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.
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)
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.
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?
Simple Rules for Solubility
1. Most nitrate (NO3) salts are soluble.
2. Most alkali (group 1A) salts and NH4+ are soluble.
3. Most Cl, Br, and I salts are soluble (NOT Ag+,
Pb2+, Hg22+)
4. Most sulfate salts are soluble (NOT BaSO4, PbSO4,
HgSO4, CaSO4)
5. Most OH salts are only slightly soluble (NaOH,
KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally
soluble)
6. Most S2, CO32, CrO42, PO43 salts are only
slightly soluble.
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.
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.
(11.5g NaOH/1.50L)(1 mol NaOH/40.00g NaOH)
= 0.192M NaOH
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
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-
Molarity Calculations
Typical blood serum is about 0.14M NaCl. What
volume of blood contains 1.0 mg of NaCl?
(1.0mg NaCl)(1g/1000mg)(1mol/58.45g)(1L/0.14mol)
= 1.2 x 10-4 L blood serum
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.
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)
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.
04_46
Rubber bulb
500 mL
(a)
(b)
Steps to dilute a stock solution.
(c)
Dilution Calculations
What volume of 16 M sulfuric acid must be used to
prepare 1.5 L of a 0.10 M H2SO4 solution?
(0.10mol H2SO4/1L)(1.5L)(1L/16mol)(1000mL/1L)
= 9.4 mL conc H2SO4
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)
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.
Simple Rules for Solubility
1. Most nitrate (NO3) salts are soluble.
2. Most alkali (group 1A) salts and NH4+ are soluble.
3. Most Cl, Br, and I salts are soluble (NOT Ag+,
Pb2+, Hg22+)
4. Most sulfate salts are soluble (NOT BaSO4, PbSO4,
HgSO4, CaSO4)
5. Most OH salts are only slightly soluble (NaOH,
KOH are soluble, Ba(OH)2, Ca(OH)2 are marginally
soluble)
6. Most S2, CO32, CrO42, PO43 salts are only
slightly soluble.
Solubility
Using the solubility rules, predict what will
happen when the following pairs of solutions
are mixed.
No precipitate forms
a) KNO3(aq) & BaCl2(aq)
b) Na2SO4(aq) & Pb(NO3)2(aq) PbSO4(s) forms
c) KOH(aq) & Fe(NO3)3(aq)
Fe(OH)3(s) forms
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.
Precipitation Calculations
When aqueous solutions of Na2SO4 &
Pb(NO3)2 are mixed, PbSO4 precipitates.
Calculate the mass of PbSO4 formed when
1.25 L of 0.0500 M Pb(NO3)2 & 2.00 L of
0.0250 M Na2SO4 are mixed.
1. Species present are Na+, SO42-, Pb2+, &
NO3-.
2. Pb2+(aq) + SO42-(aq) ----> PbSO4(s)
Precipitation Calculations
Continued
3. (1.25L)(0.500mol Pb2+/L) = 0.0625 mol Pb2+
(2.00L)(0.0250mol SO42-/L) = 0.0500 mol SO424. (0.0625mol Pb2+)(1mol SO42-/1mol Pb2+) =
0.0625 mol SO42SO42- is the limiting reactant.
5. (0.0500mol SO42-)(1mol PbSO4/1mol SO42-)
(303.3g/1mol) = 15.2 g PbSO4
Performing Calculations for
Acid-Base Reactions
1.
2.
3.
4.
5.
6.
List initial species and predict reaction.
Write balanced net ionic reaction.
Calculate moles of reactants.
Determine limiting reactant.
Calculate moles of required reactant/product.
Convert to grams or volume, as required.
Acid-Base Calculations
What volume of a 0.100M HCl solution is needed
to neutralize 25.0 mL of 0.350 M NaOH?
1. H+, Cl-, Na+, & OH2. H+(aq) + OH-(aq) ----> HOH(l)
(25.0mL)(0.350mol NaOH/1L)(1mol HCl/1mol
NaOH)(1L/0.100mol) = 87.5 mL HCl solution
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.
Titration Calculations
If 41.20 mL of NaOH is required to neutralize 1.3009 g
of KHP (KHC8H4O6), what is the concentration of
the NaOH solution? Assume there is one acidic
hydrogen in KHP.
HC8H4O6-(aq) + OH-(aq) ----> HOH(l) + C8H4O62-(aq)
(1.3009g KHP)(1mol/204.22g)(1mol NaOH/1mol
KHP)(1/41.20mL)(1000mL/1L) = 0.1546 M NaOH
Redox Reactions
Redox reactions are reactions in which
electrons are transferred.
Decomposition and synthesis reactions may be
redox.
Single replacement reactions are always redox.
Double replacement reactions are never redox.
Combustion reactions are always redox.
OIL RIG
Oxidation Is Loss.
Reduction Is Gain.
Redox
Oxidizing agent is the electron acceptor-usually a nonmetal.
Reducing agent is the electron donor--usually
a metal.
CH4(g) + 2O2(g) ----> CO2(g) + 2HOH(g)
Carbon is oxidized.
Oxygen is reduced.
CH4 is the reducing agent.
O2 is the oxidizing agent.
Rules for Assigning
Oxidation States
1. Oxidation state of an atom in an element = 0
2. Oxidation state of monatomic element = charge
3. Oxygen = 2 in covalent compounds (except in
peroxides where it = 1)
4. H = +1 in covalent compounds
5. Fluorine = 1 in compounds
6. Sum of oxidation states = 0 in compounds
Sum of oxidation states = charge of the ion
Determining Oxidation States
SF6
+6
-6 = 0
(-1 for each F)
[NO3]+5 -6 = -1
(-2 for each O)
Balancing by Half-Reaction
Method
1. Write separate reduction, oxidation
reactions.
2. For each half-reaction:
- Balance elements (except H, O)
- Balance O using H2O
- Balance H using H+
- Balance charge using electrons
Balancing by Half-Reaction
Method (continued)
3. If necessary, multiply by integer to
equalize electron count.
4. Add half-reactions.
5. Check that both elements and charges are
balanced.
Balancing By Half-Reaction
Acidic Solution
H+(aq) + Cr2O72-(aq) + C2H5OH(l) ---> Cr3+(aq) + CO2(g) + HOH(l)
Red
Ox
Cr2O72-(aq) ---> Cr3+(aq)
C2H5OH(l) ---> CO2(g)
Red 2(6e- + 14H+(aq) + Cr2O72-(aq) ---> 2Cr3+(aq) + 7HOH(l))
Ox
C2H5OH(l) + 3HOH(l) ---> 2CO2(g) + 12H+(aq) + 12e16H+(aq) + 2Cr2O72-(aq) + C2H5OH(l) ---> 4Cr3+(aq) + 11HOH(l) +
2CO2(g)
12+ = 12+
Half-Reaction Method Balancing in Base
1. Balance as in acid.
2. Add OH that equals H+ ions (both sides!)
3. Form water by combining H+, OH.
4. Check elements and charges for balance.
Balancing By Half-Reaction
Basic Solution
Ag(s) + CN-(aq) + O2(g) ---> Ag(CN)2-(aq)(Basic)
Ox CN-(aq) + Ag(s) ---> Ag(CN)2-(aq)
Red O2(g) --->
Ox 4(2CN-(aq) + Ag(s) ---> Ag(CN)2-(aq) + e-)
Red O2(g) + 4H+(aq) + 4e- ---> 2HOH(l)
8CN-(aq) + 4Ag(s) + O2(g) + 4H+(aq) ---> 4Ag(CN)2-(aq) +
2HOH(l)
Balancing By Half-Reaction
Basic Solution(Continued)
8CN-(aq) + 4Ag(s) + O2(g) + 4H+(aq) + 4OH-(aq) --->
4Ag(CN)2-(aq) + 2HOH(l) + 4OH-(aq)
8CN-(aq) + 4Ag(s) + O2(g) + 4HOH(l) ---> 4Ag(CN)2-(aq) +
2HOH(l) + 4OH-(aq)
8CN-(aq) + 4Ag(s) + O2(g) + 2HOH(l) ---> 4Ag(CN)2-(aq) +
4OH-(aq)
8- = 8-