1. Chapter 12 Solutions Copyright © 2008 by Pearson Education, Inc.
Publishing as Benjamin Cummings

2. Solute and Solvent Solutions
Are homogeneous mixtures of two or more substances.
Consist of a solvent and one or more solutes.
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3. Nature of Solutes in Solutions
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.
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4. Examples of Solutions
The solute and solvent can be a solid, liquid, and/or
a gas.
Table 12.3
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5. 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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6. Formation of a Solution
Na+ and Cl- ions
On the surface of a NaCl crystal are attracted to polar water molecules.
In solution are hydrated as several H2O molecules surround each.
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7. Equations for Solution Formation
When NaCl(s) dissolves in water, the process can be written as:
H2O
NaCl(s) Na+(aq) + Cl- (aq)
solid separation of ions
The Na+ ions are attracted to the oxygen atom ( -) of water.
The Cl- ions are attracted to the hydrogen atom (+) of water.

8. 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.

9. Water and a Polar Solute
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10. Like Dissolves Like Solvents Solutes Water (polar) Ni(NO3)2
CH2Cl2(nonpolar) (polar)
I2 (nonpolar)
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11. Electrolytes and Nonelectrolytes
Chapter Solutions
Electrolytes and Nonelectrolytes
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12. 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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13. Strong Electrolytes Strong electrolytes
Dissociate in water producing positive and negative ions.
Conduct an electric current in water.
In equations show the formation of ions in aqueous (aq) solutions.
H2O % ions
NaCl(s) Na+(aq) + Cl− (aq)
H2O
CaBr2(s) Ca2+(aq) + 2Br− (aq)

14. Weak Electrolytes A weak electrolyte
Dissociates only slightly in water.
In water forms a solution of only a few ions and mostly undissociated molecules.
HF(g) H2O(l) H3O+(aq) + F- (aq)
NH3(g) + H2O(l) NH4+(aq) + OH- (aq)
Note: Unequal lengths of the arrows

15. Nonelectrolytes Nonelectrolytes Dissolve as molecules in water.
Do not produce ions in water.
Do not conduct an electric current.
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16. Chapter 12 Solutions Solubility
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17. Solubility Solubility
Is the maximum amount of solute that dissolves in a specific amount of solvent.
Can be expressed as grams of solute in 100 grams of solvent, usually water.
g of solute
100 g water

18. Effect of Temperature on Solubility
Depends on temperature.
Of most solids increases as temperature increases.
Of gases decreases as temperature increases.
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19. Unsaturated Solutions
Contain less than the maximum amount of solute.
Can dissolve more solute.
Dissolved solute
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20. Saturated Solutions Saturated solutions
Contain the maximum amount of solute that can dissolve.
Have undissolved solute at the bottom of the container.
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21. Supersaturated Solutions

22. 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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23. Solubility Rules Soluble salts
Typically contain at least one ion from Groups 1A(1) or
NO3−, or C2H3O2− (acetate).
Table 12.3
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24. Using the Solubility Rules
The solubility rules predict if a salt
Is soluble or
Insoluble in water.
Table 12.4
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25. Formation of a Solid When solutions of salts are mixed,
A solid forms if ions of an insoluble salt are present.
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26. 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. Ions remaining in solution are spectator ions.
Pb2+(aq) + 2Cl−(aq) PbCl2(s)

27. Equations for the Insoluble Salt
STEP 1 Observe the ions in the reactants.
Pb2+(aq) + 2NO3−(aq)
2Na+(aq) + 2Cl−(aq)
STEP 2 Determine if any new ion combinations are
insoluble salts. Yes. PbCl2(s)
STEP 3 Ionic equation with insoluble salt product.
Pb2+(aq) + 2NO3−(aq) + 2Na+(aq) + 2Cl−(aq)
PbCl2(s) + 2Na+(aq) + 2NO3−(aq)
STEP 4 Net ionic equation.
Pb2+(aq) + 2Cl−(aq) PbCl2(s)

28. Chapter 12 Solutions Molarity and Dilution
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29. Molarity (M) Molarity (M) is A concentration term for solutions.
The moles of solute in 1 L solution.
moles of solute
liter of solution

30. Preparing a 1.0 Molar Solution
A 1.00 M NaCl solution is prepared
By weighing out 58.5 g NaCl (1.00 mol) and
Adding water to make 1.00 liter of solution.
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31. Calculation of Molarity
What is the molarity of L NaOH solution if it
contains 6.00 g NaOH?
STEP 1 Given g NaOH in L solution
Need molarity (mol/L)
STEP 2 Plan g NaOH mol NaOH molarity
STEP 3 Conversion factors 1 mol NaOH = g
1 mol NaOH and g NaOH
40.00 g NaOH mol NaOH

32. Calculation of Molarity (cont.)
STEP 4 Calculate molarity.
6.00 g NaOH x 1 mol NaOH = mol
40.00 g NaOH
0.150 mol = mol = M NaOH
0.500 L L

33. Molarity Conversion Factors
The units of molarity are used as conversion factors in calculations with solutions.
Table 2.6
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34. Molarity in Calculations
How many grams of KCl are needed to prepare
125 mL of a M KCl solution?
STEP 1 Given mL (0.125 L) of M KCl
Need Grams of KCl
STEP 2 Plan L KCl mol KCl g KCl

35. Molarity in Calculations (cont.)
STEP 3 Conversion factors
1 mol KCl = g
1 mol KCl and g KCl
74.55 g KCl mol KCl
1 L KCl = mol KCl
1 L and mol KCl
0.720 mol KCl L
STEP 4 Calculate grams.
0.125 L x mol KCl x g KCl = 6.71 g KCl
1 L mol KCl

36. Dilution In a dilution Water is added. Volume increases.
Concentration decreases.
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37. Comparing 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 equation
M1V1 = M2V2
initial diluted

38. Dilution Calculations
What is the molarity if L of M KOH is
diluted to a final volume of L?
STEP 1 Prepare a table:
M1= M V1 = L
M2= ? V2 = L
STEP 2 Solve dilution equation for unknown.
M1V1 = M2V2 M1V1/ V2 = M2
STEP 3 Set up and enter values:
M2 = M1V1 = (0.600 M)(0.180 L) = M
V L