Reactions in Aqueous
Solution
AP Chemistry – Chapter 4
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Reactions in Aqueous Solution
• Fundamental factor that causes reaction
involving ions to occur is the starting ions are
removed from solution.
Ways this is accomplished:
1) Formation of a precipitate
2) Formation of a gas
3) Formation of a primarily molecular species
4) Form original ions into different ions.
(oxidation - reduction)
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1) Precipitation Reactions
Use solubility Rules ---> (table 4.1 pg. 144) (desk tops or handout)
Most nitrates (NO3-) are soluble.
Most salts containing the alkali metal ions (Li+, Na+, K+,
Cs+, Rb+) and the ammonium ion (NH4+) are soluble.
3. Most chloride, bromide, and iodide salts are soluble.
1.
2.
Exceptions: Ag+, Pb2+, and Hg22+.
4.
Most sulfate salts are soluble.
Exceptions: BaSO4, PbSO4, HgSO4, and CaSO4.
5.
Most hydroxide are only slightly soluble.
Exceptions: Group I & Calcium down in Group II.
6.
Most sulfide (S2-), carbonate (CO32-), chromate (CrO42-),
and phosphate (PO43-) salts are only slightly soluble.
S2- exceptions: Group I, Group II, NH4+
3- exceptions:
+
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CO32-, CrO42-,JodiPO
Group
I,
NH
4
4
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Chemistry 6th edition
Solubility Rules
„ Put these in a special “spot.” I recommend
starting a small (1/2 inch) 3-ring binder for
only predicting reactions papers.
„ MEMORIZE! We’ll try to quiz on these rules
daily this unit.
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1) Precipitation Reactions
1st example - Mix solutions of Ba(NO3)2 and
Na2CO3. What happens?
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1) Precipitation Reactions
2nd example - Mix solutions of CuCl2 and NaOH.
What happens?
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1) Precipitation Reactions
3rd example - Mix solutions of MgCl2 and KNO3.
What happens?
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Precipitate Colors
PbI2
HgCl2
AgCl (white)
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HgS
Precipitate Colors
Cu2+(aq) + 2NH3(aq) + 3 H2O (l) <=> Cu(OH)2(s) + 2NH4+ (aq)
Cu(OH)2 (s) + 4 NH3 (aq) <=> [Cu(NH3)4]2+ (aq) + 2OH- (aq)
Cu2+ (aq) + 2OH- (aq) <=> Cu(OH)2 (s)
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2) Gas formation Reactions –
(H2S, CO2, SO2, NH3)
H2CO3
Æ H2O + CO2
_________ Æ H2O + SO2
_________ Æ H2O + NH3
_________ Æ H2S (g)
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2) Gas formation Reactions –
(H2S, CO2, SO2, NH3)
4th example - Mix solutions of Na2CO3 and HCl.
What happens?
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3) Acid-Base Reactions
„ Acids form H+ ions in solution
Difference between strong and weak acids:
* Must know 6 strong acids - any other acid is
weak
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3) Acid-Base Reactions
Strong Acid Reaction:
Weak Acid Reaction:
Ex. 1.00 M HF:
[HF] = 0.97 M,
[H+] = [F-] = 0.03 M
3% ionized
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3) Acid-Base Reactions
EXAMPLES - (Species present at highest
concentration is written as the reactant.)
A) Strong Acid - Strong Base
HCl + Ca(OH)2
Net ionic equation:
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3) Acid-Base Reactions
EXAMPLES - (Species present at highest
concentration is written as the reactant.)
B) Strong Acid - Weak Base
HCl + NH3
Net ionic equation:
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3) Acid-Base Reactions
EXAMPLES - (Species present at highest
concentration is written as the reactant.)
C) Weak Acid - Strong Base
HF + NaOH
Net ionic equation
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REVIEW OF ELECTROLYTES &
NET IONIC REACTIONS
Strong Electrolytes – exist 100% as ions in solution
Weak Electrolytes – exists partially as ions in
solution
Non-electrolytes – do NOT exist as ions
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REVIEW OF ELECTROLYTES
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Problem Set 4.1 Summary
„ Problem Set 4.1 is all about writing net ionic reactions.
„ When writing an ionic equation write everything that exists
as ions as ions.
Use Solubility Rules on
Write as Ions
pg. 141 (or rules below
Aqueous (soluble) ionic compound
the overhead) to
determine solubility.
Strong acids
Write as molecules
Insoluble ionic compounds (or any ionic cpd not in solution)
Weak acids
Molecular compounds
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Problem Set 4.1 Summary
See back
page of your
note packet!
NOTE: The strong bases are also the soluble hydroxides.
Ex. A solution of sodium hydroxide is mixed with hydrochloric acid
Total ionic:
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) Æ H2O (l) + Na+ (aq) + Cl-(aq)
Net ionic:
OH-(aq) + H+(aq) Æ H2O(l)
Ex. Magnesium hydroxide is mixed with hydrochloric acid
Total ionic:
Mg(OH)2 (s) + 2 H+ (aq) + 2 Cl- (aq) Æ 2 H2O(l) + Mg2+(aq) + 2 Cl-(aq)
Net ionic:
Mg(OH)2 (s) + 2 H+ (aq) Æ 2 H2O (l) + Mg2+ (aq)
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Problem Set 4.1 Summary
NOTE: THE CALCULATIONS ARE SIMPLY
3-STEP MOLE PROBLEMS, BUT THEY ARE
MUCH HARDER THAN MOST YOU’VE
DONE BEFORE.
(YOU’LL NEED TO WRITE BALANCED
EQUATIONS FOR MOST OF THEM.)
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Stoichiometry Flow Chart
1. Identify Reaction Type
2. Write a balanced Equation
3. Calculate moles of reactants Given
Æ V & M:
Æ Mass:
n=M•V
n = g/MM
4. Determine Limiting Reagent
5. Calculate moles of products
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Go back to
pg. 3 of
note packet
Quantitative Analysis
Quantitative analysis – the determination of how
much of a given component is present in a
sample
Titration – a process in which one reagent is added
to another (usually acids and bases) with which
it reacts.
„ An indicator is used to determine the point @
which equivalent quantities of the two reagents
have been added.
Equivalence point – point @ which just enough
titrant has been added to fully react with B.
end point – the point during a titration where an
indicator turns color
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ex. Phenolphthalein
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Quantitative Analysis
„ Ex. 30.0 mL of fruit juice is titrated with 14.85
mL of an NaOH solution which contains 1.82
mg/mL. Determine the mg of vitamin C,
C6H8O6, per milliliter of juice.
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Solution Stoichiometry Examples:
MH+V H+= M OH-VOH1.
When aqueous solutions of sodium hydroxide and
Iron (III) nitrate are mixed, a red gelatinous
precipitate forms. Calculate the mass of precipitate
formed when 50.00 mL of 0.200 M NaOH and 30.00
mL of 0.125 M Fe(NO3)3 are mixed.
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Solution Stoichiometry Examples:
2.
Barium Hydroxide reacts with acetic acid, HC2H3O2.
Calculate the concentration of all ions and
molecules except H2O after the reaction if 75.00 mL
of 0.0200 M Ba(OH)2 is added to 22.68 mL of 0.500
M HC2H3O2. (Assume the final is the sum of the
initial volumes.)
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Solution Stoichiometry Examples:
3. What volume of 0.0720 M AlCl3 is needed to
react fully with 25.0 mL of 0.364 M
Pb(NO3)2?
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Solution Stoichiometry Examples:
4. 30.0 mL of acetic acid, CH3COOH, is titrated
with 14.22 mL of LiOH. The LiOH solution is
1.42 mg/mL. Determine the mg of the acid
per mL of solution.
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