Transcript Chapter 5

Chapter 5
Reactions
in
Aqueous Solutions
Dr. S. M. Condren
Reactions in Aqueous Solution
Many reactions involve ionic compounds,
especially reactions in water — aqueous
solutions.
KMnO4 in water
K+(aq) + MnO4-(aq)
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CCR, page 177
An Ionic Compound,
CuCl2, in Water
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Aqueous Solutions
How do we know ions are
present in aqueous
solutions?
The solutions conduct
electricity!
They are called
ELECTROLYTES
HCl, CuCl2, and NaCl are
strong electrolytes.
They dissociate completely
(or nearly so) into ions.
Dr. S. M. Condren
Aqueous Solutions
HCl, CuCl2, and NaCl are strong
electrolytes. They dissociate
completely (or nearly so) into ions.
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Aqueous Solutions
Acetic acid ionizes only to a small extent,
so it is a weak electrolyte.
CH3CO2H(aq) --->
CH3CO2-(aq) + H+(aq)
Dr. S. M. Condren
Aqueous Solutions
Acetic acid ionizes only
to a small extent, so it
is a weak
electrolyte.
CH3CO2H(aq) --->
CH3CO2-(aq) + H+(aq)
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Aqueous Solutions
Some compounds
dissolve in water but
do not conduct
electricity. They are
called
nonelectrolytes.
Examples include:
sugar
ethanol
ethylene glycol
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Electrolytes
• Conduct electricity in solution due to the
presence of ions
• Strong electrolyte – completely ionized in
solution
• Weak electrolyte – partially ionized in
solution
• Non-electrolyte – nonionic solution
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Solubility Rules
1. All nitrates (NO3-1) are soluble.
2. All compounds of Group IA metals and the
ammonium ion, NH4+, are soluble.
3. All chlorides are soluble except: AgCl,
Hg2Cl2 and PbCl2.
4. All sulfates are soluble except: PbSO4,
BaSO4, and SrSO4.
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Solubility Rules
5. All hydroxides (OH-1)and sulfides (S-2)are
insoluble except those of the Group IA
metals and the ammonium ion.
6. All carbonates (CO3-2) and phosphates
(PO4-3) are insoluble except those of the
Group IA metals and the ammonium ion.
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Net Ionic Equation
Balanced Chemical Equation
Pb(NO3)2 + Na2SO4 ---> PbSO4 + 2NaNO3
Total Ionic Equation
Pb+2 + 2NO3-1 + 2Na+1 + SO4-2 
2Na+ + 2NO3 -1 + PbSO4
Net Ionic Equation
Pb+2 + SO4-2  PbSO4
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A solution of Ba(NO3)2 is added to a
solution of Na2SO4 to make a precipitate.
From a table of solubility rules, the product
is
barium sulfate, sodium nitrate
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Types of Reactions
• synthesis reactions or combination
reactions
• decomposition reactions
• precipitation reactions
• neutralization reactions
– acid
– base
• oxidation-reduction reaction
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Synthesis or Combination Reactions
Formation of a compound from simpler
compounds or elements.
2Na(s) + Cl2(g)  2 NaCl(s)
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Decomposition Reactions
Separation into constituents by chemical
reaction.
2 H2O2 (aq)
catalysis
2H2O(l) + O2(g)
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Precipitation Reactions
The process of separating a substance from a
solution as a solid.
AgNO3 + NaCl ---> AgCl + NaNO3
precipitate
Total Ionic Equation
Ag+1(aq) + NO3(aq)-1 + Na(aq)+1 + Cl(aq)-1  AgCl(s) + NO3(aq)-1 + Na(aq)+1
Net Ionic Equation
Ag+1(aq) + Cl(aq)-1  AgCl(s)
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Neutralization Reactions
acid + base ---> “salt” + water
Balanced Chemical Equation
HCl + NaOH ---> NaCl + H2O
Total Ionic Equation
H+1 + Cl-1 + Na+1 + OH-1  Na+ + Cl-1 + H2O
Net Ionic Equation
H+1 + OH-1  H2O
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Neutralization Reactions
acid + base ---> “salt” + water
Balanced Chemical Equation
H2SO4 + 2KOH ---> K2SO4 + 2H2O
Total Ionic Equation
2H+ + SO4-2 + 2K+ + 2OH-1  2H2O + SO4-2 + 2K+
Net Ionic Equation
2H+1 + 2OH-1  2H2O
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Neutralization Reactions
• acid
• base
• salt
Household acids and Bases
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Neutralization Reactions
• acid
– Any of a large class of sour-tasting
substances whose aqueous solutions are
capable of turning blue litmus indicators red,
of reacting with and dissolving certain metals
to form salts, and of reacting with bases or
alkalis to form salts.
– Substance that donates H+ ions to solution
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ACIDS
An acid -------> H+ in water
Some strong acids are
HCl
H2SO4
HClO4
HNO3
hydrochloric
sulfuric
perchloric
nitric
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HNO3
Weak Acids
WEAK ACIDS = weak
electrolytes
CH3CO2H acetic acid
H2CO3
carbonic acid
H3PO4
phosphoric acid
HF
hydrofluoric acid
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Acetic acid
ACIDS
Nonmetal oxides can be acids
CO2(aq) + H2O(liq)
---> H2CO3(aq)
SO3(aq) + H2O(liq)
---> H2SO4(aq)
and can come from burning coal
and oil.
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Neutralization Reactions
• base
– Any of a large class of compounds, including
the hydroxides and oxides of metals, having a
bitter taste, a slippery solution, the ability to
turn litmus blue, and the ability to react with
acids to form salts.
– Substance that donates a OH-1 ion to solution
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BASES
Base ---> OH- in water
NaOH(aq)
 Na+(aq) + OH-(aq)
NaOH is a strong base
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Ammonia, NH3
An Important Base
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Neutralization Reactions
• salt
– The term salt is also applied to substances
produced by the reaction of an acid with a
base, known as a neutralization reaction.
– Salts are characterized by ionic bonds,
relatively high melting points, electrical
conductivity when melted or when in solution,
and a crystalline structure when in the solid
state.
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pH Scale
@ -2 -----> @ +16
pH = - log [H3O+]
[H3O+] = -antilog pH = 10-pH
For log problems, only decimal places are
significant, and all decimal places count
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Example: What is the pH of a 1.0 M
solution of HCl?
pH = - log [H3O+]
[H3O+] = MHCl = 1.0 M
• because it is a strong electrolyte
pH = -log(1.0) = 0.00
2 significant figures
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Example: What is the pH of a 1.0 M
solution of HCl?
pH = - log [H3O+]
[H3O+] = MHCl = 1.0 M
• because it is a strong electrolyte
pH = -log(1.0) = 0.00
2 SF
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Example: What is the pH of a 0.010 M
solution of HCl?
pH = - log [H3O+]
[H3O+] = MHCl = 0.010 M
• because it is a strong electrolyte
pH = -log(0.010) = 2.00
2 SF
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Example: What is the pH of a 2.0 M
solution of HCl?
pH = - log [H3O+]
[H3O+] = MHCl = 2.0 M
• because it is a strong electrolyte
pH = -log(2.0) = -0.30
2 SF
Dr. S. M. Condren
Example: What is the [H3O+] of a
solution that has a pH = 2.30?
[H3O+] = -antilog pH = 10-pH
decimal point pacement
[H3O+] = 10-pH = 10-2.30 = 5.0x10-3M
2 SF
2 SF
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pH Scale
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Indicators
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Oxidation-Reduction Reaction
Oxidation - loss of electrons
Reduction - gain of electrons
Redox reaction
oxidizing agent - substance that causes
oxidation
reducing agent - substance that cause
reduction
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Identify the oxidizing agent in the reaction:
2Al(s) + 6 H+ ==> 2 Al3+(aq) + 3 H2(g)
Al, H+, Al3+, H2
Identify the oxidizing agent in the reaction:
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Redox Reactions
CuSO4(aq) + Zn(s)  Cu(s) + ZnSO4(aq)
4 Al(s) + 3 O2(g)  2 Al2O3(s)
2HgO(s)  2 Hg(l) + O2(g)
2 Al(s) + 3 Br2 (l)  Al2Br6 (s)
Fe2O3(s) + 2 Al(s)  2 Fe(l) + Al2O3(s)
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Solution
Solutions, in chemistry, homogeneous
mixtures of two or more substances.
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Solute
The substance that is present in smallest
quantity is said to be dissolved and is
called the solute. The solute can be either
a gas, a liquid, or a solid.
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Solvent
The substance present in largest quantity
usually is called the solvent. The solvent
can be either a liquid or a solid.
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Preparing a Solution
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Preparing a Solution by Dilution
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Molarity
The number of moles of solute per liter of
solution.
molarity => M
moles of solute
M = -------------------liter of solution
units => molar = moles/liter = M
Dr. S. M. Condren
Example: A sample of NaNO3 weighing 0.38g is
placed in a 50.0-mL volumetric flask. The flask is then
filled with water to the mark on the neck, dissolving the
solid. What is the molarity of the resulting solution?
(0.38g NaNO3)
M=
(50.0-mL soln)
= 0.089 mol/L = 0.089 molar = 0.089 M
Dr. S. M. Condren
Stoichiometric Roadmap
Grams
of A
g B x (mol A/g A)
Divide by the
molar mass
Volume
solution A
Moles
of A
Moles
of B
mol A x (mol B/mol A)
Multiply by the
stoichiometric factor
Vol A x (mol A/L A)
Multiply by molarity
mol B x (g B/mol B)
Multiply by the
molar mass
Volume
solution B
Mol B x (L B/mol B)
Diviide by molarity
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Grams
of B
Example: What mass of Na2CO3, in grams, is
required for complete reaction with 50.0mL of
0.125 M HNO3?
Na2CO3(aq) + 2 HNO3(aq)  2 NaNO`3(aq) + CO2(g) + H2O(l)
#g Na2CO3 =(50.0 mL)
= 0.331 g Na2CO3
Dr. S. M. Condren
EXAMPLE: Lye, which is sodium
hydroxide, can be neutralized by sulfuric
acid. How many milliliters of 0.200 M
H2SO4 are needed to react completely
with 25.0 mL of 0.400 M NaOH?
2 NaOH(aq) + H2SO4(aq) -----> Na2SO4(aq) + 2 H2O
(25.0 mL NaOH) (0.400 mol NaOH) (1 L) (1 mol H2SO4)
--------------------------------#mL H2SO4 = ----------------------------------------------(1 L NaOH) (1000 mL)(2 mol NaOH)
(1000 mL H2SO4)
-----------------------(0.200 mol H2SO4)
= 25.0 mL H2SO4
Dr. S. M. Condren