)

 Solution Process

 Solution Preparation

 Solution Stoichiometry

 Dissociation
  Terms
solution           molar solubility   limiting reagent
solvent            saturated          excess reagent
solute             unsaturated        actual yield
concentrated       supersaturated     theoretical yield
dilute             dissociation       decanting
aqueous            electrolyte        pipetting
miscible           non-electrolyte    dynamic equilibrium
Immiscible         filtrate
alloy              precipitate
solubility     x
   Define the terms in bold and italics from
    pp. 237 – 240.

   Solids, liquids, and gases can combine
    to produce 9 different types of solution.
    Give an example of each type.

   p. 242 #’s 5, 7, 9, & 10
Terms
  solution       alloy
  solvent        solubility
  solute         molar solubility
  concentrated   saturated
  dilute         unsaturated
  aqueous        supersaturated
  miscible       dynamic equilibrium
  immiscible

          x
 Factors Affecting Solubility (pp.243 – 254)
1.       List 3 factors that affect the rate of
         dissolving.
2.       How does each of the following affect
         solubility?
         particle size
         temperature
         pressure
Factors Affecting Solubility
3. What type of solvent will dissolve:
   polar solutes

   nonpolar solutes

   ionic solutes

4. Why do some ionic compounds have
   low solubility in water?

     p. 254 #’s 1, 2, 4 - 6
Section 7.2 (pp. 243 – 252)

 State the generalizations regarding
  solubility and solutions (in italics)
 Define terms (in bold)
rate of dissolving     dipole
ion-dipole attractions
hydrated
electrolyte.
non-electrolytes
    Rate of Dissolving

 for most solids, the rate of dissolving is
  greater at higher temperatures
 stirring a mixture or by shaking the
  container increases the rate of dissolving.
 decreasing the size of the particles
  increases the rate of dissolving.
“Like Dissolves Like”

 ionic solutes and polar covalent
  solutes both dissolve in polar solvents
 non-polar solutes dissolve in non-
  polar solvents.
Solubility
 small molecules are often more
  soluble than larger molecules.
 the solubility of most solids increases
  with temperature.
 the solubility of most liquids is not
  greatly affected by temperature.
 the solubility of gases decreases as
  temperature increases
Solubility

   An increase in pressure increases the
    solubility of a gas in a liquid.
  Applications

1. An opened soft drink goes ‘flat’ faster if
  not refrigerated.
2. Warming of pond water may not be
  healthy for the fish living in it.
3. After pouring 5 glasses of pop from a
  2 litre container, Jonny stoppered the
  bottle and crushed it to prevent the
  remaining pop from going flat.
  Molar Concentration

Review:
- Find the molar mass of Ca(OH)2

- How many moles in 45.67 g of Ca(OH)2?

- Find the mass of 0.987 mol of Ca(OH)2.
Molar Concentration

The terms concentrated and dilute are
 qualitative descriptions of solubility.
A quantitative measure of solubility
 uses numbers to describe the
 concentration of a solution.
Molar Concentration

The MOLAR CONCENTRATION of a
solution is the number of moles of solute
(n) per litre of solution (v).
  Molar Concentration

FORMULA:
 Molar Concentration = number of moles
                       volume in litres


        C= n
           V
eg. Calculate the molar concentration of:
 4.65 mol of NaOH is dissolved to
  prepare 2.83 L of solution.

   15.50 g of NaOH is dissolved to
    prepare 475 mL of solution.

p. 268 - # 19
Eg. Calculate the following:
a) the number of moles in 4.68 L of 0.100 mol/L
   KCl solution.
b) the mass of KCl in 268 mL of 2.50 mol/L KCl
   solution.
c)    the volume of 6.00 mol/L HCl(aq) that can be
      made using 0.500 mol of HCl.
d)    the volume of 1.60 mol/L HCl(aq) that can be
      made using 20.0 g of HCl.




     p. 268 #’s20-24
Dilution (p. 272)       Number of moles
                         before dilution
When a solution is diluted:
- The concentration decreases

- The volume increases

-   The number of moles remains
    the same
         ni = nf       Number of moles
                         after dilution
  Dilution (p. 272)
          ni = nf
         Ci Vi = C f Vf
eg. Calculate the molar concentration of a
 vinegar solution prepared by diluting 10.0
 mL of a 17.4 mol/L solution to a final
 volume of 3.50 L.
p. 273 #’s 25 – 27

p. 276   #’s 1, 2, 4, & 5

  DON’T SHOW UP UNLESS
      THIS IS DONE!!
    Solution Preparation & Dilution
   standard solution – a solution of known
    concentration
   volumetric flask – a flat-bottomed glass vessel
    that is used to prepare a standard solution
    delivery pipet – pipets that accurately measure
    one volume
    graduated pipet – pipets that have a series of
    lines that can be use to measure many different
    volumes
  To prepare a standard solution:
1. calculate the mass of solute needed
2. weigh out the desired mass
3. dissolve the solute in a beaker using less
  than the desired volume
4. transfer the solution to a volumetric flask
  (rinse the beaker into the flask)
5. add water until the bottom of the
  meniscus is at the etched line
    To dilute a standard solution:
 1. Rinse the pipet several times with
  deionized water.
 2. Rinse the pipet twice with the standard
  solution.
 3. Use the pipet to transfer the required
  volume.
 4. Add enough water to bring the solution
  to its final volume.
    Percent Concentration
 Concentration may also be given as a %.
 The amount of solute is a percentage of
  the total volume/mass of solution.
   liquids in liquids - % v/v

   solids in liquids - % m/v

   solids in solids - % m/m
  Percent Concentration
                   mass of solute (g)
Percent (m/v)                          x 100
                volume of solution (mL)


   p. 258 #’s 1 – 3    DSUUTID!!
                mass of solute (g)
Percent (m/m)                       x 100
                mass of solution (g)

p. 261 #’s 5 – 9


             DSUUTID!!
                 volume of solute (mL)
Percent (v/v)                          x 100
                volume of solution (mL)

p. 263 #’s 10 – 13


             DSUUTID!!
Concentration in ppm and ppb
 Parts per million (ppm) and parts per
 billion (ppb) are used for extremely
 small concentrations
            ppm x msolution
msolute           6
                10
            ppb x msolution
msolute          9
                10
eg. 5.00 mg of NaF is dissolved in
  100.0 kg of solution. Calculate the
  concentration in:
a) ppm
b) ppb
ppm = 0.005 g x 106
     100,000 g
   = 0.05 ppm

ppb = 0.005 g x 109
      100,000 g
    = 50.0 ppb
   p. 265 #’s 15 – 17

   pp. 277, 278
      #’s 11, 13, 15 – 18, 20

    DON’T SHOW UP UNLESS
        THIS IS DONE!!
Solution Stoichiometry
1. Write a balanced equation
2. Calculate moles given
   n=m/M       OR        n=CV
3. Mole ratios
4. Calculate required quantity

   n                 n
V        OR      C           OR   m  nM
   C                 V
Solution Stoichiometry
eg. 45.0 mL of a HCl(aq) solution is used
 to neutralize 30.0 mL of a 2.48 mol/L
 NaOH solution.
 Calculate the molar concentration of the
 HCl(aq) solution.

p. 304: #’s 16, 17, & 18
Worksheet
  Sample Problems
1. What mass of copper metal is needed
  to react with 250.0 mL of 0.100 mol/L
  silver nitrate solution?
2. Calculate the volume of 2.00 M HCl(aq)
  needed to neutralize 1.20 g of dissolved
  NaOH.
3. What volume of 3.00 mol/L HNO3(aq) is
  needed to neutralize 450.0 mL of 0.100
  mol/L Sr(OH)2(aq)?
  Sample Problem Solutions
 Cu(s) + 2 AgNO3(aq) → 2 Ag(s) + Cu(NO3)2(aq)

Step 2   n = 0.02500 mol AgNO3

Step 3   n = 0.01250 mol Cu

Step 4   m = 0.794 g Cu
  Sample Problem Solutions
 HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

Step 2   n = 0.0300 mol NaOH

Step 3   n = 0.0300 mol HCl

Step 4   V = 0.0150 L HCl
  Sample Problem Solutions
2 HNO3(aq) + Sr(OH)2(aq) →
                        2 H2O(l) + Sr(NO3)2(aq)
Step 2 n = 0.04500 mol Sr(OH)2

Step 3   n = 0.0900 mol HNO3

Step 4   V = 0.0300 mol/L HNO3
The Solution Process (p. 299)
 Dissociation occurs when an ionic
  compound breaks into ions as it
  dissolves in water.
 A dissociation equation shows what
  happens to an ionic compound in water.
eg. NaCl(s) → Na+(aq) + Cl-(aq)

    K2SO4(s) → 2 K+(aq) + SO42-(aq)
The Solution Process (p. 299)
 Solutions of ionic compounds conduct
  electric current.
 A solute that conducts an electric
  current in an aqueous solution is
  called an electrolyte.
The Solution Process (p. 299)
 Acids are also electrolytes.
 Acids form ions when dissolved in
  water.
 eg. H2SO4(aq) → 2 H+(aq) + SO42-(aq)

    HCl(s) → H+(aq) + Cl-(aq)
     The Solution Process (p. 299)
Molecular Compounds DO NOT
 dissociate in water.
eg. C12H22O11(s) → C12H22O11(aq)

   Because they DO NOT conduct electric
    current in solution, molecular compounds
    are non-electrolytes.
  The Solution Process (p. 299)
The molar concentration of any dissolved
 ion is calculated using the ratio from the
 dissociation equation.
eq. What is the molar concentration of
 each ion in a 5.00 mol/L MgCl2(aq)
 solution:


   5.00 mol/L      5.00 mol/L   10.00 mol/L
p. 300 #’s 7 – 9

What mass of calcium chloride is
required to prepare 2.00 L of 0.120
mol/L Cl-(aq) solution?

p. 302 # 14
p. 311 #’s 11, 12, 16, & 18

				
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