Types of Chemical Reactions
& Solution Stoichiometry
Chapter 4
Aqueous Solutions
Water is the dissolving
medium, or solvent.
Some Properties of Water
Water is able to dissolve so many substances
because:
- Water is “bent” or V-shaped.
- The O-H bonds are covalent.
- Water is a polar molecule.
- Hydration occurs when salts dissolve in
water.


04_40
H

2
105
O
H


Water is a polar molecule because it is a bent
molecule. The hydrogen end is + while the oxygen
end is -, Delta () is a partial charge--less than 1.
04_41
+



 +
+ 
 +
+ 
H
O H
+
Cation




+ 
 +
+ 
+

+

H
H O



+

+


Anion

Polar water molecules interact with the positive
and negative ions of a salt, assisting in the
dissolving process. This process is called hydration.
A Solute
-
dissolves in water (or other “solvent”)
changes phase (if different from the
solvent)
is present in lesser amount (if the same
phase as the solvent)
A Solvent
-
retains its phase (if different from the
solute)
is present in greater amount (if the same
phase as the solute)
Solubility
The general rule for solubility is:
“Like dissolves like.”
Polar water molecules can dissolve other polar
molecules such as alcohol and, also, ionic
substances such as NaCl.
Nonpolar molecules can dissolve other
nonpolar molecules but not polar or ionic
substances. Gasoline can dissolve grease.
Miscibility
Miscible -- two substances that will mix
together in any proportion to make a
solution. Alcohol and water are miscible
because they are both polar and form
hydrogen bonds.
Immiscible -- two substances that will not
dissolve in each other. Oil and vinegar are
immiscible because oil is nonpolar and
vinegar is polar.
Solubility
How does the rule “Like dissolves like.” apply to
cleaning paint brushes used for latex paint as
opposed to those used with oil-based paint?
C6H14
H20
I2
C6H14
H20
I2
NaNO3
H20
Electrolytes & Nonelectrolytes
An electrolyte is a material that dissolves in
water to give a solution that conducts an
electric current.
A nonelectrolyte is a substance which, when
dissolved in water, gives a nonconducting
solution.
Electrolytes
Strong - conduct current efficiently and are
soluble salts, strong acids, and strong bases.
NaCl, KNO3, HNO3, NaOH
Weak - conduct only a small current and are
weak acids and weak bases.
HC2H3O2, aq. NH3, tap H2O
Non - no current flows and are molecular
substances
pure H2O, sugar solution, glycerol
Power Source
04_43
+ 
+
+



+


+
+
(a)
(b)
(c)
Electrical conductivity of aqueous solutions. a) strong
electrolyte b) weak electrolyte c) nonelectrolyte in solution.
Svante Arrhenius first identified these electrical properties.
04_1529
BaCl2(s)
dissolves
= Ba2+
= Cl
When BaCl2 dissolves, the Ba2+ and Cl- ions are randomly
dispersed in the water. BaCl2 is a strong electrolyte.
Acids
Strong acids - dissociate completely (~100 %) to
produce H+ in solution
HCl, H2SO4, HNO3, HBr, HI, & HClO4
Weak acids - dissociate to a slight extent (~ 1 %)
to give H+ in solution
HC2H3O2, HCOOH, HNO2, & H2SO3
04_1530
 +
+

+
+




+

+
+

+
+


+
+ = H+

= Cl
HCl is completely ionized and is a strong electrolyte.
Bases
Strong bases - react completely with water to
give OH ions. sodium hydroxide
NaOH(s) ---> Na+(aq) + OH-(aq)
Weak bases - react only slightly with water to
give OH ions. ammonia
NH3(aq) + HOH(l) <---> NH4+(aq) + OH-(aq)
04_1531

+

+

+



+

+
+
+
+

+

-
= OH
+
= Na+

+
An aqueous solution of sodium hydroxide which is
a strong bases dissociating almost 100 %.
04_1532
Acetic acid(CH3COOH) exists in water mostly as undissociated
molecules. Only a small percent of the molecules are ionized.
Write the equation of the dissolving of the
following compounds.
CaCl2
HCl
Fe(NO3)3
KBr
(NH4)2Cr2O7
Molarity
Molarity (M) = moles of solute per volume of
solution in liters:
moles of solute
M  molarity 
liters of solution
6 moles of HCl
3 M HCl 
2 liters of solution
Molarity Calculations
Calculate the molarity of a solution prepared by
dissolving 11.5 g of solid NaOH in enough
water to make 1.50 L of solution.
(
Molarity Calculations
Calculate the molarity of a solution prepared by
dissolving 1.56 g of gaseous HCl in enough
water to make 26.8 mL of solution.
(1.56g HCl/26.8mL)(1 mol HCl/36.46g HCl)
(1000mL/1L) = 1.60M HCl
How many moles of Co(NO3)2 are
present in 25.00 mL of a 0.75 M
Co(NO3)2 solution?
Molarity Calculations
How many moles of nitrate ions are present in
25.00 mL of a 0.75 M Co(NO3)2 solution?
(25.00mL)(1L/1000mL)(0.75mol Co(NO3)2/1L)
(2 mol NO3-/1 mol Co(NO3)2) = 3.8 x 10-2 mol NO3-
mol
Calculate the moles of each of the ions in
40.00 ml of the following solution.
.20 M Na2CO3
P106 Q1,-,7,
8,9a-c
Calculate the moles of potassium ions in
50.00 ml of the following solution.
2 M K3P2
Standard Solution
A standard solution is a solution whose
concentration is accurately known.
Standard solutions are made using a volumetric flask
as follows:
• mass the solute accurately and add it to the
volumetric flask
• add a small quantity of distilled HOH
• dissolve the solute by gently swirling the flask
• add more distilled HOH until the level of the
solution reaches the mark on the neck
• invert the capped volumetric 25X to thoroughly
mix the solution.
04_44
Volume marker
(calibration mark)
Wash Bottle
Weighed
amount
of solute
(a)
(b)
(c)
(d)
Steps involved in making a standard solution.
04_46
Rubber bulb
500 mL
(a)
(b)
Steps to dilute a stock solution.
(c)
Common Terms of Solution
Concentration
Stock - routinely used solutions prepared in
concentrated form.
Concentrated - relatively large ratio of solute
to solvent. (5.0 M NaCl)
Dilute - relatively small ratio of solute to
solvent. (0.01 M NaCl)
Make a solution demo
Dilution of Stock Solutions
When diluting stock solutions, the moles of
solute after dilution must equal the moles
of solute before dilution.
Stock solutions are diluted using either a
measuring or a delivery pipet and a
volumetric flask.
Dilution Calculations
What volume of 6 M sulfuric acid must be used to
prepare 1 L of a 3.0 M H2SO4 solution?
use dilution formula
What volume of 3 M sulfuric acid must be
used to prepare .5 L of a .25 M H2SO4
solution?
Types of Solution Reactions
-
Precipitation reactions
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
-
Acid-base reactions
NaOH(aq) + HCl(aq)  NaCl(aq) + H2O(l)
-
Oxidation-reduction reactions
Fe2O3(s) + 2Al(s)  2Fe(l) + Al2O3(s)
1.
Which of the following substances would
you expect to be insoluble in water?
Barium hydroxide
Magnesium sulfate
Silver chloride
Calcium carbonate
Ammonium acetate
Sodium hydroxide
Silver nitrate
Hydrochloric acid
Ammonium nitrate
Lithium carbonate
Barium sulfate
Lead I Chloride
Ammonium nitrate
Solubility
Using the solubility rules, predict what will
happen when the following pairs of solutions
are mixed.
a) KOH(aq) & Mg(NO3)2(aq) Mg(OH)2(s) forms
b) Na2SO4(aq) & Pb(NO3)2(aq)
c) KNO3(aq) & BaCl2(aq)
PbSO4(s) forms
No precipitate forms
Describing Reactions in
Solution
1. Molecular equation (reactants and
products as compounds)
AgNO3(aq) + NaCl(aq)  AgCl(s) + NaNO3(aq)
2. Complete ionic equation (all strong
electrolytes shown as ions)
Ag+(aq) + NO3(aq) + Na+(aq) + Cl(aq) 
AgCl(s) + Na+(aq) + NO3(aq)
Describing Reactions in
Solution (continued)
3. Net ionic equation (show only
components that actually react)
Ag+(aq) + Cl(aq)  AgCl(s)
Na+ and NO3 are spectator ions.
Write the balanced complete ionic and
net ionic equations:
CuSO4(aq) + BaCl2(aq) →
P108 q 28
Sodium Sulfate and Lead II Nitrate
Old limiting problem to
compare
If 68.5 g of CO(g) is reacted with 8.60 g of
H2(g), what is the theoretical yield of
methanol that can be produced?
__H2(g) + __CO(g) ---> __CH3OH(l)
Precipitation Calculations
When aqueous solutions of Na2SO4 & Pb(NO3)2 are
mixed. Calculate the mass of the percipitate
formed when 1.25 L of 0.0500 M Pb(NO3)2 &
2.00 L of 0.0250 M Na2SO4 are mixed.
1.
04_
STOICHIOMETRY FOR REACTIONS IN SOLUTION
STEP 1
Identify the species present in the combined solution, and determine
what reaction occurs.
STEP 2
Write the balanced net ionic equation for the reaction.
STEP 3
Calculate the moles of reactants.
STEP 4
Determine which reactant is limiting.
STEP 5
Calculate the moles of product or products, as required.
STEP 6
Convert to grams or other units, as required.
3.
Precipitation Calculations
Continued
(1.25L)(0.0500mol Pb(NO3)2/1L) =
0.0625 molPb(NO3)2
(2.00L)(0.0250mol Na2SO4/1L) = 0.0500 molNa2SO4
4.(0.0625 mol Pb(NO3)2 (1mol Na2SO4/1mol Pb(NO3)2 ) =
0.0625 mol SO42Na2SO4 is the limiting reactant.
5. (0.0500mol Na2SO4)(1mol PbSO4/1mol Na2SO4)
(303.3g/1mol PbSO4) = 15.2 g PbSO4
PRACTICE
What mass of precipitate should
result when 0.550 L of 0.500 M
aluminum nitrate solution is
mixed with 0.240 L of 1.50 M
sodium hydroxide solution?
answer
9.3g
What mass of precipitate should result
when 0.350 L of 0.200 M aluminum
nitrate solution is mixed with 0.540 L
of .50 M sodium hydroxide solution?
What is the net ionic equation?
P108 29-32
What volume of 0.415 M silver
nitrate will be required to
precipitate as silver bromide
all the bromide
ion in 35.0 mL of 0.128 M
calcium bromide?
answer
2 AgNO3(aq) + CaBr2(aq)  Ca(NO3)2(aq) + 2 AgBr(s)
= 0.0216 L AgNO3
0.0350 L CaBr2 0.128 moles CaBr2 2 moles AgNO3 1 L AgNO3
1 L CaBr2
1 moles CaBr2 0.415 mole AgNO3
How many mL of 0.280 M barium nitrate are
required to precipitate as barium sulfate all
the sulfate ions from 25.0 mL of 0.350 M
aluminum sulfate?
(93.8 mL barium nitrate)
answer
Acid-Base Calculations
What volume of a 0.100M HCl solution is needed
to neutralize 25.0 mL of 0.350 M NaOH?
1. Cl-, Na+,
2. 2. H+(aq) + OH-(aq) ----> HOH(l)
(25.0mL)(0.350mol NaOH/1L)(1mol HCl/1mol
NaOH)(1L/0.100mol) = 87.5 mL HCl solution
What volume of a 0.500M HCl solution is
needed to neutralize 32.0 mL of 0.250 M
LiOH?
Key Titration Terms
Titrant - solution of known concentration used
in titration
Analyte - substance being analyzed
Equivalence point - enough titrant added to
react exactly with the analyte
Endpoint - the indicator changes color so you
can tell the equivalence point has been
reached.
Ammonium sulfate is manufactured by
reacting sulfuric acid with potassium
hydroxide. What concentration of
sulfuric acid is needed to react with
24.4 mL of a 2.20 mol/L potassium
hydroxide solution if 50.0 mL of sulfuric
acid is used?
Titration Calculations
Titration
49,a,b 53,54
practice for unit Textbook 181-183
15a
17a
Additional questions
23a
36,a,b
29a-c
41
40
35ab
18,b
39 try
12
45b
24
reminder ,Do redox in redox