The Nature of the Solutions The example of KMnO4

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The Nature of the Solutions(The example of KMnO4)

• KMnO4(aq) ? K(aq) MnO4-(aq)
• If you make a solution that is 0.30 M in KMnO4,
  this means that
• K MnO4- 0.30 M

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The Nature of a Na2CO3 Solution

• This water-soluble compound is ionic
• Na2CO3(aq) ? 2 Na(aq) CO32-(aq)
• If Na2CO3 0.100 M, then
• Na 0.200 M
• CO32- 0.100 M

3 Na2CO3
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USING MOLARITY
What mass of oxalic acid, H2C2O4, is required to
make 250 mL of a 0.0500 M solution?
Conc (M) moles/volume mol/V
moles M V

• Step 1 Calculate moles of acid required.
• (0.0500 mol/L)(0.250 L) 0.0125 mol
• Step 2 Calculate mass of acid required.
• (0.0125 mol )(90.00 g/mol) 1.13 g

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Concentrations of Solutions

• Dilution
• We recognize that the number of moles are the
  same in dilute and concentrated solutions.
• So we can dilute a concentrated solution to get
  one that is less concentrated
• MdiluteVdilute moles MconcentratedVconcentrate
  d

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PROBLEM You have 50.0 mL of 3.0 M NaOH and you
want 0.50 M NaOH. What do you do?

• Add water to the 3.0 M solution to lower its
  concentration to 0.50 M
• Dilute the solution!

How much water is added? The important point is
that ? moles of NaOH in ORIGINAL solution
moles of NaOH in FINAL solution

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PROBLEM You have 50.0 mL of 3.0 M NaOH and you
want 0.50 M NaOH. What do you do?

• Moles of NaOH in original solution
• M V (3.0 mol/L)(0.050 L) 0.15 mol
  NaOH
• Therefore, moles of NaOH in final solution must
  also equal 0.15 mol NaOH
• (0.15 mol NaOH)(1 L/0.50 mol) 0.30 L
• or 300 mL volume of final solution
• add enough water to make the initial 50.0 mL of
  3.0 M NaOH to a final volume of 300 mL and the
  final molarity will be 0.50 M NaOH.

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Preparing Solutions by Dilution
A shortcut

• Minitial Vinitial Mfinal Vfinal

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Preparing a solution by dilution.Example 4-10
A particular analytical chemistry procedure
requires 0.0100 M K2CrO4. What volume of 0.250 M
K2CrO4 should we use to prepare 0.250 L of 0.0100
M K2CrO4?
Calculate
1.000 L
0.0100 mol
VK2CrO4 0.2500 L
0.0100 L
0.250 mol
1.00 L
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SOLUTION STOICHIOMETRY An Example

• Zinc reacts with acids to produce H2 gas. If you
  have 10.0 g of Zn, what volume of 2.50 M HCl is
  needed to convert the Zn completely?

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Zinc reacts with acids to produce H2 gas. If you
have 10.0 g of Zn, what volume of 2.50 M HCl is
needed to convert the Zn completely?

• Step 1 Write the balanced equation
• Zn(s) 2 HCl(aq) ? ZnCl2(aq) H2(g)
• Step 2 Calculate moles of Zn

Step 3 Use the stoichiometric factor to
calculate moles of HCl
2 mol HCl/1 mol Zn
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Zinc reacts with acids to produce H2 gas. If you
have 10.0 g of Zn, what volume of 2.50 M HCl is
needed to convert the Zn completely?

• Step 3 Use the stoichiometric factor to
  calculate moles of HCl

Step 4 Calculate volume of HCl required
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Reactions Involving aLIMITING REACTANT

• Definition In a given reaction, there is not
  enough of one reagent to use up the other reagent
  completely.
• The reagent in short supply LIMITS the quantity
  of product that can be formed.
• The stoichiometric coefficients are used to
  determine the limiting reagent

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LIMITING REACTANTS
React solid Zn with 0.100 mol HCl (aq) The
reaction Zn 2 HCl ? ZnCl2 H2

• Rxn 1 Rxn 2 Rxn 3
• mass Zn (g) 7.00 3.27 1.31

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LIMITING REACTANTS
React solid Zn with 0.100 mol HCl (aq) The
reaction Zn 2 HCl ? ZnCl2 H2

• Rxn 1 Rxn 2 Rxn 3
• mass Zn (g) 7.00 3.27 1.31
• mol Zn 0.107 0.050 0.020
• mol HCl 0.100 0.100 0.100
• mol HCl/mol Zn 0.93 2.00 5.00

Limiting reagent determines reaction yield
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2 Al 3 Cl2 ? Al2Cl6
Reaction to be Studied
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PROBLEM Mix 5.40 g of Al with 8.10 g of Cl2.
How many grams of Al2Cl6 can form?
Convert lab units into chemical units and then
back
Stoichiometric factor
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2 Al 3 Cl2 ? Al2Cl6
Step 1 of LR problem compare actual mole ratio
of reactants to theoretical mole ratio.

• The ideal mole ration of reactants is

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Determining the Limiting Reactant
2 Al 3 Cl2 ? Al2Cl6

• If

then there is not enough Al to use up all the
Cl2, and the limiting reagent is Al
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Determining the Limiting Reactant
2 Al 3 Cl2 ? Al2Cl6

• If

then there is not enough Cl2 to use up all the
Al, and the limiting reagent is Cl2
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Step 2 of LR problem Calculate moles of each
reactant
We have 5.40 g of Al and 8.10 g of Cl2
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Find mole ratio of reactants
This should be 3/2 or 1.5/1 if reactants are
present in the exact stoichiometric ratio.
Limiting reagent is Cl2
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• Mix 5.40 g of Al with 8.10 g of Cl2. What mass
  of Al2Cl6 can form?

2 Al 3 Cl2 ? Al2Cl6
Limiting reactant Cl2 Base all calculations on
Cl2
grams Cl2
moles Cl2
moles Al2Cl6
Another stoichiometric factor
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CALCULATIONS calculate mass of Al2Cl6 expected.
Step 1 Calculate moles of Al2Cl6 expected based
on LR. Step 2 Calculate mass of Al2Cl6
expected based on LR.
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How much of which reactant will remain when
reaction is complete?

• Cl2 was the limiting reactant. Therefore, Al was
  present in excess. But how
  much?
• First find how much Al was required.
• Then find how much Al is in excess.

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Calculating Excess Al
2 Al 3 Cl2
products
0.114 mol LR
0.200 mol
Excess Al Al available - Al required 0.200
mol - 0.0760 mol 0.124 mol Al in excess
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Where are we?

• Key concepts from Chapters 2 3
• Definition of atomic mass units (amu)
• Conversion between amu and g
• Avogadros number
• Conversions between mass, moles, and of atoms
• Determination of atomic mass and molar mass
• Determination of composition by mass
• Determination of empirical formula from
  composition
• Determination of molecular formula from empirical
  formula and molar mass.

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Where are we?

• Key concepts from Chapter 4
• Writing and balancing chemical equations
• Law of conservation of mass
• Define stoichiometric coefficient
• Determine quantities of products formed when
  there is a limiting reagent
• Calculation of yield
• Calculating stoichiomteric amount in reactions
• Using solutions rather than solids
• Limiting reagent and calculation of yields.

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Introduction to Reactions in Aqueous Solutions
Chapter 5This Chapter is a general overview to
the types of reactions that we will see for the
remainder of the course
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IONIC COMPOUNDSCompounds in Aqueous Solution

• Many reactions involve ionic compounds,
  especially reactions in water aqueous
  solutions.
• Ions dissociate in water.
• In solution, each ion is surrounded by water
  molecules.

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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, KMnO4, MgCl2, and NaCl are strong
  electrolytes. They dissociate completely (or
  nearly so) into ions.

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General Properties of Aqueous Solutions

• Strong and Weak Electrolytes
• Strong electrolytes completely dissociate in
  solution.
• For example
• Weak electrolytes produce a small concentration
  of ions when they dissolve.
• These ions exist in equilibrium with the
  unionized substance.
• For example

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Aqueous Solutions

• Some compounds dissolve in water but do not
  conduct electricity. They are called
  nonelectrolytes.
• If there are no ions in solution, there is
  nothing to transport electric charge.
• Examples include

• sugar
• ethanol
• ethylene glycol (in antifreeze)

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Representation of Electrolytes using Chemical
Equations
A strong electrolyte

• MgCl2(s) ? Mg2(aq) 2 Cl-(aq)

A weak electrolyte
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Calculating Ion Concentrations in a Solution of a
Strong Electrolyte.
What are the aluminum and sulfate ion
concentrations in 0.0165M Al2(SO4)3?.
Balanced Chemical Equation
Al2(SO4)3 (s) ? 2 Al3(aq) 3 SO42-(aq)
Aluminum Concentration
0.0330 M Al3
Sulfate Concentration
0.0495 M SO42-
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COMMON REACTIONS

• precipitation
• acid-base
• oxidation/reduction (redox)