Chemistry 101 : Chap. 4
Aqueous Reactions and Solution Stoichiometry
(1) General Properties of Aqueous Solutions
(2) Precipitation Reactions
(3) Acid-Base Reactions
(4) Concentration of Solutions
(5) Solution Stoichiometry and Chemical Analysis
Aqueous Solutions
 Solution : A homogeneous mixture of two or more substance
 Solvent : The substance present in the largest amount
 Solute(s) : The other substance in the solution
Aqueous Solutions : Solutions in which water is the dissolving
medium (or solvent)
Ionic Compounds in Water
When ionic compounds dissolve in water, they dissociate completely
into ions. The ions are surrounded by water molecules
-
+
+
-
-
+
+
-
NaCl (s)
+ H2O
-
+
+
NaCl (aq) or Na+ (aq) + Cl- (aq)
[ Solvation of Na+ and Cl- ]
Molecular Compounds in Water
When molecular compounds dissolve in water, they do not
dissociate into ions in general. The molecules remain intact and
are surrounded by water molecules
+ H2O
CH3OH (l)
CH3OH (aq)
Molecular Compounds in Water
Some molecular compounds, when dissolved in water, dissociate
(ionoized) into ions.
 Strong Acid, such as hydrochloric acid, dissociate completely
HCl (aq)

H+ (aq)
+
Cl- (aq)
 Weak Acid, such as acetic acid, dissociate only partially
CH3COOH (aq)
H+ (aq)
+ CH3COO- (aq)
NOTE: All three species are present in solutions
Electrolyte vs. Nonelectrolyte
 Electrolyte : A substance whose aqueous solutions contains ions
 Ionic compounds
strong electrolyte
 Strong acids
strong electrolyte
 Weak acids
weak electrolyte
 Nonelectrolyte : A substance that does not form ions in solution
NOTE: Don’t confuse the extent to which an electrolyte
dissolves with whether it is strong or weak
Ex. CH3COOH is extremely soluble, but it is a weak electrolyte.
Ba(OH)2 is not very soluble, but it is a strong electrolyte.
 the dissolved Ba(OH)2 completely dissociate
Electrolyte Solutions
Electrolyte solutions conduct electricity !!
 Electrical conductivity is the flow of charged particles through
substance, such as the flow of electrons through a wire or
the flow of ions through a solution.
 Testing electrical conductivity of electrolyte solutions
Electrolyte Solutions
Precipitation Reactions
 Recall : Three types of chemical reactions studied in Ch. 3
 Combination Reaction : A + B  C
 Decomposition Reaction : A  B + C
 Combustion Reaction : CxHy + O2  CO2 + H2O
 Exchange Reaction (Metathesis Reaction)
AB + CD
 AD
+ CB
Reactants “exchange partners” to form the products
Precipitation Reactions
 Precipitation Reactions : Reactions that result in the formation
of insoluble products
Pb(NO3)2 (aq) + 2KI (aq)  PbI2 (s) + 2KNO3 (aq)
yellow ppt
AgNO3 (aq) + NaCl (aq)  AgCl (s) + NaNO3 (aq)
white ppt
Mg(NO3)2 (aq) + 2NaOH (aq)  Mg(OH)2 (s) + 2NaNO3 (aq)
white ppt
Precipitation Reactions
What is happening in molecular level (nanoscale view)?
AgNO3 (aq) + NaI (aq)  AgI (s) + NaNO3 (aq)
brown ppt
+
NOTE: Water molecules are omitted in these pictures
Solubility Guidelines
 Question: How can you tell if a precipitation will form during an
exchange reaction?
Solubility Guidelines
 Example: Which of the following compounds is likely soluble
in water?
Fe(NO3)3
AgCl
CaCO3
(NH4)2SO4
CuSO4
Cu(OH)2
Solubility Guidelines
 Empirical Rule : Salts containing following ions are always soluble
 Group 1A ion (alkali metal):
Li+, Na+, K+, …
 Ammonium ion :
 Nitrate ion :
NH4+
NO3-
 Acetate :
CH3COO-
 Insoluble Ionic compounds :
Ag+, Hg2+, Pb2+
+
Halogen anions (Cl-, Br-, I-)
Predicting Precipitation Reactions
 Example : Complete the following equation and determine whether
or not a precipitation reaction (exchange reaction) will
occur. If a reaction will occur, balance the equation
Pb(NO3)2 (aq) + KI (aq) 
(1) Determine the products of exchange reaction:
(2) Determine if there is any insoluble product
(3) Balance the chemical equation
Predicting Precipitation Reactions
 Example : Complete the following equation and determine whether
or not a precipitation reaction (exchange reaction) will
occur. If a reaction will occur, balance the equation
Mg(NO3)2 (aq) + KCl (aq) 
(1) Determine the products of exchange reaction:
(2) Determine if there is any insoluble product
(3) Balance the chemical equation (?)
Net Ionic Equation
 Molecular equation :
Pb(NO3)2 (aq) + 2 KI (aq)  PbI2 (s) +2 KNO3 (aq)
 (complete) Ionic equation:
Pb2+(aq) + 2NO3-(aq) + 2K+(aq) + 2I- (aq) 
PbI2 (s) + 2K+(aq) + 2NO3-(aq)
spectator ions
 Net ionic equation :
Pb2+(aq) + 2I- (aq)  PbI2 (s)
NOTE: Mg(NO3)2 (aq) + 2 KCl (aq)  All ions are spectators !
No net reaction !!
Net Ionic Equation
 Summary:
 Spectator Ions : Ions present in solution before and after
the reaction.
 Net Ionic Equation : Spectators are omitted and only the ions
involved in forming the precipitation are shown
NOTE: Net ionic equation must be mass balanced and charge
balanced.
 Writing net ionic equation :
(1) Write a balanced molecular equation for the reaction
(2) Write the complete ionic equation [precipitation or not]
(3) Identify and remove the spectator ions
Net Ionic Equation
 Example : Write the net ionic equation for the precipitation reaction
that occurs when solutions of nickel nitrate and
sodium hydroxide are mixed.
(1) Write the balanced molecular equation
(2) Write the complete ionic equation
(3) Write the net ionic equation
Net Ionic Equation
 Example : Write the net ionic equation for the precipitation reaction
that may occur when solutions of calcium nitrate and
sodium acetate are mixed.
(1) Write the balanced molecular equation
(2) Check the solubility of products
(3) Net ionic equation:
Acids and Bases
 Properties of Acids
 Acids ionize to produce protons: HCl (aq)  H+(aq) + Cl-(aq)
 Turn blue litmus red
 Taste sour
 Aqueous solution has a pH less than 7
 React with metal to produce H2 gas
 Properties of Bases
 Bases produce OH- in water: NaOH (aq)  Na+(aq) + OH-(aq)
 Turn red litmus blue
 Bitter taste
 Aqueous solution has a pH more than 7
Acids and Bases
 Strong acids and bases
Strong acids and bases ionize completely in aqueous solution.
Strong acids and bases are strong electrolytes.
HCl (aq)  H+(aq) + Cl-(aq)
100% ionized
NaOH (aq)  Na+(aq) + OH-(aq) 100% ionized
HCl, HBr, HI, HNO3, H2SO4, HClO4, HClO3 strong acids
group 1A hydroxide [LiOH, NaOH, KOH]
strong bases
group 2A hydroxide [Ca(OH)2, Sr(OH)2, Ba(OH)2]
Acids and Bases
 Weak acids and bases
Weak acids and bases ionize partially in aqueous solution.
Weak acids and bases are weak electrolytes.
HF (aq)  H+(aq) + F-(aq)
~10% ionized
HF, CH3COOH, HNO2, …
NH3, …
weak acids
weak bases
NOTE: NH3 do not have OH. However, they produce OH- by
accepting a proton from H2O.
NH3 (aq) + H2O (l)  NH4+ (aq) + OH- (aq)
Acid-Base Reactions
 acid-base neutralization reactions:
An exchange reaction in which a molecular compound (H2O)
is formed.
molecular equation:
HCl(aq) + NaOH(aq)  H2O(l) + NaCl(aq)
complete ionic equation :
H+(aq) + Cl-(aq) + Na+(aq) + OH-(aq)  H2O(l) + Na+(aq) + Cl-(aq)
net ionic equation :
H+(aq) + OH-(aq)  H2O(l)
Acid-Base Reactions
NOTE: (1) The products of simple acid-base reactions are water
and ionic compounds called “salts”.
HNO3(aq) + NaOH (aq)  H2O(l) + NaNO3(aq)
acid
base
water
salt
(2) The formation of water is what drives the acid-base
neutralization reaction
H+(aq) + OH-(aq)
H2O(l)
possible, but much less
favorable than the other way
Acid-Base Reactions
 Example : Complete and balance the following equation for an
acid-base neutralization reaction: H3PO4(aq) + KOH(aq) 
What is the net ionic equation?
Acid-Base Reactions
 Example : Complete and balance the following equation for an
acid-base neutralization reaction: Mg(OH)2(s) + HCl(aq) 
What is the net ionic equation?
Acid-Base Reactions
 acid-base reactions with gas formation :
Acids reacting with certain weak bases produce a salt (ionic
compound) and gas (molecular compound). The most common
reaction of this type involves carbonates (CO32-).
2 HCl(aq) + Na2CO3(aq)  H2CO3 (aq) + 2NaCl(aq)
unstable
H2CO3(aq)  H2O (l) + CO2(g)
The overall reaction will be
2 HCl(aq) + Na2CO3(aq)  H2O(l) + 2NaCl(aq) + CO2(g)
Acid-Base Reactions
 Example : What is the net ionic equation for the reaction between
Na2CO3(aq) and HCl(aq)?
Acid-Base Reactions
 Example: Complete and balance the following equation.
HCl(aq) + CaCO3(s) 
What is the net ionic equation?
Concentrations of Solutions
 Concentration: The amount of solute dissolved in a given amount
of solvent or solution
 Molarity: The most common way of expressing concentration
in chemistry
Molarity (M) 
moles solute
volume of solution in liter
A 1.00 molar solution (= 1.00M) contains 1.00 mol of
solute in every liter of solution.
Concentrations of Solutions
1 mol
1 liter
1M
1 mol
? liter
1M
Concentrations of Solutions
 Example : Calculate the molarity of a solution made by dissolving
23.4 g of Na2SO4 (MW= 142) in enough water to form
125 mL of solution.
Concentrations of Electrolyte
1L
1 mole of
Na2SO4 (s)
1M solution
of Na2SO4
2Na+(aq) + SO42-(aq)
Each formula unit of Na2SO4 that dissolve produces two Na+ ions
and one SO42- ion.  The concentration of Na+ (2M) will be twice
the SO42- concentration (1M).
Concentrations
 Example: How many grams of Na2SO4 (MW=142) are required
to make 0.350L of 0.500M Na+(aq)?
Dilutions
add solvent
Mconc , Vconc
Mdilute , Vdilute
number of solute
molecules before dilution
number of solute
molecules after dilution
number of moles of
solute before dilution
=
=
Mconc  Vconc =
number of moles of
solute after dilution
Mdilute  Vdilute
Dilutions
 Example : What is the molarity of the resulting solution when
15 mL of 0.65M solution is diluted to 315mL?
Solution Stoichiometry
Example : What volume of a 0.100 M HNO3 solution is needed
to completely neutralize 0.1 g of Ca(OH)2 ?
Stratagy: (1) What is the chemical reaction involving HNO3(aq)
and Ca(OH)2 (s)?
(2) How many moles of HNO3 is needed to neutralize
0.1 g of Ca(OH)2?
(3) What is the volume of 0.100M HNO3 solution
neutralizing that many moles of Ca(OH)2?
Solution Stoichiometry
aA + bB  cC + dD