Download Chapter 4: Reactions in Aqueous Solution

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Chapter 4: Chemical Quantities & Aqueous Reactions
I) Interpreting a Balanced Chemical Reaction
1) Reactions can be viewed from the microscopic level as well as the macroscopic
level.
2) The coefficients in a chemical reaction specify the relative amounts in moles of
each substance involved in the reaction.
II) Stoichiometry
1) Stoichiometry deals with the quantity of materials consumed and or produced
in chemical reactions.
2) Stoichiometry deals specifically with mass-mole relationships between
reactants and products found in a chemical equation.
3) Fundamental Principles
i) ALWAYS have the balanced chemical equation
ii) MOLES are the vehicle that carry us everywhere we want to go
in chemistry. [AXIOM #3]
4) Handy Dandy 5 Step Method for Solving Stoichiometry Problems:
Step 1:
Write out the balanced chemical equation. Write givens and
unknowns above items in the equation.
Step 2:
Calculate Molar Masses for items needed in reaction.
Step 3:
Convert given quantities into moles. Remember axiom #3.
Step 4:
Use balanced equation to set up mole ratios between given
and unknown. Convert given mole quantity into moles of
what we are after.
Step 5:
Convert final quantity moles to units desired in answer.
5) Some Examples
How many grams of oxygen are required for the combustion of 96.1 g of propane
(C3H8)?
Step 1:
96.1 g ? g
?g
?g
C3H8 + 5O2 --------> 3 CO2 + 4 H2O
Step 2: MM (g/mol)
44.094 32.00
44.01
18.016
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Note: The calculation of molar mass values are shown below.
C3H8 (3 x 12.01 amu) + (8 x 1.008 amu) = 44.094 amu
44.094 g C3H8 = 1 mole C3H8
O2
32.00 g O2 = 1 mole O2
(2 x 16.00 amu) = 32.00 amu
Step 3:96.1 g C3H8 x (1 mol C3H8 /44.094 g C3H8) = 2.179 mole C3H8
Note: The underlined digit (2.179 mole C3H8) represents the first
uncertain digit and is used to remind us of the proper number of
significant figures to that point in the calculation.
Step 4:2.179 mole C3H8 x (5 mol O2 /1 mol C3H8 ) = 10.89 mole O2
Step 5:10.89 mole O2 x (32.000 g O2 / 1 mol O2) = 348.7 g O2 = 349 g O2
Check: Answer has correct units and seems reasonable.
(For Homework find # g of two products produced).
Answers:
288 g CO2
157 g H2O
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III) Yields of Chemical Reactions
1) Most reactions do not go to completion (100% conversion of reactants into
products).
2) In addition to the specified reaction, side reactions may also occur.
3) Instead we normally speak of the yield of a reaction (the amount of product
actually formed).
4) Percent Yield Calculations
% yield = (actual yield / theoretical yield) * 100 %
actual yield – amount obtained by investigator during analysis
theoretical yield – amount obtained assuming 100 % complete reaction
Example Problem
Rodney J. Smith is in the laboratory synthesizing aspirin C9H8O4. Aspirin
is prepared by the following chemical reaction. Rodney starts out by
reacting 11.56 g of salicylic acid (C7H6O3) with an excess of acetic
anhydride (C4H6O2). If Rodney isolated 10.92 g of aspirin (C9H8O4), what
is the percent yield for his experiment?
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Apply the Handy Dandy Five Step Method to find theoretical yield.
Step 1:
Step 2: MM in g/mole
Step 3:
11.56 g excess
10.92 g = actual yield
C7H6O3 + C4H6O2  C9H8O4 + C2H4O
138.12
180.15
The acetic anhydride is in excess; that means there is more than enough
acetic anhydride present to consume ALL of the salicylic acid.
11.56 g C7H6O3 x ( 1 mole C7H6O3 / 138.12 g C7H6O3) = 0.083695 moles C7H6O3
Step 4:
0.083695 moles C7H6O3 x ( 1 mole C9H8O4 / 1 mole C7H6O3 ) = 0.083695 moles C9H8O4
Step 5:
0.083695 moles C9H8O4 x ( 180.15 g C9H8O4 / 1 mole C9H8O4 ) = 15.077 g C9H8O4
% yield = (10.92 g / 15.077 g) * 100 % = 72.42 % yield = Answer
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IV) Calculations Involving Limiting Reactants
1) Limiting Reactant (reagent): reactant that is consumed FIRST in a
chemical reaction and, therefore, limits amount of products that form.
2) Examples:
A) What happens if ratio of reactants is 1:1 ?
- the reactant with the smallest # of moles is limiting reactant
B) Suppose ratio of reactants is NOT 1:1 ?
- must calculate number of moles product that forms from each
reactant. Results establish which reactant is limiting (it produces
the least amount of product).
CH4 (g) + 2 O2 (g) ------> CO2 (g) + 2 H2O (g)
0.500 mol 2.50 mol
? mol
0.500 mol CH4 x ( 1 mol CO2 / 1 mole CH4) = 0.500 mol CO2
2.50 mol O2 x (1 mol CO2 / 2 mol O2) = 1.25 mol CO2
CH4 is limiting reactant; it produces the fewest moles of product.
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- Another example
N2 (g) + 3 H2 (g) ------> 2 NH3 (g)
5 mol
9 mol
? mol
Ans. = 6 mol
Let us construct a table to visualize what is going on here.
N2 (g) + 3 H2 (g) ------> 2 NH3 (g)
initial
1st
2nd
3rd
5 mol
4 mol
3 mol
2 mol
9 mol
6 mol
3 mol
0 mol
0 mol
2 mol
4 mol
6 mol = Answer
H2 is limiting reactant so when it is used up the reaction STOPS !!!
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3) Production Problem
Step 1:
Step 2: MM (g/mole)
(Real Life Situation)
25.0 kg 5.00 kg
? kg
N2 (g) + 3 H2 (g) ------> 2 NH3 (g)
28.02
2.016
17.03
Step 3: Convert starting material quantities into moles and determine limiting reactant.
25.0 kg N2 x (1000 g / 1 kg) x ( 1 mol N2 / 28.02 g N2 ) = 8.922 x 102 mol N2
5.00 kg H2 x (1000 g / 1 kg) x ( 1 mol H2 / 2.016 g H2 ) = 2.480 x 103 mol H2
Note: Since reaction is NOT 1:1, we must make a comparison assuming a 1:1
ratio in order to find limiting reactant.
We see from the balanced chemical equation that 1 mol nitrogen reacts with 3 mol
hydrogen, so we now find the EXACT # moles hydrogen needed for a complete
reaction (stoichiometric reaction).
8.922 x 102 mol N2 x ( 3 mol H2 / 1 mol N2 ) = 2.676 x 103 mol H2
Find the limiting reactant
Since 2.676 x 103 mol H2 (exact quantity needed) is >> 2.480 x 103 mol H2
(calculated quantity), H2 is limiting reactant .
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Steps 4 & 5: Solve the problem in the usual "handy dandy" fashion !
2.480 x 103 mol H2 x (2 mol NH3 / 3 mol H2 ) = 1.653 x 103 mol NH3
1.653 x 103 mol NH3 x (17.03 g NH3 / 1 mol NH3) = 28,158 g NH3
Simply converting to kg gives final answer
28,158 g NH3 x (1 kg / 1000 g) = 28.1 kg NH3
(Value is reasonable).
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V) Introduction to Solutions
1) A solution is a homogeneous mixture consisting of two components.
A) Solute – substance being dissolved in the solution.
B) Solvent – dissolving medium. This component is always in greatest
amount.
C) Most chemical reactions are carried out in the liquid state or in
solution. This is due to the requirement that reactant molecules or ions be
highly mobile so they can interact with each other.
2) Although solutions can be described as being concentrated or dilute; it is
often necessary to quantify a solution’s concentration.
A) There are many ways of expressing solution concentration, but for the
moment we will only focus on molarity.
B) Molarity = M = moles solute / Volume of solution in liters
C) To find the number of moles of solute one uses the following
relationship
Volume of solution in liters x Molarity = moles of solute
D) To find the volume of solution one need only manipulate the above
equation
Volume of solution in liters = moles of solute / Molarity
3) Solutions are often prepared by adding a weighed quantity of solute to a
volumetric flask and dissolving the contents in solvent. When the flask is filled to
the mark with solvent, the concentration of the solution is accurately known.
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4) Example Molarity Calculation
What is the molarity of a solution prepared by adding 11.5 g of NaOH to
1.50 L of water? (Note: The molar mass of NaOH is 40.00 g/mole)
Unknown:
Given:
? M (NaOH)
11.5 g NaOH + 1.50 L water
Rel. Info:
M = # moles solute / Vol. soln. (L)
Solution:
# moles NaOH = 11.5 g NaOH x (1 mole / 40.00 g NaOH)
Water is solvent!
# moles NaOH = 0.288 moles NaOH
Therefore,
M = 0.288 mol NaOH / 1.50 L = 0.192 M
(Reasonable)
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5) Diluting Concentrated Solutions
A) Dilution - process by which water (or another solvent) is added to a
solution of known concentration to achieve a new solution of lower
concentration.
B) Dilution does not alter the number of moles of solute.
# MOLES SOLUTE AFTER = # MOLES SOLUTE BEFORE
DILUTION
DILUTION
C) Dilution Equation
M1V1 = M2V2
VI) Solution Stoichiometry
1) Solving stoichiometry problems with one or both reactants in solution involves
applying the same 5 step method. The only “change” usually involves getting the
number of moles of reactant(s) that are in solution.
2) To obtain the number of moles of reactant, one need only apply the equation
Volume of solution in liters x Molarity = moles of solute
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VII) Introduction to Titrations
1) A titration is a technique that allows one to determine the concentration of a
solution by allowing a carefully measured volume to react with another substance
whose concentration is accurately known.
2) The solution of known concentration, called the titrant, is dispensed from a
buret.
3) An indicator is used as a visual means of determining when the titration is
finished. The end of a titration is usually observed when the indicator undergoes a
dramatic color change
VIII) Water, The Universal Solvent
1) Water is a very common solvent due to its wide availability and low cost
(most of our world is water).
2) Many reactions take place in aqueous solution. The term aqueous means
dissolved in water.
3) Hydration of solids in Water
A) Solid dissolves (falls apart) through interaction of ions with water.
B) Newly formed ions are now “FREE TO MOVE”.
C) “Like Dissolves Like”
i) polar materials dissolve in polar solvents
ii) nonpolar materials dissolve in nonpolar solvents.
4) For now, consider that chemical reactions involve a “driving force” that
account for reactants being converted into products.
5) Three general categories of aqueous solutions.
A) Precipitation reactions
B) Acid-Base Neutralization reactions
C) Oxidation-Reduction (Redox) reactions
IX) Electrolytes in Aqueous Solutions
1) Electrical conductivity is a property associated with solutions.
2) Substances (e.g. NaCl, KBr, etc.) that produce ions when dissolved in water are
called electrolytes.
3) The concept of electrolytes was first proposed by Arrhenius.
4) An electrolyte solution conducts electricity because the ions are free to when
dissolved in water.
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5) Types of Electrolytes
A) strong electrolytes: completely ionize in solution
Examples:
SA
(HCl, HNO3, HClO4, H2SO4)
SB
(NaOH, LiOH, KOH)
soluble salts (NaCl, KF, BaCl2)
Arrhenius:
acids- source of H+ ions
bases - source of OH- ions
HCl

NaOH 
sour taste
feel slippery
H+ (aq) + Cl- (aq)
Na+ (aq) + OH- (aq)
B) weak electrolytes: incompletely ionize in solution
Examples
WA
WB
C) nonelectrolytes:
Examples
acetic acid
ammonia
HoAc <-----> H+ + oAcNH3 <--------> NH4+ + OH-
do not produce ions when dissolved in water.
EtOH and sucrose dissolved in water
EtOH (l) -------> EtOH (l)
sucrose (s) ---------> sucrose (l)
X) Precipitation Reactions & Solubility Rules
1) A precipitate is an insoluble solid formed by mixing two or more solutions
together.
2) Precipitates are NEUTRAL in charge.
3) Precipitate formation is possible since dissolved ions move freely in solution.
4) Beaker Analogy – used to determine identity of products in solution.
Example 1:
Ba(NO3)2 (aq) + K2CrO4 (aq) ---------> products
colorless
yellow soln
yellow ppt observed
Ba2+ (aq) + 2 NO3 - (aq) + 2 K+ (aq) + CrO4 2- (aq) ---------> products
Check solubility (Review Table 4.1). Know the general trends for solubility.
From picture, KNO3 and BaCrO4 are products that are possible. Other two are reactants !!
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- Also yellow soln in reactants leads us to believe yellow ppt. in product contains
ion from that reactant...
KNO3 is soluble salt and BaCrO4 is an insoluble salt (must be identity of yellow solid)...
Describing Precipitation Reactions
1) Molecular Equation - gives overall solution stoichiometry...
Ba(NO3)2 (aq) + K2CrO4 (aq) ---------> 2 KNO3 (aq) + BaCrO4 (s)
2) Complete Ionic Equation - strong electrolytes represented as free moving ions...
Ba2+ (aq) + 2 NO3- (aq) + 2 K+ (aq) + CrO42- (aq) -----> 2 K+ (aq) + 2 NO3- (aq) +
BaCrO4 (s)
3) Net Ionic Equation - gets rid of spectator ions (ions that do not DIRECTLY
participate in the reaction). ONLY include species which directly take part in the
reaction.
Ba2+ (aq) + CrO4 2- (aq) ---------> BaCrO4 (s)
Example 2:
(yellow solid)
AgNO3 (aq) + KCl (aq) ---------> products
colorless
colorless
(white solid observed)
AgNO3 (aq) + KCl (aq) ---------> AgCl (s) + KNO3 (aq)
Ag+ (aq) + NO3- (aq) + K+ (aq) + Cl- (aq) ---------> AgCl (s) + K+ (aq) + NO3 - (aq)
Ag+ (aq) + Cl- (aq) ---------> AgCl (s)
KEY TO SUCCESS:
DRAW PICTURE. DETERMINE LIKELY PRODUCTS & KNOW TRENDS. (Review
Table 4.1)
Slightly soluble means same thing as insoluble
(Why the distinction?)
Slightly soluble means a tiny portion of solid dissolves but is not noticeable to
observer
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XI) Acid/Base Reactions
1) There are three systems for defining acids & bases.
A) Arrhenius
i) Acids contain H+ in chemical formula.
ii) Bases contain OH- in chemical formula.
H+  proton
B) Bronsted-Lowry
i) Acids are proton donors.
ii) Bases are proton acceptors.
C) Lewis (CHE 124)
2) Mixing Acids and Bases
[Neutralization Reaction]
Acid + Base  salt + water
KEY: Write down species present and
figure out reaction being studied...
(always)
[SA + SB -------> Neutral Salt + Water]
[WA + SB ------> Basic Salt + Water]
[WB + SA ------> Acidic Salt + Water]
XII) Oxidation/Reduction Reactions (Redox Rxns)
1) Redox reactions are characterized by electrons being transferred between
species.
2) Number line pneumonic
3) Fundamental Definitions KNOW THESE
A) oxidation - loss of electrons (moving to right on number line)
B) reduction - gain of electrons (moving to left on number line)
C) oxidizing agent - species being reduced (electron acceptor)
D) reducing agent - species being oxidized (electron donor)
4) Oxidation States (Oxidation Numbers)
A) Electron Accounting
B) Learn rules for assigning oxidation numbers (see page ).
Designate charged ions as n+ or m-
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Designate oxidation states as +n or -m
5) Oxidation and reduction reactions always occur together.